RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1

Other Exercises

Question 1.
Fill in the blanks using the correct word given in brackets :
(i) All circles are …………… (congruent, similar).
(ii) All squares are ………… (similar, congruent).
(iii) All …………….. triangles are similar (isosceles, equilaterals) :
(iv) Two triangles are similar, if their corres-ponding angles are ……… (proportional, equal)
(v) Two triangles are similar, if their corres-ponding sides are ……………. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are …………. and (b) their corres-ponding sides are …………….. (equal, proportional).
Solution:
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilaterals triangles are similar.
(iv) Two triangles are similar, if their corresponding angles are equal.
(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Question 2.
Write the truth value (T/F) of each of the following statements :
(i) Any two similar figures are congruent.
(ii) Any two congruent figures are similar.
(iii) Two polygons are similar, if their corresponding sides are proportional.
(iv) Two polygons are similar if their corresponding angles are proportional.
(v) Two triangles are similar if their corresponding sides are proportional.
(vi) Two triangles are similar if their corresponding angles are proportional.
Solution:
(i) False : In some cases, the similar polygons can be congruent.
(ii) True.
(iii) False : Its corresponding angles must be equal also.
(iv) False : Angle are equal not proportional.
(v) True.
(vi) False : Sides should be proportional and corresponding angles should be equal.

 

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RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Other Exercises

Question 1.
If the circumference and the area of a circle are numerically equal, then diameter of the circle is
(a) π/2
(b) 2π
(c) 2
(d) 4
Solution:
Let r be the radius of the circle, then Circumference = 2πr
and area = πr²
But 2πr= πr²
∴ 2r = r²
⇒ r = 2
Diameter = 2r = 2 x 2 = 4 (d)

Question 2.
If the difference between the circum-ference and radius of a circle is 37 cm., then using π =  \(\frac { 22 }{ 7 }\)  the circumference (in cm) of the circle is
(a) 154
(b) 44
(c) 14
(d) 7 [CBSE 2013]
Solution:
Difference between circumference and radius of a circle = 37 cm
∴  2πr-r = 37
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 1

Question 3.
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be
(a) 3520 cm²
(b) 6400 cm²
(c) 7744 cm²
(d) 8800 cm²
Solution:
Radius of a circular wire (r) = 56 cm
Circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 56 cm = 352 cm
Now perimeter of square = 352 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 2

Question 4.
 If a wire is bent into the shape of a square, then the area of the square is 81 cm². When wire is bent into a semicircular shape, then the area of the semi-circle will be
(a) 22 cm²
(b) 44 cm²
(c) 77 cm²
(d) 154 cm²
Solution:
Area of a square wire = 81 cm²
∴ Side of square = \(\sqrt { Area } \)  = \(\sqrt { 81 } \)  cm = 9 cm ard per in eret of square =4a = 4 x 9 = 36cm
Perimeter of semicircular wire whose bent = 36 cm
Let r be the radius, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 3

Question 5.
A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is
(a) 20 m
(b) 21 m
(c) 22 m
(d) 24 m
Solution:
Let R and r be the radii of the outer and inner circles of the park, then 2πR – 2πr = 132
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 4

Question 6.
The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be
(a) 2800
(b) 4000
(c) 5500
(d) 7000
Solution:
Radius of the wheel (r) = 0.25 m = 25 cm
Circumference of the wheel
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 5

Question 7.
The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is
(a) 55 m
(b) 110 m
(c) 220 m
(d) 230 m
Solution:
Ratio in the outer and inner perimeter of a circular path = 23 : 22
Width of path = 5 m
Let R and r be the radii of outer and inner path then R- r = 5 m ….(i)
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 6
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 7

Question 8.
The circumference of a circle is 100 cm. The side of a square inscribed in the circle is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 8
Solution:
Circumference of a circle (c) = 100 cm
Diagonal of square which is inscribed in the circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 9

Question 9.
The area of the incircle of an equilateral triangle of side 42 cm is :
(a) 22 73 cm²
(b) 231 cm²
(c) 462 cm²
(d) 924 cm²
Solution:
Side of an equilateral triangle (a) = 42 cm
Radius of inscribed circle = \(\frac { 1 }{ 3 }\) x altitude
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 10

Question 10.
The area of incircle of an equilateral triangle is 154 cm2. The perimeter of the triangle is
(a) 71.5 cm
(b) 71.7 cm
(c) 72.3 cm
(d) 72.7 cm
Solution:
Area of incircle of an equilateral triangle = 154 cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 11

Question 11.
The area of the largest triangle that can be inscribed in a semi-circle of radius r. is
(a) r²
(b) 2 r²
(c) r³
(d) 2r³
Solution:
The largest triangle inscribed in a semi-circle of radius r, can be ΔABC as shown in the figure, whose base = AB = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 12

Question 12.
The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is
(a) 70 cm²
(b) 140 cm²
(c) 210 cm²
(d) 420 cm²
Solution:
The perimeter of a triangle = 30 cm
and circumference of its incircle = 88 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 13

Question 13.
The area of a circle is 220 cm², the area of a square inscribed in it is
(a) 49 cm²
(b) 70 cm²
(c) 140 cm²
(d) 150 cm²
Solution:
Area of a circle = 220 cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 14

Question 14.
If the circumference of a circle increases from 4π to 8π, then its area is
(a) halved
(b) doubled
(c) tripled
(d) quadrupled
Solution:
In first case circumference of a circle = 4π
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 15

Question 15.
If the radius of a circle is diminished by 10%, then its area is diminished by
(a) 10% 
(b) 19%
(c) 20%
(d) 36%
Solution:
Let in first case radius of a circle = r
Then area = πr²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 16
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 17

Question 16.
If the area of a square is same as the area of a circle, then the ratio of their perimeter, in terms of 7t, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 18
Solution:
Let side of square = a
Perimeter = 4 a
Then area = a²
∴ Area of circle = a²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 19

Question 17.
The area of the largest triangle that can be inscribed in a semi-circle of radius r is
(a) 2r
(b) r²
(c) r
(d) \(\sqrt { r } \)
Solution:
Radius of semicircule = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 20

Question 18.
The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is
(a) π : \(\sqrt { 2 } \)
(b) π : \(\sqrt { 3 } \)
(c) \(\sqrt { 3 } \) :π
(d) \(\sqrt { 2 } \) : π
Solution:
Let side of equilateral triangle = a
Then area = \(\frac { \sqrt { 3 } }{ 4 }\) a²
Diameter of circle = a
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 21

Question 19.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then r + r
(a) >r²
(b) = r²
(c) < r²
(d) None of these
Solution:
Sum of area of two circles with radii r1  and r2
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 22

Question 20.
If the perimeter of a semi-circular protractor is 36 cm, then its diameter is
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
Perimeter of a semicircle = 36 cm
Let d be its diameter, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 23

Question 21.
The perimeter of the sector OAB shown in the fiugre, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 24
Solution:
Radius of sector of 60° = 7 cm
∴ Perimeter = arc AB + 2 r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 25

Question 22.
 If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is
(a) 58 cm²
(b) 52 cm²
(c) 25 cm²
(d) 56 cm²
Solution:
Radius of a sector = 6.5 cm
and perimeter = 29 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 26

Question 23.
If the area of a sector of a circle bounded by an arc of length 5K cm is equal to 20K cm², then its radius is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Let r be the radius, then
Length of the arc of sector of θ angle = 5π
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 27

Question 24.
The area of the circle that can be inscribed in a square of side 10 cm is
(a) 40π cm²
(b) 30π cm²
(c) 100π cm²
(d) 25π cm²
Solution:
Side of square = 10 cm
∴ Diameter of the inscribed circle = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 28

Question 25.
If the difference between the circumference
(a) 154 cm²
(b) 160 cm²
(c) 200 cm²
(d) 150 cm²
Solution:
Let r be the radius of a circle then circum-ference = 2πr
∴  2πr-r = 37
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 29

Question 26.
The area of a circular path of uniform width h surrounding a circular region of radius r is
(a) π (2r + h) r
(b) π (2r + h) h
(c) π (h + r)r
(d) π (h + r) A
Solution:
Let r be the radius of inner circle h is the width of circular path
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 30

Question 27.
 If AB is a chord of length 5\(\sqrt { 3 } \)  cm of a circle with centre O and radius 5 cm, then area of sector OAB is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 31
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 32
Solution:
Radius of the circle (r) = 5 cm
AB is a chord = 5\(\sqrt { 3 } \)
Draw OM ⊥ AB which bisects the chord AB at M
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 33

Question 28.
The area of a circle whose area and circumference are numerically equal, is
(a) 2π sq. units
(b) 4π sq. units
(c) 6π sq. units
(d) 8π sq. units
Solution:
Let radius of the circle = r
∴ Area = πr²
and circumference = 2πr
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 34

Question 29.
If diameter of a circle is increased by 40%, then its area increases by
(a) 96%
(b) 40%
(c) 80%
(d) 48%
Solution:
Let the diameter of a circle in first case = 2r
Then radius = r
Area = πr²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 35

Question 30.
In the figure, the shaded area is
(a) 50 (π – 2) cm²
(b) 25 (π – 2) cm²
(c) 25 (π + 2) cm²
(d) 5 (π – 2) cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 36
Solution:
In the figure, ∠AOB = 90°
and radius of the circle = 10 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 37

Question 31.
In the figure, the area of the segment PAQ is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 38
Solution:
a is the radius of the circle arc PAQ subtends angle 90° at the centre
∴ Area of segment PAQ
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 39

Question 32.
In the figure, the area of segment ACB is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 40
Solution:
r is the radius of the circle and arc ACB subtends angle of 120° at the centre
Area of segment ACB = r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 41

Question 33.
If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20rc cm², then the radius of the circle is
(a) 12 cm
(b) 16 cm
(c) 8 cm
(d) 10 cm
Solution:
Length of arc = 5π cm
area of sector = 20π cm²
Let the angle at the centre be θ
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 42

Question 34.
In the figure, the ratio of the areas of two sectors S1 and S2 is
(a) 5 : 2
(b) 3 : 5
(c) 5 : 3
(d) 4 : 5
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 43
Solution:
Let r be the radius of the circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 44
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 45

Question 35.
If the area of a sector of a circle is \(\frac { 5 }{ 18 }\) of the area of the circle, then the sector angle is equal to
(a) 60°
(b) 90°
(c) 100°
(d) 120°
Solution:
Area of sector of a circle = \(\frac { 5 }{ 18 }\) x area of circle
Let θ be its angle at the centre and r be radius
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 46

Question 36.
If the area of a sector of a circle is \(\frac { 7 }{ 20 }\) of the area of the circle, then the sector angel is equal to
(a) 110°
(b) 130°
(c) 100°
(d) 126°
Solution:
Area of sector of a circle = \(\frac { 7 }{ 20 }\) of the area of the circle
Let r be the radius and θ be its angle at the centre
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 47

Question 37.
In the figure, if ABC is an equilateral triangle, then shaded area is equal to?
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 48
Solution:
ΔABC is an equilateral triangle inscribed in a circle with centre O and radius r
BO and CO are joined
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 49

Question 38.
 In the figure, the ara of the shaded region is
(a) 3π cm²
(b) 6π cm²
(c) 9π cm²
(d) 7π cm²
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 50
Solution:
In the figure, ∠B = ∠C = 90°, ∠D = 60?
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 51
∴ ∠A= 360° – (90° + 90° + 60°) = 360° – 240° = 120°
Radius of the sector = 3 cm
∴Area of shaded portion
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 52

Question 39.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 13 : 22
(b) 14 : 11
(c) 22 : 13
(d) 11 :14
Solution:
Let side of square = a units
∴ Area = a² sq. units
and perimeter = 4a units
Now perimeter of circle = 4a units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 53
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 54

Question 40.
The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 55
Solution:
Radius of circle (R) = 20 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 56
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 57

Question 41.
The area of a sector whose perimeter is four times its radius r units, is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 58
Solution:
Radius of sector = r
Perimeter = 4r
and length of arc = 4r – 2r = 2r
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 59

Question 42.
If a chord of a circle of radius 28 cm makes an angle of 90° at the centre, then the area of the major segment is
(a) 392 cm²
(b) 1456 cm²
(c) 1848 cm²
(d) 2240 cm²
Solution:
A chord AB makes an angle of 90° at the centre
Radius of the circle = 28 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 60
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 61

Question 43.
If the area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the trianlge is
(a) 17\(\sqrt { 3 } \) units
(b) 36 units
(c) 72 units
(d) 48\(\sqrt { 3 } \)  units
Solution:
Area of a circle inscribed in an equilateral triangle = 48π sq. units
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 62

Question 44.
The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is
(a) 2.75 cm²
(b) 5.5 cm²
(c) 11 cm²
(d) 10 cm²
Solution:
Length of hour hand of a clock (r) = 6 cm
Time 11.20 am to 11.55 am = 35 minutes
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 63
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 64

Question 45.
ABCD is a square of side 4 cm. If ? is a point in the interior of the square such that ΔCED is equilateral, then area of ΔACEis
(a) 2 (\(\sqrt { 3 } \) – 1) cm²
(b) 4 (\(\sqrt { 3 } \) -1) cm²
(c) 6(\(\sqrt { 3 } \)-1)cm²
(d) 8(\(\sqrt { 3 } \)-1)cm²
Solution:
Side of square ABCD = 4 cm
and side of equilateral ΔCED = 4 cm
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 65
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 66

Question 46.
If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the larger circle (in cm) is
(a) 34
(b) 26

(c) 17
(d) 14

Solution:
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 67

Question 47.
If π is taken as 22/7, the-distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is (a) 2.2
(b) 1.1
(b) 9.625
(d) 96.25   [CBSE 2013]
Solution:
Diameter of a wheel = 35 cm = \(\frac { 35 }{ 100 }\) m
Circumference of the wheel = πd
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 68

Question 48.
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3). The length of one of its diagonals is
(a) 5
(b) 4

(c) 3
(d) 25 [CBSE 2014]
Solution:
Three vertices of a rectangle ABCD are B (4,0), C (4, 3) and D (0, 3) length of one of its diagonals
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 69

Question 49.
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 70
Solution:
Take a point C on the circumference of the semi-circle and join it by the end points of diameter A and B.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 71

Question 50.
If the sum of the areas of two circles with radii r1 and r2 is equal to the area of a circle of radius r, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 72
Solution:
According to the given condition,
Area of circle = Area of first circle + Area of second circle.
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 73

Question 51.
If the sum of the circumference of two circles with radii r, and r2 is equal to the circumference of a circle of radius r, then
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 74
Solution:
According to the given condition, Circumference of circle = Circumference of first circle + Circumference of second circle
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 75

Question 52.
 If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle < Area of the square
(c) Area of the circle > Area of the square

(d) Nothing definite can be said
Solution:
According to the given condition, Circumference of a circle = Perimeter of square 2 πr = 4 a
[where, r and a are radius of circle and side of square respectively]
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 76

Question 53.
 If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Solution:
Let radius of circle be r and side of a square be a
According to the given condition,
Perimeter of a circle = Perimeter of a square
RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS 77

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3

RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3

Other Exercises

Question 1.
For the following arithmetic progressions write the first term a and the common difference d :
(i) -5, -1, 3, 7, …………
(ii) \(\frac { 1 }{ 5 }\) , \(\frac { 3 }{ 5 }\) , \(\frac { 5 }{ 5 }\) , \(\frac { 7 }{ 5 }\) , ……
(iii) 0.3, 0.55, 0.80, 1.05, …………
(iv) -1.1, -3.1, -5.1, -7.1, …………..
Solution:
(i) -5, -1, 3, 7, …………
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 1
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 2

Question 2.
Write the arithmetic progression when first term a and common difference d are as follows:
(i) a = 4, d = -3
(ii) a = -1, d= \(\frac { 1 }{ 2 }\)
(iii) a = -1.5, d = -0.5
Solution:
(i) First term (a) = 4
and common difference (d) = -3
Second term = a + d = 4 – 3 = 1
Third term = a + 2d = 4 + 2 x (-3) = 4 – 6 = -2
Fourth term = a + 3d = 4 + 3 (-3) = 4 – 9 = -5
Fifth term = a + 4d = 4 + 4 (-3) = 4 – 12 = -8
AP will be 4, 1, -2, -5, -8, ……….
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 3
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 4

Question 3.
In which of the following situations, the sequence of numbers formed will form an A.P?
(i) The cost of digging a well for the first metre is ₹ 150 and rises by ₹ 20 for each succeeding metre.
(ii) The amount of air present in the cylinder when a vacuum pump removes each time \(\frac { 1 }{ 4 }\) of the remaining in the cylinder.
(iii) Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …, and so on. [NCERT Exemplar]
Solution:
(i) Cost of digging a well for the first metre = ₹ 150
Cost for the second metre = ₹ 150 + ₹ 20 = ₹ 170
Cost for the third metre = ₹ 170 + ₹ 20 = ₹ 190
Cost for the fourth metre = ₹ 190 + ₹ 20 = ₹ 210
The sequence will be (In rupees)
150, 170, 190, 210, ………..
Which is an A.P.
Whose = 150 and d = 20
(ii) Let air present in the cylinder = 1
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 5
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 6
(iii) Amount at the end of the 1st year = ₹ 1100
Amount at the end of the 2nd year = ₹ 1210
Amount at the end of 3rd year = ₹ 1331 and so on.
So, the amount (in ₹) at the end of 1st year, 2nd year, 3rd year, … are
1100, 1210, 1331, …….
Here, a2 – a1 = 110
a3 – a2 = 121
As, a2 – a1 ≠ a3 – a2, it does not form an AP

Question 4.
Find the common difference and write the next four terms of each of the following arithmetic progressions :
(i) 1, -2, -5, -8, ……..
(ii) 0, -3, -6, -9, ……
(iii) -1, \(\frac { 1 }{ 4 }\) , \(\frac { 3 }{ 2 }\) , ……..
(iv) -1, – \(\frac { 5 }{ 6 }\) , – \(\frac { 2 }{ 3 }\) , ………..
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 7
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 8
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Question 5.
Prove that no matter what the real numbers a and b are, the sequence with nth term a + nb is always an A.P. What is the common difference ?
Solution:
an = a + nb
Let n= 1, 2, 3, 4, 5, ……….
a1 = a + b
a2 = a + 2b
a3 = a + 3b
a4 = a + 4b
a5 = a + 5b
We see that it is an A.P. whose common difference is b and a for any real value of a and b
as a2 – a1 = a + 2b – a – b = b
a3 – a2 = a + 3b – a – 2b = b
a4 – a3 = a + 4b – a – 3b = b
and a5 – a4 = a + 5b – a – 4b = b

Question 6.
Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 12
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 13
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Question 7.
Find the common difference of the A.P. and write the next two terms :
(i) 51, 59, 67, 75, …….
(ii) 75, 67, 59, 51, ………
(iii) 1.8, 2.0, 2.2, 2.4, …….
(iv) 0, \(\frac { 1 }{ 4 }\) , \(\frac { 1 }{ 2 }\) , \(\frac { 3 }{ 4 }\) , ………..
(v) 119, 136, 153, 170, ………..
Solution:
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 19
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 20
RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 21
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Hope given RD Sharma Class 10 Solutions Chapter 5 Arithmetic Progressions Ex 5.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

A Question of Trust Extra Questions and Answers Class 10 English Footprints Without Feet

A Question of Trust Extra Questions and Answers Class 10 English Footprints Without Feet

Here we are providing A Question of Trust Extra Questions and Answers Class 10 English Footprints Without Feet, Extra Questions for Class 10 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-english/

A Question of Trust Extra Questions and Answers Class 10 English Footprints Without Feet

A Question of Trust Extra Questions and Answers Short Answer Type

A Question Of Trust Extra Questions Question 1.
What did Horace Danby hear from the doorway?
Answer:
Horace Danby heard a voice from the doorway. It was the voice of a lady. As Horace Danby has sneezed loudly, therefore the lady asked what it was and he replied that it was due to hay fever.

A Question Of Trust Extra Question Answer Question 2.
How did the lady in red convince Horace Danby to open the lock?
Answer:
The lady told Horace Danby that she had promised her husband to take her jewels to the hank but she left them in the safe. She had forgotten the numbers to open the safe and wanted to wear the jewels to a party. Horace Danby believed her and opened the safe for her.

A Question Of Trust Questions And Answers Question 3.
Was Horace Danby a typical thief?
Answer:
Horace Danby was not a typical thief. He made locks and was successful in his business. He loved rare and expensive books. To purchase these books he used to rob only one safe every year.

A Question Of Trust Class 10 Extra Questions Question 4.
What advice did the lady give Horace Danby about his. hay fever?
Answer:
The lady advised him that he could cure his hay fever with a special treatment, if he could find out just. what plant gave him the disease. She said sympathetically that he had better see a doctor if he was serious about his work.

Question Of Trust Extra Questions Question 5.
Why did Horace Danby feel sure of his success in that year’s robbery?
Answer:
Horace Danby felt sure of his success in that year’s robbery as he had been studying room, paths and gardens of the house at Shot over Grange for two weeks. He knew that the family was in London and two servants who lived in the house had gone to watch a movie that afternoon.

A Question Of Trust Question Answer Question 6.
Did Horace Danby get the jewels from the Grange safe? Then why did the Police arrest him?
Answer:
Horace Danby did not get the jewels from the Grange safe but he was arrested by the police as his fingerprints were traced on the lighter which he gave to the lady to light the cigarette. He was eager to please the lady to win her favour.

A Question Of Trust Short Question Answer Question 7.
What story did the lady tell Horace Danby to get the jewels?
Answer:
The lady made up a story that before going to London, she promised her husband to take her jewels to their bank, but she left them there in the safe. She wished to put on the jewels to a party that night. Above all she had forgotten the numbers to open the safe. Thus she convinced Horace Danby to open the safe.

Extra Questions Of A Question Of Trust Question 8.
Why was it not difficult for Horace to open the safe?
Answer:
It was not difficult for Horace Danby to open the safe because he had lived with locks and safes all his life. The burglar alarm was poorly built. He cut the wire without any difficulty.

A Question Of Trust Question And Answers Question 9.
What does the author tell us about Horace Danby’s life?
Answer:
Horace Danby was unmarried and fifty year old and lived with a housekeeper who bothered about his health. He had hay fever. He made locks and was successful in his business. He was good and respectable but not honest.

A Question Of Trust Extra Questions And Answers Question 10.
How often did Horace Danby commit a robbery every year? What did he do with the stolen money?
Answer:
Horace Danby committed only one robbery every year. He was fond of collecting rare and expensive books. He used to buy these books with stolen money through an agent secretly.

The Question Of Trust Questions And Answers Question 11.
What does Horace Danby like to collect?
Answer:
Horace Danby likes to collect rare and expensive books.

Question Of Trust Question Answer Question 12.
Why does he steal every year?
Answer:
He used to steal every year so that he could buy the rare and expensive books that he loved to collect. He planned meticulously before attempting a burglary, stole enough to last twelve months and secretly bought the books through an agent.

The Question Of Trust Extra Questions Question 13.
Who is speaking to Horace Danby?
Answer:
A lady standing in the doorways is speaking to Horace Danby. She is young and pretty, and is dressed in red. She said she had come just in time, or else her family would have been robbed by Horace. She, thus, pretended to be one of the members of the family living at Shotover Grange.

Class 10 A Question Of Trust Extra Questions Question 14.
Who is the real culprit in the story?
Answer:
The real culprit is the woman who pretended to be a member of the family living at Shotover Grange. She tricked Horace into believing her, and cleverly took away all the jewels that were kept in the safe.

A Question of Trust Extra Questions and Answers Long Answer Type

Question 1.
What do you think is the meaning of the phrase ‘honour among thieves’? Which of the two thieves lack the honour?
OR
Which of the two thieves lacked ‘honour’ in the story ‘A Question of Trust’?
Answer:
“There is honour among Thieves” means that ‘dishonest people may have certain standards of behaviour which they will respect’. The young lady pretended to be the owner of the house and innocently asked Horace Danby to-open the safe as she would need the jewels to wear to the party that day. Horace Danby found himself trapped in a tricky situation and could not make out how to escape. He readily opened the safe unwittingly leaving his fingerprints. He was arrested and nobody believed that the lady of the house asked him to open the safe.

Question 2.
At times we keep on planning things but our plans fail, we are not always responsible for the failure. Explain with reference to the story ‘A Question of Trust’.
OR
In the story ‘A Questions of Trust’, Horace Danby carefully planned his theft, but was outwitted by another thief‘The lady in red’. Would you agree that over confidence may prove fatal one day? Discuss.
Answer:
It is true that we keep on planning things in our life. We wish thing would materialise as we had planned but Man proposes God disposes. Horace Danby was a very careful planner. He made all plans very carefully before the robbery. He drew picture of wirihg of electricity and even took every efforts of befriending pets. But fate had planned something different. Lady in red turned out to be smarter and duped him. Horace, without any suspicion, opened the safe without gloves. This mistake landed him in prison. So one lesson that we learn that we should not trust anyone.

Question 3.
“Lying and stealing are next door neighbours”. Comment.
Answer:
A path of truthfulness and honesty is always better than an easy path of deceit. Horace Danby was a thief, he robbed safe every year. But, in the end he was be fooled by another thief and was arrested by the police. Stealing or lying cannot bring happiness or success. A truthful and honest person always leads a happy and peaceful life. He does not have to live with any kind of guilt. Stealing or lying robs one of peace. Those who are honest do not live a life of tension and are able to get peace of mind too.

Question 4.
Our sins never go unpunished. Horace Danby ultimately had to go to prison. This shows that inspite of planning a crime intelligently and carefully a criminal can’t escape the law— Explain / discuss.
Answer:
A path of truthfulness and honesty is always better than an easy path of deceit. Horace Danby was a thief, he robbed safe every year. But, in the end he was befooled by another thief and was arrested by the police. Stealing or lying cannot bring happiness or success. A truthful and honest person always leads a happy and peaceful life. He does not have to live with any kind of guilt. Stealing or lying robs one of peace. Those who are honest do not live a life of tension and are able to get peace of mind too.

Question 5.
“Horace Danby was clever but the young woman was cleverer”. Discuss.
OR
How did the lady in red turn out to be smarter than Horace Danby?
Answer:
Horace Danby was a cleaver thief. He lived a life of a respectable and successful businessman. He used to make locks. He was fond of rare and expensive books. He did not have enough money to buy these books. He used to rob a safe for this once in an year. He was very clever. He made his plan in such a way that he was never caught. He spent a lot of time in planning his mission. He never left any sign of theft but the lady who was another thief proved herself cleverer than Danby. She was a great actress as well.

She pretended to be the owner of the house. She tricked Danby who opened the safe for her and ‘ handed over the jewels. Horace was so nervous that he did not realise that he had no gloves in his hand. The lady had no tools, no gloves but managed to rob the safe without leaving any evidence against her. Thus, we can say that the lady was cleverer than Danby.

Question 6.
Did you begin to suspect, before the end of the story, that the lady was not the person Horace Danby took her to be? If so, at what point did you realize this, and how?
Answer:
Yes, one does begin to suspect before the end of the story that the lady was not the person Horace Danby took her to be. She was unusually calm on seeing Horace. This seemed strange enough. When she did not call the police, and instead asked Horace to break open the safe and take out all the jewels from it. It seemed suspicious. Moreover, it also seemed unlikely that she would forget the code to open the safe. Therefore, it was evident, before the story ended, that she was not the person Horace had taken her to be.

Question 7.
What are the subtle ways in which the lady manages to deceive Horace Danby into thinking she is the lady of the house? Why doesn’t Horace suspect that something is wrong?
Answer:
Her confident walk, her act of touching up her make-up and the ease with which she picks cigarette at the right place, are enough to deceive anybody. Horace was too frightened to think properly so he didn’t suspect anything.

Question 8.
“Horace Danby was good and respectable but not completely honest”. Why do you think this description is apt for Horace? Why can’t he be.categorized as a typical thief?
Answer:
“Horace Danby was good and respectable—but not completely honest”. This description is apt for Horace. He was about fifty years old. He robbed only from rich people. His purpose of stealing money was only to buy rare and expensive books. He stole only once every year. His intention to use the booty in buying books was good. However, the fact that he stole to achieve this intention showed that he was not completely honest.

He cannot be categorized as a typical thief because he did not steal to eat or drink and was not a regular offender. He did not harm anybody during the act. He had a house. He made locks, had two people to help him, and was successful in his business. He only stole enough money to buy the books. For a couple of days, he even kept his promises to the lady he met at Shotover Grange by not stealing or planning any robbery.

Question 9.
Horace Danby was a meticulous planner but still he faltered. Where did he go wrong and why?
Answer:
Horace Danby failed to get enough information about real occupants of the house. He seemed to be too occupied with collecting other little details and information about house map, wiring and location of valuable things. Although he was smart enough to know the dog’s actual name but overlooked getting identity of each and every occupant of the house. Once he was in problem then probably his clever mind gave way to carelessness leading him to open the safe without wearing gloves.

Question 10.
Do you think Horace Danby was unfairly punished, or that he deserved what he got?
Answer:
He deserved what he got. A crime is a crime no matter what the thief does with the booty. Whether it is committed a hundred times or just once, or even if nobody gets harmed still it is a punishable act.

Question 11.
Do intentions justify actions? Would you, like Horace Danby, do something wrong if you thought your ends justified the means? Do you think that there are situations in which it is excusable to act less than honestly?
Answer:
“Ends do not justify means”, this is a very old saying and has been tested many a times. Nobody should harm others for his own benefit. But this world doesn’t function on idealism. There are many examples of people duping others for quick gains and earning easy money. But crime is crime even if done for something good. These acts should be deplored and dealt with severely.

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Here we are providing Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-10-social-science/

The Nationalist Movement in Indo-China Class 10 Extra Questions History Chapter 2

Question 1.
How many nations are parts of Indo China?
Answer:
Three: Cambodia, Vietnam and Laos.

Question 2.
When was French Indo-China formed?
Answer:
Though the French troops landed in Vietnam in 1858, and the region came under the French control in 1880s, it was only after the French-Chinese war that Indo-China was formed in 1857.

Question 3.
Mention three barriers in Vietnam which were to be overcome as suggested by Paul Bernard.
Answer:

  1. High population level,
  2. low agricultural productivity,
  3. extensive indebtedness

Question 4.
What was the main object the French colonisers in introducing education in Vietnam?
Answer:
To create Asiatic France solidly tied to European France; the education would make the Vietnamese respect the French culture.

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Question 5.
Why did the French follow the deliberate policy of failing students in Vietnam?
Answer:
The policy was to fail the students so that they do not become eligible for jobs, in 1977 out of a population of 170 lakhs, only 400 students were passed.

Question 6.
Name the political party the students began forming.
Answer:
In 1920s, the students formed Party of Young Annan, and also a journal the Annanesc Student.

Question 7.
What were Vietnam’s religious beliefs?
Answer:
The Vietnamese religious beliefs were mixture or Buddhism, Confucianism, leading to a general uprising in Ngu An and Ha Tien where a thousand Catholics were killed.

Question 9.
What do you know about Hoa Hao movement?
Answer:
A religious anti-imperialist movement inspired by Huynh Phu So. The movement was launched in 1939 in Mekong area. The French government attempted to suppress it by sending the leader to exile to Laos, and his supporters to concentration camps.

Question 10.
What is ‘go-east movement;?
Answer:
‘Go-east movement’ was one movement through which Vietnamese students, inspired by Japan, wanted to drive out the French, overthrow the puppet monarchy supported by the French.

Question 11.
When was the Communist Party of Vietnam established? Who was its leader?
Answer:
The Communist Party of Vietnam (later renamed as the Indo-Chinese Communist Party) was established in 1930. Ho Chi Minh was its leader.

Question 12.
When was the Democratic Republic of Vietnam formed and who was its Chairman?
Answer:
The Democratic Republic Vietnam was formed in 1945 and Ho Chi Minh became its chairman.

Question 13.
Following the defeat of France, what were the results of the Geneva negotiations?
Answer:
Following the defeat of France in May 1954, the Vietnamese were forced to accept the division of Vietnam at the Geneva negotiation. The North Vietnam came under Ho Chi Minh at the South Vietnam, under Bao Dai.

Question 14.
What were the consequences of the US entry in Vietnam?
Answer:
The US entry in Vietnam, between 1965 and 1974, proved costly for both the Vietnamese as well as the American. The death toll was 47244 and those who were wounded were 303,704.

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Question 15.
Mention briefly women’s role in the Vietnam war.
Answer:
Women, during the Vietnamese war, were represented not only as warriors but also as workers. They used to have a rifle in one hand, and hammer, in the other hand. They worked as nurses.

Question 16.
What does the 1902 plague and measures to control it tell us about the French colonial attitude towards questions of health and hygiene.
Answer:
The attitude of the French government towards questions of health and hygiene and Vietnamese people was indifferent, irresponsible and careless.

Question 17.
How did the French establish their grip on Vietnam?
Answer:
The French troops landed in Vietnam in 1858 and by mid-1880s, they had established a firm grip over the northern region of the country. After the Franco-Chinese war, the French assumed control of Tonkin and Annam and, in 1887, French Indo-China. was formed. In the following decades the French sought to consolidate their position, and people in Vietnam began reflecting on the nature of the loss that Vietnam was suffering, nationalist resistance developed out of this reflection.

Question 18.
Why the French thought the colonies are necessary?
Answer:
Colonies were considered essential to supply natural resources and other essential goods. Like other Western nations, France also thought it was the mission of the ‘advanced’ European countries to bring the benefits of civilisation to backward people. For extracting maximum profit, they began building canals and draining lands for cultivating rice. They also developed infrastructure projects to help transport goods and for trade.

Question 19.
What was the immediate effect of expanding areas for irrigation?
Answer:
The French began irrigating lands in Mekong delta. The area under rice cultivation went up from 274,000 hectares in 1873 to 1.1 million hectares in 1900 and 2.2 million in 1930. Vietnam exported two-thirds of its rice production and by 1931 had become the third-largest exporter of rice in the world.

Question 20.
Mention the infrastructure projects which the French undertook in Vietnam.
Answer:
The French began infrastructure projects to help transports goods for trade, move military garrisons and control the entire region. Construction of a trans-Indo-China rail network that would link the northern and southern parts of Vietnam and China was begun. This final link with Yunan in China was completed by 1910. The second line was also built, linking Vietnam to Siam (as Thailand was then called), via the Cambodian capital of Phnom Penh.

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Question 21.
How did the French project themselves and the Vietnamese in school textbooks?
Answer:
School textbooks glorified the French and justified colonial rule. The Vietnamese were represented as primitive and backward, capable of manual labour but not of intellectual reflection; they could work in the fields but not rule themselves; they were ‘skilled copyists’ but not creative. School children were told that only French rule the Vietnamese peasant no longer lives in constant terror of pirates Calm is complete, and the peasant can work with a good heart.

Question 22.
How do you describe Phan Boi Chau as a nationalist?
Answer:
Phan Boi Chau (1869-1940) was an important nationalist of Vietnam. In fact, he became a major figure in the anti-colonial resistance from the time he formed the Revolutionary Society (Duy Ten Hoi) in 1903, with Prince Cuong De as the head Phan Boi Chau met the Chinese reformer Liang Qichao (1873-1929) in Yokohama in 1905. Phan’s most influential book, The History of the loss of Vietnam was written under the strong influence and advice of Qichao.

It became a widely read bestseller in Vietnam and China and was even made into a play. The book focuses on two connected themes; the loss of sovereignty and the severing of ties with China-ties that bound the elites of the two countries within a shared culture.

Question 23.
Give a brief account of Phan Chu Trinh.
Answer:
Phan Chu Trinh (1871-1926) was intensely hostile to the monarchy and opposed to the idea of resisting the French with the help of the court. His desire was to establish a democratic republic, Profoundly influenced by the democratic ideals of the West, he did not want to wholesale rejection of Western civilisation.

He accepted the French revolutionary ideal of liberty but charged the French for not abiding by the ideal. He demanded that the French set up legal and educational institutions, and develop agriculture and industries.

Question 24.
When was the Democratic Republic of Vietnam formed?
Answer:
After establishing the Communist Party of Vietnam, Ho Chi Minh became an important leader of the country. During the Second World War, the Japanese occupied Vietnam. The nationalist now were faced with two enemies: the Japanese and the French.

The League for the Independence of Vietnam (Vietnam Doc Lap Dong Mirth), which came to be known as the Vietnam, fought the Japanese occupation and recaptured Hanoi in September 1945. The Democratic Republic of Vietnam was formed and Ho Chi Minh became Chairman.

Question 25.
Give an account of the major event after the formation of the Republic of Vietnam in 1945.
Answer:
The new republic faced a number of challenges. The French tied to regain control by using the emperor, Bao Dai, as their puppet. Faced with the Faced offensive, the Vietnam were forced to retreat to the hills. After eight years of fighting the French were defeated in 1954 at Dien Bien Phu. The entire commanding staff, including a general, India And Contemporary World-II /17 16 colonels and 1749 officers were taken, prisoners.

In the peace negotiations in Geneva that followed the French”defeat, the Vietnamese were persuaded to accept the division of”the country. North and South were split: Ho Chi Minh and the”put in power in the south.” communists took power in the north while Boi Dai’s regime was”account of the US bombing in Vietnam.”

Question 26.
What was the deadly poison ’Agent Orange? Give an account of the US bombing in Vietnam.
Answer:
Agent Orange is a defoliant, a plant killer, so-called “because it was stored in drums marked with an orange band” Between 1961 and 1971, some 11 million gallons of this chemical” was sprayed from cargo planes by US forces. Their plan was to”destroy forests and fields, so that would be easier to kill if there”was no jungle cover for people to hide in.

Over 14 per cent of the country’s farmland was affected by this poison. The tonnage of”bombs, including chemical arms, used during the US intervention”(mostly against civilian targets) in Vietnam exceeds that used” throughout the Second World War.

Question 27.
What do you know about the Ho Chi Minh trail?
Answer:
The Ho Chi Minh trail, a network of footpaths and roads, “was used to transport men and materials from the north to the”south. The trail was improved from the late 1950s, and from 1967″ about 20,000 North Vietnamese troops came south each month on”this trail.

The trail had support bases and hospitals along the way. In”some parts supplies were transported in trucks, but mostly they”were carried by porters, who were mainly women. These porters”carried about 25 kilos on their backs or about 70 kilos on their bicycles.

Most of the trail was outside Vietnam in neighbouring Laos” and Cambodia with branch lines extending into South Vietnam “The US regularly bombed this trail trying to disrupt supplies, but”efforts to destroy this important supply line by intensive bombing” failed because they were rebuilt very quickly.

Ho Chi Minh (1890-1969)
Little is known about his early life mostly because Minh chose to”downplay his personal background and identify himself with the cause” of Vietnam, Probably born as Nguyen Van Thanh in Central Vietnam,” he studied at French schools that produced leaders such as Ngo Dinh” Diem, Vo Nguyen Giap ane, Pham Van Dong He briefly taught in I “1910, and in 1911, learnt baking and took a job on a French liner on”the Saigon-Marseilles run Minh became an active member of the”Comintern, meeting Lenin and other leaders in May 1941, after 30” years abroad in Europe, Thailand and China, Minh finally returned to”Vietnam In 1943 he took the name. Ho Chi Minh (Fie who Enlightens).

“He became president of the Vietnam Democratic Republic, Ho Chi” Minh died on 3 September 1969. He led the party successfully for over”40 years, struggling to preserve Vietnamese autonomy.

Question 28.
What dilemmas were the French faced with regard to the” spread of education in Vietnam?
Answer:
Education was seen as one way to civilise the ‘native .” But in order to educate them, the French had a dilemma. How far”were the Vietnamese to be educated? The French needed an”educated local labour force but they feared that education might”create problems. Once educated, the Vietnamese may begin to question colonial domination.

Moreover, French citizens living in Vietnam (called colons) began fearing that they might lose their”jobs – as teachers, shopkeepers, policemen – to the educated”Vietnamese. So they opposed policies that would give the”Vietnamese full access to French education.

The French were faced with yet another problem in the sphere” of education: the elites in Vietnam were powerfully influenced by”Chinese culture. To consolidate their power, the French had to”counter the Chinese influence. So they systematically dismantled”the traditional educational system and established French schools ‘for the Vietnamese, But this was not easy.

The solution which came up was that the Vietnamese be taught”at the lower classes, and the French at the higher ones. The French”wanted Asiatic France solidly tied to European France. However,” who learnt the French culture could acquire French citizenship.

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Question 29.
How was the Hoa Hao movement opposed to the colonial”rule in Vietnam. Discuss briefly.
Answer:
Numerous movements were opposed to the French colonialism One such movement was the Hoa Hao movement. It began in 1935 and gained great popularity in the fertile Mekong” uprisings of the nineteenth century, “delta area. It drew on religious ideas popular in anti-French uprisings of the nineteenth century.

The founder of Hoa Hao was a man called Huynh Phu So. He performed miracles and helped j the poor. His criticism against”useless I expenditure had a wide appeal. He also opposed the sale” of child brides, gambling and the use of alcohol and opium.

The French tried to suppress the movement inspired by Huynh Phu So. They declared him mad, called him the Mad Bonze, and”put him in a mental asylum. Interestingly, the doctor who had to”prove him insane became his follower, and finally in 1941, even”the French doctors declared that he was sane. The French” authorities exiled him to Laos and sent many of his followers to”concentration camps.

Question 30.
Give the story of the Vietnamese plague and the”subsequent hunt for rats.
Answer:
The French, in their effort to modernize Vietnam, took” measures to beautify Hanoi,, its ! French part while keep the ’native'” part as dirty I as before. In the French part, a well-laid out sewer” system was introduced. The refuse from the native part of the city” began draining towards the French part of Hanoi. The result was”the spread of plague. The sewers, as a great transport system, “allowing the rats to move around the city without any problem. “And rats began to enter the well-cared-for homes of the French” through I the sewage pipes. What was to be done.

To stem this invasion, a rat hunt was started m 1902. The French” hired Vietnamese workers and paid them for each rat they caught.”Rats began to be caught in thousands: on 30 May, for instance,”20,000 were caught but still there seemed to be no end. For the”Vietnamese the rat hunt seemed to provide an early lesson in the”success of collective bargaining.

Question 31.
How did the Vietnamese nationalists S look to Japan and”China for inspiration? Explain. !
Answer:
Early Vietnamese nationalists had a dose relationship” with Japan and China. They provided models for those looking to change, a refuge for those who were escaping French police, and a location where a wider Asian network of revolutionaries could be”established. In the first decade of the twentieth century, a ‘go east movement’ became popular. In 1907-08 some 300 Vietnamese” students went to Japan to acquire modern education.

For many of them, the primary objective was to drive out the French from Vietnam, overthrow the purple emperor and re-establish the”Nguyen dynasty that had been deposed by the French. These”nationalists looked for foreign arms and help. They appealed to”the Japanese as fellow Asians, Japan had modernised itself and”had resisted colonisation by the West, Besides, its victory over, “Russia in 1907 proved its military capabilities.

Developments in China also inspired, Vietnamese nationalists. In 1911, the long-established monarchy in China was overthrown by a popular movement under Sun Yat-sen, and j a Republic was set up. Inspired by these developments, Vietnamese students organised the Association for the Restoration of Vietnam (VietNam Quan Phuc Hoi). Now the nature of the anti-French”independence movement changed. The objective was no longer to”set up a constitutional monarchy but a democratic republic/’society.

Question 32.
Mention how the new woman arose in the Vietnamese society.
Answer:
Women in Vietnam traditionally enjoyed greater equality than in China, particularly among the lower classes, but they had only limited freedom to determine their future and played no role in public life.
As the nationalist movement grew, a new image of womanhood emerged.

Writers and political thinkers began”idealising women who rebelled against social norms. In the 1930s, a famous novel by Nhat Linh caused a scandal because it showed”a woman leaving a forced marriage and marrying someone of her choice, someone who was involved in nationalist politics. This rebellion against social conventions marked the arrival of the new* woman in Vietnamese society.

Rebel women of the past were similarly celebrated In 1913, the nationalist Phan Boi Chau wrote a play based on the lives of the Trung sisters who had fought against Chinese domination in 39-43 CE. In this play, he depicted these sisters as patriots fighting to save the Vietnamese nation from the Chinese.

They were depicted in paintings, plays and novels as “representing the indomitable will and the intense patriotism of the Vietnamese We are told that they gathered a force of over 30,000, resisted the Chinese for two years, and when ultimately defeated,” they committed suicide, instead of surrendering to the enemy.

Question 33.
How did the war come to an end in Vietnam?
Answer:
The prolongation of the war created strong reactions even” within the US. It was clear that the US had failed to achieve its objectives: the Vietnamese resistance had not been crushed; the support of the Vietnamese people for US action had not been won. In the meantime, thousands of young US soldiers had lost their lives, and countless Vietnamese civilians had been killed. This was a war that has been called the first television war.

Battle scenes were shown on the daily news programmes. Many became disillusioned with what the US was doing and writers such as Mary McCarthy, and actors Intake Jane Fonda even visited North Vietnam and praised their heroic defence of the country. The scholar Noam Chomsky called the war the greatest threat to peace, to national self-determination, and to international cooperation.

The widespread questioning of government policy strengthened moves to negotiate an end to the war. A peace” settlement was signed in Paris in January 1974. This ended conflict with the US but fighting between tire Saigon regime and the NLF continued, The NLF occupied the presidential palace in Saigon on 30 April 1975 and unified Vietnam.

Objective Type Questions

Question 1.
Choose true (✓) and false (✗) in the following:
(i) After the Franco-Chinese war, the French took control” of Hanoi.
(ii) Before the French occupation of Vietnam, the country” was ruled by Nguyen dynasty,
(iii) Tonkin Free School began functioning in 1907.
(iv) Plague struck Hanoi in 1899.
(v) The English intervened in the Vietnam war in 1974.
Answer:
(i) ✗,
(ii) ✓,
(iii) ✓,
(iv) ✗
(v) ✗

Class 10 History Chapter 2 Extra Questions and Answers The Nationalist Movement in Indo-China

Question 2.
Choose the most appropriate alternative
(i) Cochinchina (the South) became a French colony in
(a) 1867
(b) 1877
(c) 1888
(d) 1898
Answer:
(a) 1867

(ii) The Communist Party of Vietnam was formed by
(a) Phan Boj Chau
(b) Ho Chi Minh
(c) Than Chu Trinh
(d) Hugnu Phu So
Answer:
(b) Ho Chi Minh

(iii) The New Republic of Vietnam in the north was established in
(a) 1955
(b) 1934
(c) 1956
(d) 1957
Answer:
(b) 1934

(iv) Phan Boj Chau wrote the following book
(a) The History of the loss of Vietnam
(b) The History of Vietnam
(c The Loss of Vietnam
(d) None of the above
Answer:
(a) The History of the loss of Vietnam

(v) The name of the plant killer used by the USA in Vietnam war was
(a) Agent Mange
(b) Agent Pear
(c) Agent Apple
(d) Agent Orange
Answer:
(d) Agent Orange.

Extra Questions for Class 10 Social Science

Mijbil the Otter Extra Questions and Answers Class 10 English First Flight

Here we are providing Mijbil the Otter Extra Questions and Answers Class 10 English First Flight, Extra Questions for Class 10 English was designed by subject expert teachers.

Mijbil the Otter Extra Questions and Answers Class 10 English First Flight

Mijbil the Otter Extra Questions and Answers Very Short Answer Type

Mijbil The Otter Extra Question Answer Question 1.
Where did Gavin Maxwell live?
Answer:
Gavin Maxwell lived in Camusfeama.

Mijbil The Otter Question Answer Question 2.
What did the author decide to experiment?
Answer:
The author decided to have an otter as a pet instead of a dog.

Mijbil The Otter Class 10 Extra Questions And Answers Question 3.
Why was his place a suitable place for that experiment?
Answer:
The author’s place was a suitable place for that experiment because otters love to see flowing water and feed on fish.

Mijbil The Otter Extra Questions Question 4.
What did the author’s friend advise him?
Answer:
The author’s friend advised him to get an otter from the Tigris marshes.

Question 5.
Why did the author go to Basra?
Answer:
The author went to Basra to the Consulate—General to collect and answer his mail from Europe.

Question 6.
Why did the author cable to England?
Answer:
The author cabled to England to get information about the status of his mail.

Question 7.
Why did the author wait for three days?
Answer:
The author waited to telephone for three days as there were problems with it.

Question 8.
When did the author receive his mail?
Answer:
Five days later, the author received his mail.

Question 9.
What did the author find in his bedroom?
Answer:
The author found two Arabs in his bedroom.

Question 10.
What did the author see in the bathroom?
Answer:
The author found Mijbil upon the bath tub.

Question 11.
Where did Mijbil go?
Answer:
Mijbil went inside the bathroom to have fun with water.

Question 12.
What did the otter want to do?
Answer:
Mijbil, the otter was trying to turn the tap so that water could come out of it.

Question 13.
How did Mijbil spend his time?
Answer:
Mijbil spent his time in playing with rubber ball.

Question 14.
Who was in the box.
Answer:
Mij, the otter was in the box.

Question 15.
What did the author find in the box?
Answer:
The author found complete silence in the box.

Question 16.
What was the condition of Mijbil in the box?
Answer:
Mijbil’s condition was miserable.

Question 17.
How did Mijbil spend his time in London?
Answer:
Mijbil spent his time playing with a selection of toys—ping-pong balls, rubber fruit and terrapin shell.

Question 18.
What did Mijbil invent?
Answer:
Mijbil invented a new game with the ping-pong ball.

Question 19.
How did he play with his new game?
Answer:
Mijbil placed the ball on the light end and tried to grab it from the other end.

Question 20.
What did the labourer want to know?
Answer:
The labourer wanted to know the species of Mijbil.

Mijbil the Otter Extra Questions and Answers Short Answer Type

Question 1.
What special characteristic of Mijbil did Maxwell learn after he took it to the bathroom?
Answer:
When Maxwell took Mijbil to the bathroom, he learnt that he went wild with joy in water, plunging and rolling in it and splashed water.

Question 2.
Why did the writer stay in Basra?
Answer:
The writer and his friend were going to Basra to the Consulate—General to collect and answer their mails. They found that the mail of his friend was available but the author had to stay in Basra for five days.

Question 3.
What did the otter look like?
Answer:
The otter looked like a very small dragon. Its appearance was very pretty. It was coated with pointed scales of mud on its back. It spread mud all round.

Question 4.
How did Maxwell get the otter?
Answer:
Maxwell wanted to keep an otter as a pet instead of dog. He expressed his wish before his friend who managed to get an otter and sent it to Maxwell through two Arabs. Thus Maxwell got his tamed otter.

Question 5.
Why was the otter called Maxwell’s otter?
Answer:
The otter was not a common wild creature. It belonged to an unknown race of otters that was later christened by zoologists as Lutrogale perspicillata Maxwelli; As it came into light as Maxwell’s pet, it was named as Maxwell’s otter.

Question 6.
Why did Maxwell call the airhostess “the very queen of her kind”?
Answer:
Maxwell was worried about Mijbil’s welfare. Mijbil was exhausted and wounded. His condition was very miserable in the box. Maxwell wanted Mijbil to be free. The airhostess allowed him to have Mijbil on his knee. Out of gratitude for her sympathy and kindness, the author said so.

Question 7.
What happened when Maxwell decided to transport Mijbil to England by air?
Answer:
When Maxwell decided to transport Mijbil to England, Mijbil was kept into a box. He tore off the box and wounded himself. Then Mijbil disappeared down the aircraft, terrorizing the other passengers. After that, he settled on Maxwell’s knee and gave no further inconvenience to others.

Question 8.
Write a short paragraph from the text to show that Mijbil was an intelligent animal.
Answer:
One day Mijbil escaped from Maxwell’s room and went to the bathroom. When the writer reached there he found that the otter had turned the tap in less than a minute and he had managed to get full flow of water for a bath.

Question 9.
What game did Mijbil invent?
Answer:
Mijbil discovered that if he placed the ball on the high end of the damaged suitcase, it would run down the length of the suitcase. He would dash around to the other end to ambush its arrival, hide from it,. crouching, to spring up and take it by surprise, grab it and trot off with it to the high end once more.

Question 10.
What guesses did the Londoners make about what Mijbil was?
Answer:
The writer was not surprised to know that the Londoners could not recognize an otter. They thought he was a baby seal, a squirrel, a walrus, a hippo, a beaver, a bear cub, a leopard.

Question 11.
What ‘experiment’ did Maxwell think Camusfearna would be suitable for?
Answer:
Maxwell was staying in a cottage in Camusfearna. There was a lot of water around the cottage. He thought to have an otter as a pet instead of a dog. An otter likes flowing water. So the author thought that Camusfearna would be suitable spot for that experiment.

Question 12.
Why does he go to Basra? How long does he wait there, and why?
Answer:
He goes the Basra to the Consulate—General to collect and answer his mail from Europe. His mail did not arrive in time. He cabled to England, and when, three days later, nothing had happened, he tried to telephone. The call had to be booked twenty-four hours in advance. On the first day the line was out of order; on the second day, exchange was closed for a religious holiday. On the third day there was another breakdown. His mail arrived five days later. He had to wait for five days.

Question 13.
How does he get the otter? Does he like it? Pick out the words that tell you this?
Answer:
Maxwell expressed his desire to have an otter as a pet to his friend. His friend kept it in mind and sent an otter through two

Question 14.
Why was the otter named ‘Maxwell’s otter’?
Answer:
This race of animals was unknown to the scientists in the beginning. After its discovery it was christened by zoologists Lutrogale perspicillata maxwelli or Maxwell’s otter.

Question 15.
Tick the right answer. In the beginning, the otter was

  • aloof and indifferent
  • friendly
  • hostile

Answer:
In the beginning the otter was aloof and indifferent.

Question 16.
What happened when Maxwell took Mijbil to the bathroom? What did it do two days after that?
Answer:
When Maxwell took Mijbil to the bathroom he went wild with joy in the water, plunging and rolling in it, shooting pp and down the length of the bathtub underwater, and making enough slosh and splash for a hippo. Two days after, Mijbil escaped from the bedroom and entered the bathroom. He struggled with the chromium tap till it had a full flow.

Question 17.
How was Mij to be transported to England?
Answer:
The British airline to London did not permit to fly animals. The author had to book a flight to Paris on another airline. The airline insisted that Mij should be packed into a box not more than eighteen inches square. The author acted accordingly and transported Mij to England.

Question1 8.
What did Mij do to the box?
Answer:
The author put Mij into the box an hour before he started for the airport so that Mij would become accustomed to it and left for a meal. When he came back he found complete silence in the box. He saw blood stains around the airholes. Mij had tom the lining of the box to shreds. He got himself hurt.

Question 19.
Why did Maxwell put the otter back in the box? How do you think he felt when he did this?
Answer:
Maxwell put the otter back in the box because he was already late for the airport. It was just ten minutes time for the flight, and the airport was five miles distant. He felt bad in doing so. He did not want to keep the’miserable otter in the box but there was no other option due to time constraint.

Question 20.
Why does Maxwell say the airhostess was “the very queen of her kind”?
Answer:
The author told the airhostess about the incident that took place half an hour before at home and took her into his confidence. The airhostess was a considerate lady. She cooperated with the author and suggested him to keep his pet on his knee. This made the author call her “ the very queen of her kind”.

Question 21.
What happened when the box was opened?
Answer:
The airhostess suggested the author that he could keep the pet on his knee. The author opened the box. Mij was out of the box in a flash. He disappeared at high speed down the aircraft. There was chaos all around in the plane.

Question 22.
What game had Mij invented?
Answer:
Mij had invented a game of his own with a ping-pong ball. During the journey, the suitcase of the author was damaged. It got a slope on one end. Mij discovered that if he placed the ball on the high end it would run down the length of the suitcase.

Question 23.
What are ‘compulsive habits’? What does Maxwell say are the compulsive habits of
(i) school children
(ii) Mij ?
Answer:
Compulsive habits are those acts which result from an irresistible urge. School children on their way to and from school must place their feet squarely on the centre of each paving block; must touch every seventh upright of the iron railings, or pass to the outside of every second lamp post. Mij had also developed certain compulsive habits. There was a single-storied primary school opposite to the author’s home. There was a two feet high wall also. On his way to the home, Mij would tug the author to that wall, jump on to it, and gallop the full length of its thirty yards.

Question 24.
What groups of animals do otters belong to?
Answer:
Otters belong to a comparatively small group of animals called ‘Mustellines’. The zoologists call it Lutrogale perspicillata maxwelli.

Question 25.
What guesses did the Londoners make about what Mij was?
Answer:
The average Londoner did not recognise an otter. They made a lot of guesses about what Mij was. For some it was a baby seal, a walrus, a hippo, a beaver and for others a bear cub, a leopard or a brontosaur etc. For them Mij was anything but an otter.

Mijbil the Otter Extra Questions and Answers Long Answer Type

Question 1.
“The airhostess was the very queen of her kind” Do you agree? Comment.
Answer:
The airhostess was the very queen of her kind. I fully agree with the author’s statement. The chief duty of an airhostess is to make the passengers feel comfortable during the journey. The airhostess on Maxwell’s plane does all she can do for him. That is why Maxwell calls her the very queen of her kind.

She calmly listens to the author, allows him to keep the animal on his knees and tries to catch it when it escapes. In doing so she does not loose her patience. The author also tries to catch the animal and in doing so, he lands himself on a passenger. His face is covered with curry. Once again the airhostess comes to Maxwell’s help. Thus we see that the airhostess was really a nice lady/kind lady.

Question 2.
What do you know about Mijbil’s journey to London?
Answer:
The author had to come back to London from Basra. The British Airlines does not allow to fly animals so he had to book his ticket in another airline. The airline authorities insisted the author to pack Mijbil in a box. The author had a box and put Mijbil in it an hour before he started for the airport so that Mijbil . would become accustomed to it and left for a hurried meal. When he came back he found Mijbil wounded in the box. He had ten minutes left to catch the flight, so he kept it back to the box.

In the flight, the author told the airhostess about the miserable condition of Mijbil and took her into confidence. She was a considerate lady and suggested that he could keep Mijbil on his knee. The author opened the box. Mijbil was out of the box in a flash and disappeared down the aircraft. The author dived to catch Mijbil and he missed; but he found his mouth covered with curry. The airhostess helped in the search. The author came back to his seat and found Mijbil near his knee.

Question 3.
If you were in place of the airhostess and some incidents might have happened with you in the flight, how would you describe your experience?
Answer:
If I were in place of the airhostess and the same incidents might have happened with me in the flight, my experience might be the same as of the airhostess. But I would like to say something in this matter. As an airhostess, I would never permit to open the box of the otter (animal). I would be punctual for my duties. I know that kindness has a unique importance in our life. But duty is above all. I would think about the bad impacts of a free otter in the plane. It might be a great risk for all the passengers. In that condition, my reaction might be different. When the author would ask me to let him open the box, I would make him understand about the risk. In the name of rules and regulations basic values are ignored but people like the airhostess in “Mijbil the Otter” as a ray of hope.

Question 4.
What things does Mij do which tell you that he is an intelligent, friendly and fun-loving animal who needs love?
Answer:
Mij was an intelligent, friendly and fun-loving animal. When the author received it, for the first twenty- four hours it remained aloof and indifferent but later on he took interest in his surroundings. He became friendly to the author. He enjoyed his bath. One day he disappeared from the bedroom and entered the bathroom where he tried to open the tap.

His intelligence is further revealed when he invented a game of his own of ping- pong ball. Mij discovered that if the ball was placed on the high end of the suitcase, it would rather run down the length of the suitcase. He enjoyed playing with ball and marbles. Mij had developed certain compulsive habits like school children. He used to gallop at full speed on the thirty yards wall of a primary school.

Question 5.
What are some of the things we come to know about otters from this text?
Answer:
Some of the things we come to know about otters from this text are as under:
(i) they are intelligent, friendly and fun-loving creatures.
(ii) they love to see water flowing.
(iii) they don’t,like static water.
(iv) they love galloping and jumping.
(v) they like to be in water.

Question 6.
Why is Mij’s species now known to the world as Maxwell’s otter?
Answer:
Maxwell brought an otter back from Iraq and raised it in Scotland. He took the otter, called Mijbil, to the London zoological society, where it was decided that this was a previously unknown sub-species of smooth coated otter, and it was named after him. Since then Mij’s species is known as Maxwell’s otter.

Mijbil the Otter Extra Questions and Answers Reference to Context Questions

Read the following passages and answer the questions that follow:

Question 1.
Early in the new year of 1956 I travelled to Southern Iraq. By then it had crossed my mind that I should like to keep an otter instead of a dog, and that Camusfeama, ringed by water a stone’s throw from its door, would be an eminently suitable spot for this experiment. (Page 104)
(i) When did the author travel to Southern Iraq?
(ii) What did the author desire to keep instead of a dog?
(iii) Pick out the words from the passage that mean same as ‘thought, came to mind’.
(iv) What did the writer decide to experiment?
Answer:
(i) In the new year of 1956 the author had travelled to southern Iraq.
(ii) The author desired to keep an otter instead of a dog.
(iii) ‘crossed my mind’.
(iv) The writer decided to have an other as a pet instead of a dog.

Question 2.
When I casually mentioned this to a friend, he as casually replied that I had better get one in the Tigris marshes, for there they were as common as mosquitoes, and were often tamed by the Arabs. We were going to Basra to the Consulate – General to collect and answer out mail from Europe. At the Consulate-General we found that my friend’s mail had arrived but that mine had not.
(i) Who is T in this passage?
(ii) What did the author mention to a friend?
(iii) What did his friend advise him?
(iv) Pick out the word from the passage that means same as “Post”.
Answer:
(i) T here in the passage is used for Gavin Maxwell.
(ii) The author mentioned that he wanted to keep an otter as a pet.
(iii) His friend advised him to get an otter from Tigris marshes.
(iv) ‘Mail’.

Question 3.
I cabled to England, and when, three days later, nothing had happened, I tried to telephone. The call had to be booked twenty-four hours in advance. On the first day the line was out of order; on the second the exchange was closed for a religious holiday. On the third day there was another breakdown. My friend left, and I arranged to meet him in a week’s time. Five days later, my mail arrived.
(i) What did the author cable to England?
(ii) What happened after three days?
(iii) When did the author receive the mail?
(iv) Pick out the word from the passage that means same as ‘sent a message by telegram’.
Answer:
(i) The author cabled to England to get information about, his mail.
(ii) The author did not receive any mail after three days.
(iii) Five days later the author received the mail.
(iv) cabled.

Question 4.
The creature that emerged from this sack on to the spacious tiled floor of the Consulate bedroom resembled most of all a very small, medievally conceived, dragon. From the head to the tip of the tail he was coated with symmetrical pointed scales of mud armour, between whose tips was visible a soft velvet fur like that of a chocolate-brown mole. He shook himself, and I half expected a cloud of dust, but in fact it was not for another month that I managed to remove the last of the mud and see the otter, as it were, in his true colours.
(i) Write the name of creature which came out from the sack?
(ii) What was there on its tips?
(iii) How did the creature/otter look?
(iv) Pick out the word from the passage that means same as—‘clean’.
Answer:
(i) The creature was an otter which came out of the saSk.
(ii) There was a small chocolate brown mole.
(iii) The creature looked like a small medievally conceived dragon.
(iv) ‘remove’.

Question 5.
Mijbil, as I called the otter, was, in fact, of a race previously unknown to science, and was at length christened by zoologists Luthrogale perspicillata maxwelli, or Maxwell’s otter. For the first twenty-four hours Mijbil was neither hostile nor friendly; he was simply aloof and indifferent, choosing to sleep on the floor as far from my bed as possible. The second night Mijbil came on to my bed in the small hours and remained asleep in the crook of my knees until the servant brought tea in the morning, and during the day he began to lose his apathy and take a keen, much too keen, interest in his surroundings.
(i) How did the zoologists name Mijbil?
(ii) How did Mijbil behave in the beginning hours?
(iii) Where did Mijbil sleep on the second night?
(iv) Pick out the word from the passage that means the same as— ‘enemy’.
Answer:
(i) Zoologists name Mijbil as Lutrogale perspicillata maxwelli.
(ii) Mijbil behaved in a manner which was neither friendly nor hostile in the beginning hours.
(iii) Mijbil slept on the second night on the author’s bed in the crook of his knees.
(iv) ‘hostile’.

Question 6.
Two days later, Mijbil escaped from my bedroom as I entered it, and I turned to see his tail disappearing round the bend of the corridor that led to the bathroom. By the time I got there he was up on the end of the bathtub and fumbling at the chromium taps with his paws. I watched, amazed; in less than a minute he had turned the tap far enough to produce a trickle of water, and after a moment or two achieved the full flow.
(i) What did Mijbil do two days later?
(ii) Where did Mijbil go?
(iii) What was Mijbil doing.
(iv) Pick out the word from the passage that means the same as—‘trying to do something in a clumsy manner’.
Answer:
(i) Mijbil escaped from the author’s bedroom two days later.
(ii) Mijbil went inside the author’s bathroom to have fun with water.
(iii) Mijbil was fumbling at the chromium taps with his paws.
‘(iv) fumbling’.

Question 7.
Very soon Mij would follow me without a lead and come to me when I called his name. He spent most of his time in play. He spent hours shuffling a rubber ball round the room like a four-footed soccer player using all four feet to dribble the ball, and he could also throw it, with a powerful flick of the neck, to a surprising height and distance. But the real play of an otter is when he lies on his back and juggles with small objects between his paws. Marbles were Mij’s favourite toys for this pastime: he would lie on his back rolling two or more of them up and down his wide, flat belly without ever dropping one to the floor.
(i) Identify ‘Mij’ and T.
(ii) How did Mij spend his time?
(iii) How did Mij look like while playing?
(iv) Pick out the word from the passage that means the same as—“moving with dragging”.
Answer:
(i) ‘Mij’ was the author’s pet and T refers to Gavin Maxwell.
(ii) Mij spent most of his time in play.
(iii) Mij looked like a four footed soccer player while playing.
(iv) ‘shuffling’.

Question 8.
The days passed peacefully at Basra, but I dreaded the prospect of transporting Mij to England, and to Camusfeama. The British airline to London would not fly animals, so I booked a flight to Paris on another airline, and from there to London. The airline insisted that Mij should be packed into a box not more than eighteen inches square, to be carried on the floor at my feet. I had a box made, and an hour before we started, I put Mij into the box so that he would become accustomed to it, and left for a hurried meal. (Page 107)
(i) What did the author dread?
(ii) Name the airline mentioned in the passage.
(iii) How did the writer go to London?
(iv) Pick out the word from the passage that means same as -‘terribly afraid of.
Answer:
(i) The author dreaded the prospect of transporting Mijbil to England.
(ii) It was the British Airline.
(iii) The writer went to London via Paris on another airline.
(iv) ‘Dreaded’.

Question 9.
When I returned, there was an appalling spectacle. There was complete silence from the box, but from its airholes and chinks around the lid, blood had trickled and dried. I whipped off the lock and tore open the lid, and Mij, exhausted and blood spattered, whimpered and caught at my leg. He had tom the lining of the box to shreds; when I removed the last of it so that there were no cutting edges left, it was just ten minutes until the time of the flight, and the airport was five miles distant. I put the miserable Mij back into the box, holding down the lid with my hand.
(i) Who was in the box?
(ii) What did the author find in the box?
(iii) Why did the author put him back in the box?
(iv) Pick out the word from the passage that means the same as—‘shocking’
Answer:
(i) Mijbil, the otter, was in the box.
(ii) The author found complete silence in the box.
(iii) The author put him back in the box because his condition was miserable and there were only ten minutes left for his flight.
(iv) ‘appalling’.

Question 10.
I sat in the back of the car with the box beside me as the driver tore thought the street of Basra like a ricochetting bullet. The aircraft was waiting to take off; I was rushed through to it by infuriated officials. Luckily, the seat booked for me was at the extreme front. I covered the floor around my feet with newspapers, rang for the air hostess, and gave her a parcel of fish (for Mij) to keep in a cool place. I took her into my confidence about the events of the last half hour. I have retained the most profound admiration for that air hostess; she was the very queen of her kind. She suggested that I might prefer to have my pet on my knee, and I could have kissed her hand in the depth of my gratitude. But, not knowing otters, I was quite unprepared for what followed.
(i) Who is ‘I’ and with what did he sit in the car?
(ii) What did the air hostess suggest to the author?
(iii) What was the author completely unprepared for?
(iv) Pick out the word from the passage that means the same as—“very angry’
Answer:
(i) I, means the author Maxwell, sat in his car with a box which contained Mijbil.
(ii) She suggested the author that he could have his pet on his knee, if he so preferred.
(iii) The author was completely unprepared for Mijbil terrifying the co-passengers and disappearing.
(iv) ‘infuriated’.

Question 11.
Mij was out of the box in a flash. He disappeared at high speed down the aircraft. There were squawks and shrieks, and a woman stood up on her seat screaming out, “A rat! A rat!” I caught sight of Mij’s tail disappearing beneath the legs of a portly white-turbaned Indian. Diving for it, I missed, but found my face covered in curry. “Perhaps,” said the air hostess with the most charming smile, “it would be better if you resumed your seat, and I will find the animal and bring it to you.”
(i) Where was Mij? Where did it disappear?
(ii) Why did the woman scream?
(iii) What did the air hostess say?
(iv) Pick out the word from the passage that means the same as—‘invisible’.
Answer:
(i) Mij was out of the box. He disappeared down the aircraft.
(ii) The woman screamed that a rat had entered in the aircraft.
(iii) The air hostess requested me to resume my seat so that she would find the animal.
(iv) ‘disappeared’.

Question 12.
Mij and I remained in London for nearly a month. He would play for hours with a selection of toys, ping- pong balls, marbles, rubber fruit, and a terrapin shell that I had brought back from his native marshes. With the ping-pong ball he invented a game of his own which could keep him engrossed for up to half an hour at a time. A suitcase that I had taken to Iraq had become damaged on the journey home, so that the lid, when closed, remained at a slope from one end to the other. Mij discovered that if he placed the ball on the high end it would run down the length of the suitcase. He would dash around to the other end to ambush its arrival, hide from it, crouching to spring up and take it by surprise, grab it and trot off with it to the high end once more.
(i) How did Mij spend his time in London?
(ii) What did Mij invent?
(iii) From where was terrapin shell obtained?
(iv) Pick out the word from the passage that means the same as—“completely interested in”.
Answer:
(i) Mij spent his time in playing with a selection of toys – ping-pong balls, marbles, rubber fruit, etc.
(ii) Mij invented a new game with the Ping-pong ball.
(iii) It was brought back from the native place of the otter.
(iv) engrossed.

Question 13.
But the question for which I awarded the highest score came from a labourer digging a hole in the street. I was still far from him when he laid down his tool, put his hands on his hips, and began to stare. As I drew nearer I saw his expression of surprise and affront, as though he would have me know that he was not one upon whom to play jokes. I came abreast off him; he spat, glared, and then growled out, “Here, Mister—what is that supposed to be?”
(i) Why did thp author face questions from the labourer?
(ii) Who asked the question in the passage?
(iii) Why did he have an expression of surprise and affront?
(iv) Pick out the word from the passage that means the same as—grumble.
Answer:
(i) The author faced question from the labourer because the people of London were unknown to otter.
(ii) The labourer asked the questions in this passage.
(iii) The labourer had never seen that animal (otter). He looked at it with an expression of surprise.
(iv) ‘growled’.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 8
Chapter Name Introduction to Trigonometry
Exercise Ex 8.3
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 1
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 2

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .\(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)
= 1 = RHS

(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = \(\frac { 108° }{ 3 }\)
∴ ∠ A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B ⇒ tan A = tan (90° – B) [ ∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B ⇒ A + B = 90° Proved

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = \(\frac { 110° }{ 5 }\)
A = 22°
∴ ∠ A = 22°

Question 6.
If A, Band Care interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 3

Question 7.
Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°

We hope the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 13

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 13
Chapter Name Surface Areas and Volumes
Exercise Ex 13.2
Number of Questions Solved 8
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of n.
Solution:
Radius of the hemisphere = 1 cm
Volume of the hemisphere = \(\frac { 2 }{ 3 }\)πr³ = \(\frac { 2 }{ 3 }\)π(1)³ = \(\frac { 2 }{ 3 }\)πcm³
Radius of base of the cone = 1 cm
Height of the cone = 1 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 1
Volume of the cone = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)πr² x 1 = \(\frac { 1 }{ 3 }\)π cm³
Total volume of the solid = Volume of the hemisphere + Volume of the cone
= \(\frac { 2 }{ 3 }\)π cm³ + \(\frac { 1 }{ 3 }\)π cm³ = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Volume of air contained in the model = Total volume of the solid
Diameter of base of each cone = 3 cm
∴ Radius of base of each cone = \(\frac { 3 }{ 2 }\)
Height of each cone = 2 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 2

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 3
Solution:
Volume of one piece of gulab jamun
= Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8
Radius of each hemispherical portion = \(\frac { 2.8 }{ 2 }\) = 1.4 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 4
Volume of both hemispherical ends = 2 x 5.74 cm³ = 11.48 cm³
Height of the cylindrical portion = (total height) – (radius of both hemispherical ends)
= 5 cm – 2(1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Radius of the cylindrical portion = 1.4 cm
Volume of the cylindrical portion of gulab jamun = πr²h
= \(\frac { 22 }{ 7 }\) x (1.4)² x 2.2cm³
= \(\frac{22 \times 1.4 \times 1.4 \times 2.2}{7}\)cm³ = 13.55 cm³
Total volume of one gulab jamun = Volume of the two hemispherical ends + Volume of the cylindrical portion
= 11.48 cm³ + 13.55 cm³ = 25.03 cm³
Volume of sugar syrup = 30% of volume of gulab jamun
= \(\frac { 30 }{ 100 }\) x 25.03 cm³ = 7.50 cm³
∴ Volume of sugar syrup in 45 gulab jamuns
= 45 (volume of sugar syrup in one gulab jamun)
= 45 x 7.50 cm³ = 337.5 cm³ = 338 cm³ approx.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.
Find the volume of wood in the entire stand (see figure).
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 5
Solution:
Radius of one conical depression = 0.5 cm
Depth of one conical depression = 1.4 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 6
Volume of cuboidal box = l x b x h
= 15 x 10 x 3.5 cm³
= 525 cm³
Remaining volume of box = Volume of cubical box – Volume of four conical depressions
= 525 cm³ – 1.464 cm³ = 523.5 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 7

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 8
Solution:
Given: radius of 1st cylinder = 12 cm
and height of 1st cylinder = 220 cm
∴ Volume of 1st cylinder = πr²h
= π(12)² (220) cm³
= 144 x 220π cm³
= 144 x 220 x 3.14 cm³
= 99475.2 cm³ … (i)
Given: radius of 2nd cylinder = 8 cm
and height of 2nd cylinder = 60 cm
∴ Volume of 2nd cylinder = πr²h
= π(8)² (60) cm³ = 64 x 60π cm³
= 64 x 60 x 3.14 cm³
= 12057.6 cm³ … (ii)
Total volume of solid = Volume of 1st cylinder + Volume of 2nd cylinder
= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³
Given: mass of 1 cm³ of iron = 8 g
∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g
= 892262.4 g = 892.262 kg

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of hemisphere = 60 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 9
When solid (hemisphere + conical) is kept in cylindrical solid, then volume of water left in cylinder
= Volume of cylinder – (Volume of hemisphere + Volume of cone)
= [π(60)² x 180 – π(60)² x 80] cm³
= π(60)² [180 – 80]cm³ = π x 3600 x 100 cm³ = 1130400 cm³ = 1.130 m³

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)
Radius of the cylindrical part = \(\frac { 2 }{ 2 }\) = 1 cm
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 10
Height of the cylindrical part = 8 cm
∴ Volume of the cylindrical part = πr²h
= 3.14 x (1)² x 8 cm³
Diameter of the spherical part Radius = 8.5 cm
∴ Radius = \(\frac { 8.5 }{ 2 }\) cm = \(\frac { 85 }{ 20 }\)
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 11

Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
= 25.12 cm³ + 321.39 cm³ = 346.51 cm³
Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 12
Chapter Name Areas Related to Circles
Exercise Ex 12.3
Number of Questions Solved 16
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Solution:
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 1

Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 400.
Solution:
∠AOC = 40° (given)
Radius of the sector AOC = 14 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 2

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
ABCD is a square
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 2a
Given: side of the square = 14 cm
∴ Area of the square = (side)² = (14)² = 196 cm²
Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm
Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²
Similarly, area of the semicircle BPC = 77 cm²
Total area of both the semicircles = 77 + 77 = 154 cm²
Area of the shaded region = Area of square – area of both semicircles
= 196 – 154 = 42 cm²

Question 4.
Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Solution:
Area of the equilateral triangle OAB
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 3

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 4
Solution:
Given: side of the square ABCD = 4 cm
Area of the square ABCD = 4 x 4 = 16 cm²
Radius of the quadrant at corner = 1 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 5
∴ Area of the remaining portion = Area of the square – Area to be cut from square
= 16 – (\(\frac { 44 }{ 7 }\)) = 16 – \(\frac { 44 }{ 7 }\)
= \(\frac { 112-44 }{ 7 }\) = \(\frac { 68 }{ 7 }\)cm²

Question 6.
In a circular table cover of the radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 6
Solution:
Radius of the circle(r) = 32cm
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) × 32 × 32 = \(\frac { 22528 }{ 7 }\)cm²
∴ An equilateral triangle is formed in the circle as shown
Angle subtended by et ch side at centre
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 7

Question 7.
In the figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 8
Solution:
Side of the square ABCD = 14 cm
Area of the squat e = (side)² = 14 x 14 = 196²
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 9

Question 8.
The given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 10
(i) the distance around the track along its inner edge.
(ii) the area of the track.
Solution:
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 11
Area of the tracks at both semicircular ends = 2 x 1100 = 2200 cm²
Area of the 2 rectangular portions = 2 x l x h = 2 x 106 x 10 = 2120 cm²
Total area of the track = area of the track at semicircular ends + area of the rectangular portions
= 2200 + 2120 = 4320 cm²

Question 9.
In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 12
Solution:
Given: OA = 7 cm
Radius of the semicircle ABC = OA = 7 cm
Area of the semicircle ABC = \(\frac { 1 }{ 2 }\)πr² = \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 7 x 7 = 11 x 7 = 77 cm²
Diameter AB = 2(OA) = 2 x 7 = 14 and OA = OC = 7 cm (radius)
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 13

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.
(Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 14
Solution:
Given: area of an equilateral triangle ABC = 17320.5 cm²
Let side of the triangle AB’C be ‘a’
∴ Area of the ∆ABC = \(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
\(\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\) = 17320.5
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 15
Now,
area of the all 3 sectors = \(\frac { 3×31400 }{ 6 }\) = 15700 cm²
∴ Area of the shaded portion = Area of the equilateral triangle
– Area of the three sectors formed at each vertex)
= 17320.5 – 15700 = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 16
Solution:
Radius of the one circular design = 7 cm
Area of the one circular design = πr² = \(\frac { 22 }{ 7 }\) x 7 x 7 = 154 cm²
Now, area of the 9 circular designs = 9 x 154 = 1386 cm²
Diameter of the circular design = 7 x 2 = 14 cm
Side of the square = 3(diameter of one circle) = 3 x 14 = 42 cm
Area of the square = 42 x 42 = 1764 cm²
Area of the remaining portion of handkerchief
= Area of the square – (Area of the 9 circular designs)
= 1764 – 1386
= 378 cm²

Question 12.
In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 17
Solution:
(i) Radius of the quadrant OACB = 3.5 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 18
(ii) OD = 2cm and OB = 3.5 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 19

Question 13.
In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use re = 3.14)
Solution:
Given: side of the square OABC = OA = 20 cm
Area of the square = 20 x 20 = 400 cm²
(Diagonal of the square)² = (side of the square)² + (side of the square)² (By pythagoras theorem)
Diagonal of the square = \(\sqrt{2}\) x (side of the square)
= \(\sqrt{2}\) x (20) = 20\(\sqrt{2}\)cm
Radius of the quadrant of circle = Diagonal of square = 20\(\sqrt{2}\)
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 20

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB=30°, find the area of the shaded region.
Solution:
Given: ∠AOB = 30°
Radius of the sector AOB = 21 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 21

Question 15.
In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 22
Solution:
Radius of the quadrant of the circle = 14 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 23

Area of the shaded region = Area of the semicircular region – Area of the region I
= 154 cm² – 56 cm² = 98 cm²

Question 16.
Calculate the area of the designed region in the figure common between the two quadrants of the circles of the radius 8 cm each.
Solution:
Side of the square = 8 cm
Area of the square = 8 x 8 = 64 cm²
Radius of the quadrant (formed at vertex) = 8 cm
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 24

We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 13
Chapter Name Surface Areas and Volumes
Exercise Ex 13.3
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Given: radius of metallic sphere = 4.2 cm
∴ Volume = \(\frac { 4 }{ 3 }\)π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 1

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radius of 1st metallic sphere = 6 cm
∴ Volume of 1st metallic sphere = \(\frac { 4 }{ 3 }\)π(6)³ cm³
Radius of 2nd metallic sphere = 8 cm
∴ Volume of 2nd metallic sphere = \(\frac { 4 }{ 3 }\)π(8)³ cm³
Radius of 3rd metallic sphere = 10 cm
∴ Volume of 3rd metallic sphere = \(\frac { 4 }{ 3 }\)π(10)³ cm³
Volume of all three metallic spheres = \(\frac { 4 }{ 3 }\)π(6³+8³+10³) cm³
∵ 3 spheres are melted and recast into a new metallic sphere of radius r.
∴ Volume of new metallic sphere = \(\frac { 4 }{ 3 }\)πr³
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 2

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Given: diameter of the well = 7 m Radius = \(\frac { 7 }{ 2 }\)m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 3

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Given: diameter of the well = 3 m
⇒ Radius = \(\frac { 3 }{ 2 }\)m
Depth of the well = 14 m
Volume of the earth taken out from the well = πr²h
= π(\(\frac { 3 }{ 2 }\))² x 14 = \(\frac { π×9×14 }{ 4 }\) = \(\frac { 63 }{ 2 }\)πm³
∵ Earth taken out from the well evenly spread to form an embankment having height h and width of embankment around the well is 4 m.
∴ External radius (R) = radius of well + width of the embankment
= \(\frac { 3 }{ 2 }\)m + 4m = \(\frac { 11 }{ 2 }\)m
Internal radius = \(\frac { 3 }{ 2 }\)m = radius of well
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 4

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 5
Let total number of ice cream cones are n.
∴ All ice cream cones are filled from ice cream in the cylinder.
Total volume of n number of ice cream cones = Volume of ice cream in the cylinder
n x π x 54 = π(36)15
⇒ n x 54 = 36 x 15
⇒ n = \(\frac { 36 × 15 }{ 54 }\) = 10

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:
Given: diameter of each coin = 1.75 cm ⇒ radius = \(\frac { 1.75 }{ 2 }\)cm
and thickness of each coin = 2 mm
Let n number of coins are melted to form a cuboid.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 6

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm
∴ Volume of the cylindrical bucket = π²h = π(18)² x 32 cm³
Let radius of the conical heap = r cm
Given: height of the conical heap = 24 cm
∴ Volume of the conical heap = \(\frac { 1 }{ 3 }\)π(r²) 24cm³
According to question,
Volume of the cylinderical bucket = Volume of the conical heap
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 7

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water – 10 km/h
Volume of the water flowing in 30 minutes = \(\frac { 6×1.5×30×10 }{ 60 }\)km³
= \(\frac { 6×1.5×10×1000×30 }{ 10×60 }\)km³ = 45000 m³
We require water for standing up to height = 8 cm = \(\frac { 8 }{ 100 }\) m
Let the required area he A
∴ Volume of water required = A(\(\frac { 8 }{ 100 }\))m³
According to question. 45000 = \(\frac { A×8 }{ 100 }\)
⇒ \(\frac { 45000×100 }{ 8 }\) = A ⇒ A = 562500 m²

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Given: diameter of the pipe = 20 cm ⇒ radius of the pipe = 10 cm
Water flowing from the pipe at rate = 3 km
Let it filled the tank in ‘t’ hours.
Volume of the water flowing in ‘t’ hours.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 8

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4.

Board CBSE
Textbook NCERT
Class Class 10
Subject Maths
Chapter Chapter 8
Chapter Name Introduction to Trigonometry
Exercise Ex 8.4
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
From trigonometric identity, cosec² A – cot² A = 1, we get
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 1

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
Since sin² A + cos² A = 1, therefore
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 2

Question 3.
Evaluate:
(i)
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 3
(ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 4
(ii) sin 25° cos 65° + cos 25° sin 65° = sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25° = 1

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A =
(A) 1
(B) 9
(C) 8
(D) 0

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ ) =
(A) 0
(B) 1
(C) 2
(D) -1

(iii) (sec A + tan A) (1 – sin A) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\)
(A) sec² A
(B) -1
(C) cot² A
(D) tan² A
Solution:
(i) 9 sec² A – 9 tan² A = 9(sec² A – tan² A) = 9 x 1 = 9
Correct option is (B)
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 5

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 6
Solution:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 7
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 8

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 9

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