Category: CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23.

Question 1.
Solution:
(i) Here, total number of trials = 300
Number of heads got = 136.
P(E) = \(\frac { 136 }{ 300 }\) = \(\frac { 34 }{ 75 }\)
(ii) Total number of trials = 300
Number of tails got = 164
P(E) = \(\frac { 164 }{ 300 }\) = \(\frac { 41 }{ 75 }\)

Question 2.
Solution:
Number times, the two coins were tossed = 200
Number of times got two heads = 58
Number of times got one head = 83
and number of times got no head = 59
(i) Probability of getting 2 heads : P(E) = \(\frac { 58 }{ 200 }\) = \(\frac { 29 }{ 100 }\)
(ii) Probability of getting one head : P(E) = \(\frac { 83 }{ 200 }\)
(iii) Probability of getting no head : P(E) = \(\frac { 59 }{ 200 }\)

Question 3.
Solution:
Number of times, a dice was thrown = 100
(i) Number of times got 3 = 18
Probability will be
P(E) = \(\frac { 18 }{ 100 }\) = \(\frac { 9 }{ 50 }\)
(ii) Number of times got 6 = 9
Probability will be
P(E) = \(\frac { 9 }{ 100 }\)
iii) Number of times got 4 = 15
Probability will be
P(E) = \(\frac { 15 }{ 100 }\) = \(\frac { 3 }{ 20 }\)
(iv) Number of times got 1 = 21
Probability will be
P(E) = \(\frac { 21 }{ 100 }\)

Question 4.
Solution:
Total number of ladies = 100
Number of ladies also like coffee = 36.
Number of ladies who dislike coffee = 64
(i) Probability of lady who like coffee :
P(E) = \(\frac { 36 }{ 100 }\) = \(\frac { 9 }{ 25 }\)
(ii) Probability of lady who dislikes coffee:
P(E) = \(\frac { 64 }{ 100 }\) = \(\frac { 16 }{ 25 }\)

 

Hope given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23 are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22.

Question 1.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph representing x-axis and y-axis.
(ii) Along OX, mark subjects and along y-axis, mark, number of marks
(iii) Take one division = 10 marks.
Now we shall draw bar graph as shown.

Question 2.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph paper. These two lines represent x-axis and y-axis respectively.
(ii) Along OX, write sports and along OY, number of students choosing on division equal to 10 students.
(iii) Now draw the bars of various heights according to the no. of students as shown on the graph.

Question 3.
Solution:
(i) Draw a horizontal line OX and a vertical line OY. These represent x-axis and y-axis respectively on the graph paper.
(ii) Along OX, write years and along OY, no. of students choosing one division = 200 students.
(iii) Draw bars of various heights according to the number of students given.
This is the required bar graph as shown.

Question 4.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, no. of scooters
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to given data as shown on the graph.

Question 5.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along OX write countries and along OY, take Birth rate per thousand.
(iii) Choose 1 division = 10.
(iv) Now draw bars of different heights according to the given data as shown on the graph.

Question 6.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis on the graph paper.
(ii) Along x-axis write states and along y-axis population in lakhs.
(iii) Choose one division = 200 (Lakhs)
(iv) Draw bars of different heights according to the data given as shown on the graph.

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis Interest in thousand cores rupees.
(iii) Choose one division = 20 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 8.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write city and along y-axis the distance (in km).
(iii) Choose one division = 200 km.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 9.
Solution:
(i) Draw a horizontal line OX and a vertical line OY represent x-axis and y-axis respectively on the graph paper.
(i) Along x-axis write countries and along y-axis life expectancy (in years)
(ii) Choose one division = 10 years
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 10.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis imports in thousand crores rupees.
(iii) Choose one division = 50 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.

Question 11.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write months and along y-axis average rainfall in cm.
(iii) Choose one small division = 5cm.
(iv) Draw different bars of different heights according to the given data as shown on the graph.

Question 12.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write Brand and along y-axis, percentage of buyers.
(iii) Choose one division = 5% of buyers.
(iv) Draw bars of different heights according to the given data, as shown on the graph.

Question 13.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write week and along y-axis. Rate per 10gm in rupees.
(iii) Choose and division = 1000
(iv) Draw bars of different heights according to the data given as shown on the graph.

Question 14.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write mode of transport and on the y-axis is number of students.
(iii) Choose one division =100 students
(iv) Draw bars of different heights according to given data as shown on the graph.

Question 15.
Solution:
(i) The bar graph shows the marks obtained by a student in different subjects.
(ii) The student is very good in Mathematics.
(iii) The student is very poor in Hindi.
(iv) Average marks

Question 16.
Solution:
(i) The bar graph shows the number of members in each of the 85 families.
(ii) 40 families have 3 number each.
(iii) Number of people living alone is nil.
(iv) The families having 3 members each is most common.

Question 17.
Solution:
(i) The highest peak is Mount Everest whose heighest is 8800 m.
(ii) The required ratio between the highest peak and the next heighest peak = 8800 : 8200 or 44 : 41
(iii) Arranging the heights of peaks in descending order are : 8800 m, 8200 m, 8000 m, 7500 m, 6000 m

Question 18.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale : 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are :
Mon. 350 and 200; Tues. 400 and 450; Wed. 500 and 300; Thurs. 450 and 250; Fri. 550 and 100 and Sat. 450 and 50.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) The number of readers in the library was maximum on TUESDAY.
(iii) Total Number of magazine readers in a week = 200 + 450 + 300 + 250 + 100 + 50 = 1350
Mean Number of readers per day = \(\frac { 1350 }{ 6 }\) = 225

Question 19.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 4
(d) The heights of various pairs of bars in terms of the number of small divisions are:
VI 95 and 92; VII 90 and 85; VIII 82 and 78; IX 75 and 69; X 68 and 62.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) For class VII, total Number of Students = 90
Number of students present = 85

Question 20.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
January 2 and 1.5 ; February 3.25 and 3; March 4 and 3.5; April 4.5 and 6; May 7.75 and 5.5; June 8 and 6.5.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) June
(iii) January

Question 21.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y’-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
Town A 640000 and 750000; Town B 830000 and 920000; Town C 460000 and 630000; Town D 290000 and 320000
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.

(ii) Town B
(iii) Town D

 

Hope given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22 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.

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
In parallelogram ABCD,
Base AB = 32cm
Height DL = 16.5cm.

Area = Base x height = 32 x 16.5 cm² = 528 cm²

Question 2.
Solution:
Base of parallelogram = 1 m 60m = 160 cm
and height = 75 cm
Area = Base x height = 160 x 75 = 12000 cm²
= \(\frac { 12000 }{ 10000 }\) m² = 1.2m²

Question 3.
Solution:
Base of parallelogram = 14dm = 140cm
and height = 6.5 dm = 65cm
Area (in cm²) = Base x height = 140 x 65 = 9100 cm²
Area (in m²) = \(\frac { 140 }{ 100 }\) x \(\frac { 65 }{ 100 }\) = \(\frac { 9100 }{ 10000 }\)
= 0.91 m²

Question 4.
Solution:
Area of parallelogram = 54 cm²
Base = 15 cm

Question 5.
Solution:
Area of parallelogram ABCD = 153 cm²

Question 6.
Solution:
In parallelogram ABCD
AB || DC and AD || BC and AB = DC, AD = BC
AB = DC = 18cm, BC = 12cm
Area of parallelogram ABCD = Base x altitude = DC x AL = 18 x 6.4cm2 = 115.2 cm²
and area of parallelogram ABCD = BC x AM
⇒ 115.2 = 12 x AM
⇒ AM = 9.6 cm

Question 7.
Solution:
In parallelogram ABCD
AB = DC = 15 cm
BC = AD = 8 cm.

Distance between longer sides AB and DC is 4cm
i.e. perpendicular DL = 4cm.
DM ⊥ BC.
Area of parallelogram = Base x altitude = AB x DL = 15 x 4 = 60 cm²
Again let DM = x cm
area ABCD = BC x DM = 8 x x = 8x cm²
8x cm² = 60 cm²
⇒ x = 7.5 cm
Distance between shorter lines = 7.5 cm

Question 8.
Solution:
Let Base of the parallelogram = x

⇒ x² = 108 x 3 = 324 = (18)²
⇒ x = 18
Base = 18 cm
and altitude = \(\frac { 1 }{ 3 }\) x 18 = 6 cm

Question 9.
Solution:
Area of parallelogram = 512 cm²
Let height of the parallelogram = x
Then base = 2x
Area = Base x height
⇒ 512 = 2x x x
⇒ 2x² = 512
⇒ x² = 256 = (16)²
⇒ x = 16
Base = 2x = 2 x 16 = 32 cm
and height = x = 16 cm

Question 10.
Solution:
(i) Each side of rhombus = 12 cm
height = 7.5 cm

Area = Base x height = 12 x 7.5 = 90 cm²
(ii) Each side = 2 cm = 20 cm
Height = 12.6 cm
Area = Base x height = 20 x 12.6 = 252 cm²

Question 11.
Solution:
(i) Diagonals of rhombus ABCD are 16 cm and 28 cm

Question 12.
Solution:
In rhombus ABCD, diagonals AC and BD intersect each other at right angles at O.
AO = OC and BO = OD

AO = \(\frac { 1 }{ 2 }\) x AC = \(\frac { 1 }{ 2 }\) x 24 cm = 12cm
Let OB = x
Each side of rhombus = 20cm
In right ∆AOB
AO² + OB² = AB² (Pythagoras Theorem)
⇒ (12)² + OB² = (20)²
⇒ 144 + OB² = 400
⇒ OB² = 400 – 144 = 256 = (16)²
⇒ OB = 16
But BD = 2 BO = 2 x 16 = 32cm
Now, area of rhombus = \(\frac { Product of diagonals }{ 2 }\)
= \(\frac { 24 x 32 }{ 2 }\) cm2 = 384 cm²

Question 13.
Solution:
Area of rhombus = 148.8 cm²
one diagonal = 19.2 cm
Let second diagonal = x

Question 14.
Solution:
Area of rhombus = 119 cm²
Perimeter = 56 cm
Its side = \(\frac { 56 }{ 4 }\) = 14 cm

Question 15.
Solution:
Area of rhombus = 441 cm²
Height = 17.5 cm

Question 16.
Solution:
Base of a triangle = 24.8 cm
Corresponding height = 16.5 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C 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.

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
Outer length of plot (L) = 75 m
and breadth (B) = 60 m
Width of path inside = 2 m

Inner length (l) = 75 – 2 x 2 = 75 – 4 = 71 m
and width (b) = 60 – 2 x 2 = 60 – 4 = 56 m
Area of the path = L x B – l x b = (75 x 60 – 71 x 56) m² = 4500 – 3976 = 524 m²
Rate of constructing it = Rs. 125 per m²
Total cost = Rs. 524 x 125 = Rs. 65500

Question 2.
Solution:
Outer length of the plot (L) = 95 m
and breadth (B) = 72 m
Width of path = 3.5 m

Inner length (l) = 95 – 2 x 3.5 = 95 – 7 = 88 m
and breadth = 72 – 2 x 3.5 = 72 – 7 = 65 m
Outer area = L x B = 95 x 72 m² = 6840 m²
and inner area = l x b = 88 x 65 m² = 5720 m²
Area of path = outer area – inner area = 6840 – 5720 = 1120 m²
Rate of constructing it = Rs. 80 per m²
Total cost = Rs. 1120 x 80 = Rs. 89600
and rate of laying grass = Rs. 40 per m²
Total cost = Rs. 40 x 5720 = Rs. 228800
Total cost = Rs. 89600 + Rs. 228800 = Rs. 318400

Question 3.
Solution:
Length of saree (L) = 5 m
and breadth (B) = 1.3 m
Width of border = 25 cm
Inner length (l) = 5 – \(\frac { 2 x 25 }{ 100 }\) = 5 – 0.5 = 4.5 m
and inner breadth (b) = 1.3 – \(\frac { 2 x 25 }{ 100 }\) = 1.3 – 0.5 = 0.8 m
Now area of the boarder = L x B – l x b
= (5 x 1.3 – 4.5 x 0.8) m²
= 6.50 – 3.60
= 2.90 m²
= 2.90 x 100 x 100 = 29000 cm²
Cost of 10 cm² = Re. 1
Total cost = Rs. \(\frac { 29000 x 1 }{ 10 }\) = Rs. 2900

Question 4.
Solution:
Inner length of lawn (l) = 38 m
and breadth (b) = 25 m.
Width of path = 2.5 m.
Outer length (L) = 38 + 2 x 2.5 = 38 + 5 = 43 m
and outer breadth (B) = 25 + 2 x 2.5 = 25 + 5 = 30 m
Area of path = Outer area – Inner area
= (43 x 30 – 38 x 25) m²
=(1290 – 950) m² = 340 m²
Rate of gravelling the path = Rs. 120 per m²
Total cost = Rs. 120 x 340 = Rs. 40800

Question 5.
Solution:
Length’ of room (l) = 9.5m
Breadth (b) = 6m
Width of outer verandah = 1.25m
Outer length (L) = 9.5 + 2 x 1.25 = 9.5 + 2.5 = 12.0 m
and breadth (B) = 6 + 2 x 1.25 = 6 + 2.5 = 8.5 m
Area of verandah = Outer area – Inner area = L x B – l x b
= (12.0 x 8.5 – 9.5 x 6) m² – (102.0 – 57.0) m² = 45 m²
Rate of cementing = Rs. 15 per m²
Total cost = Rs. 80 x 45 = Rs. 3600

Question 6.
Solution:
Each side of square bed (a) = 2m 80 cm = 2.8 m
Width of strip = 30cm
Outer side (A) = 2.8 m + 2 x 30cm = 2.8 + 0.6 = 3.4m
Outer area = (3.4 m)² = 11.56 m²

Inner area = (2.8)² = 7.84 m²
Area of increased bed flower = 11.56 – 7.84 = 3.72 m²

Question 7.
Solution:
Ratio in length and breadth of the park = 2 : 1
Its perimeter = 240 m
Let length = 2x
then breadth = x
Perimeter = 2 (2x + x) = 2 x 3x = 6x
6x = 240
⇒ x = \(\frac { 240 }{ 6 }\) = 40
Length = 2x = 2 x 40 = 80m
and breadth = x = 40m
Area = L x B = 80 x 40 m² = 3200 m²
Width of path inside the park = 2m
Inner length (l) = 80 – 2 x 2 = 80 – 4 = 76 m
and breadth (b) = 40 – 2 x 2 = 40 – 4 = 36 m
Inner area = 76 x 36 = 2736 m²
Area of path = Outer area – Inner area = 3200 – 2736 = 464 m²
Rate of paving the path = Rs. 80 per m²
Total cost = Rs. 80 x 464 = Rs. 37120

Question 8.
Solution:
Length of hall (l) = 22m
Breadth (b) = 15.5 m
Space left along the walls = 75m = \(\frac { 3 }{ 4 }\) m
Inner length (l) = 22 – 2 x \(\frac { 3 }{ 4 }\) = 20.5 m
Inner breadth (b) = 15.5 – 2 x \(\frac { 3 }{ 4 }\) = 15.5 – 1.5 = 14m
Area of carpet = Inner area = 20.5 x 14 m² = 287 m²
Outer area = 22 x 15.5 = 341 m²
Area of strip left out = 341 – 287 = 54 m²
Width of carpet = 82 cm = \(\frac { 82 }{ 100 }\) m
Length of carpet = 287 ÷ \(\frac { 82 }{ 100 }\)
= \(\frac { 287 x 100 }{ 82 }\) = 350m
Rate of carpet = Rs. 60 per metre
Total cost = Rs. 60 x 350 = Rs. 21000

Question 9.
Solution:
Area of path = 165 m²
Width of path = 2.5m.
Let side of square lawn = x m.

Outer side = x + 2 x 2.5 = (x + 5) m
Area of path = (x + 5)² – x²
⇒ x² + 10x + 25 – x² = 165
⇒ 10x = 165 – 25 = 140
⇒ x = \(\frac { 140 }{ 10 }\) = 14m
Side of lawn = 14m
and area of lawn = (14)² m² = 196 m²

Question 10.
Solution:
Ratio in length and breadth of a park = 5 : 2
Width of path outside it = 2.5 m
Area of path = 305 m2
Let Inner length (l) = 5x
and breadth (b) = 2x
Outer length (L) = 5x + 2 x 2.5 = (5x + 5) m
Width (B) = 2x + 2 x 2.5 = (2x + 5) m
Area of path = Outer area – Inner area
⇒ (5x + 5) (2x + 5) – 5x x 2x = 305
⇒ 10x² + 10x + 25x + 25 – 10x² = 305
⇒ 35x = 305 – 25 = 280
⇒ x = 8
Length of park = 5x = 5 x 8 = 40m
and breadth = 2x = 2 x 8 = 16m
Dimensions of park = 40m by 16m

Question 11.
Solution:
Length of lawn (l) = 70 m
Breadth (b) = 50 m
Width of crossing roads = 5m

Area of roads = 70 x 5 + 50 x 5 – (5)²
= 350 + 250 – (5)²
= 600 – 25 = 575 m²
Cost of constructing = Rs. 120 per m²
Total cost Rs. 120 x 575 = Rs. 69000

Question 12.
Solution:
Length of lawn (l) = 115m
and breadth (b) = 64m.
Width of road parallel to length = 2m
and width of road parallel to breadth= 2.5m

Area of roads = (115 x 2 + 64 x 2.5 – 2 x 2.5) m²
= (230 + 160 – 5) m² = (390 – 5) m² = 385 m²
Cost of gravelling = Rs. 60 m²
Total cost = Rs. 60 x 385 = Rs. 23100

Question 13.
Solution:
Length of field (l) = 50 m
and breadth (b) = 40 m

Width of road parallel to length = 2 m
and width of road parallel to breadth = 2.5 m
Area of roads = 50 x 2 + 40 x 2.5 – 2.5 x 2 = (100 + 100 – 5) m² = 195 m²
and area of remaining portion = 50 x 40 – 195 = 2000 – 195 = 1805 m²

Question 14.
Solution:
(i) Outer length = 43m
and breadth = 27m
Area = 43 x 27 = 1161 m²
Inner length = 43 – 2 x 1.5 = 43 – 3 = 40m
and breadth = 27 – 2 x 1 = 27 – 2 = 25m
Inner area = 40 x 25 = 1000 m²
Area of shaded portion = 1161 – 1000 = 161 m²
(ii) Side of square (a) = 40m
Area = (a)² = 40 x 40 = 1600 m²
Area of larger road = 40 x 3 = 120 m²
and area of shorter road = 40 x 2 = 80 m²
Area of roads = (120 + 80) – 3 x 2 = 200 – 6 = 194 m²
Area of shaded portion = (1600 – 194) m² = 1406 m²

Question 15.
Solution:
(i) Outer length = 24m
and breadth = 19m
Area = 24 x 19 = 456 m²
Length of unshaded portion = 24 – 4 = 20m
and breadth = 16.5m
Area of unshaded portion = 20 x 16.5 m² = 330.0 m²
Area of shaded portion = 456 – 330 = 126 m²
(ii) Dividing the figure an shown
Area of rectangle I = 15 x 3 cm² = 45 cm²
Area of rectangle II = (12 – 3) x 3 = 9 x 3 = 27 cm²
Area of rectangle III = 5 x 3 = 15 cm²

and area of rectangle IV = (12 – 3) x 3 = 9 x 3 = 27 cm²
Total area of shaded portion = 45 + 27 + 15 + 27 = 114 cm²

Question 16.
Solution:
Dividing the figure an shown
Area of rectangle I = 3.5 x 0.5 m² = 1.75 m²
Area of rectangle II = (3.5 – 2 x 0.5) x 0.5 = (3.5 – 1) x 0.5 = 2.5 x 0.5 = 1.25 m²
Area of rectangle III = (2.5 – 1) x 0.5 = 1.5 x 0.5 = 0.75 m²
Area of rectangle IV = (1.5 – 1.0) x 0.5 x 0.5 = 0.25 m²

Total area of shaded portion = (1.75 + 1.25 + 0.75 + 0.25) m² = 4 m²

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B 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.

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
Length of rectangle (l) = 24.5 m
Breadth (b) = 18 m

Area = l x b = 24.5 x 18 m² = 441 m²
(ii) Length of rectangle (l) = 12.5 m
Breadth (b) = 8 cm = 0.80 m
Area = l x b = 12.5 x 0.80 m² = 10 m²

Question 2.
Solution:
Length of rectangular plot (l) = 48m
and its diagonal = 50m

Question 3.
Solution:
Ratio in the sides of a rectangle = 4 : 3
Area = 1728 cm²
Let length = 4x,
then breadth = 3x
Area = l x b
1728 = 4x x 3x
⇒ 12x² = 1728
⇒ x² = 144 = (12)²
⇒ x = 12
Length = 4x = 4 x 12 = 48 m
and breadth = 3m = 3 x 12 = 36m
Now perimeter = 2 (l + b) = 2 (48 + 36) m = 2 x 84 = 168 m
Rate of fencing = Rs. 30 per metre
Total cost= 168 x 30 = Rs. 5040

Question 4.
Solution:
Area of rectangular field = 3584 m²
Length = 64 m
Area = 3584
Breadth = \(\frac { Area }{ Length }\) = \(\frac { 3584 }{ 64 }\) = 56 m
Now perimeter = 2 (l + b) = 2 (64 + 56) m = 2 x 120 = 240 m
Distance covered in 5 rounds = 240 x 5 = 1200 m
Speed = 6 km/h
Time take = \(\frac { 1200 }{ 1000 }\) x \(\frac { 60 }{ 6 }\) = 12 minutes (1 hour = 60 minutes)

Question 5.
Solution:
Length of verandah (l) = 40m
Breadth (b) = 15m
Area = l x b = 40 x 15 = 600m²
Length of one stone = 6dm = \(\frac { 6 }{ 10 }\) m
and breadth = 5 dm = \(\frac { 5 }{ 10 }\) m
Area of one stone = \(\frac { 6 }{ 10 }\) x \(\frac { 5 }{ 10 }\)

Question 6.
Solution:
Length of a room = 13 m
Breadth = 9 m
Area of floor = l x b = 13 x 9 m² = 117 m²
or area of carpet = 117 m²
Width = 75 cm = \(\frac { 75 }{ 100 }\) = \(\frac { 3 }{ 4 }\) m
Length of carpet = Area ÷ Width
= 117 ÷ \(\frac { 3 }{ 4 }\)
= 117 x \(\frac { 4 }{ 3 }\) m
= 39 x 4 = 156 m
Rate = Rs. 105 per m
Total cost = Rs. 156 x 105 = Rs. 16380

Question 7.
Solution:
Cost of carpeting a room = Rs. 19200
Rate = Rs. 80 per m
Length of carpet = \(\frac { 19200 }{ 80 }\) m = 240 m
Width of carpet = 75 cm = \(\frac { 75 }{ 100 }\) = \(\frac { 3 }{ 4 }\) m
Area of carpet = 240 x \(\frac { 3 }{ 4 }\) = 180 m²
Length of a room = 15 m
Width = \(\frac { Area }{ Length }\) = \(\frac { 180 }{ 15 }\) = 12m

Question 8.
Solution:
Ratio in length and breadth of a rectangular piece of land = 5 : 3
Cost of fencing = Rs. 9600
and rate = Rs. 24 per m
Perimeter = \(\frac { 9600 }{ 24 }\) = 400 m
Let length = 5x
Then breadth = 3x
Perimeter = 2 (l + b)
⇒ 400 = 2 (5x + 3x)
⇒ 400 = 2 x 8x = 16x
⇒ 16x = 400
⇒ x = 25
Length of the land = 5x = 5 x 25 = 125 m
and width = 3x = 3 x 25 = 75 m

Question 9.
Solution:
Length of hall (l) = 10m.
Breadth (b) = 10m
and height (h) = 5m
Longest pole which can be placed in it

Question 10.
Solution:
Side of square (a) = 8.5m
Area = a² = (8.5)² = 8.5 x 8.5 m² = 72.25 m²

Question 11.
Solution:
(i) Length of diagonal of square = 72 cm.
Let length of side = a
Then diagonal = √2 a.
√2 a = 72

= 1.2 x √2 x 1.2 x √2 m²
= 1.44 x 2 = 2.88 m²

Question 12.
Solution:
Area of a square = 16200 m²
Side = √16200 m = √(8100 x 2) m = 90√2 m
Length of diagonal = √2 (side) = √2 x 90√2 = 180 m

Question 13.
Solution:
Area of square = \(\frac { 1 }{ 2 }\) hectare
= \(\frac { 1 }{ 2 }\) x 10000 m² = 5000 m²
side (a) = √Area = √5000 m = √(2500 x 2) = 50√2 m
Length of diagonal = √2 (side) = √2 x 50√2 = 100 m

Question 14.
Solution:
Area of sphere plot = 6084 m²
Side (a) = √Area = √6084 m = 78m
Perimeter = 4a = 4 x 78 = 312 m
Length of boundary four times = 312 x 4 = 1248 m

Question 15.
Solution:
Side of a square wire = 10 cm
Perimeter = 4a = 4 x 10 cm = 40 cm
or perimeter of rectangle = 40 cm
Length of rectangle = 12 cm
Breadth = \(\frac { 40 }{ 2 }\) – 12 = 20 – 12 = 8 cm
Now area of square = a² = (10)² = 100 cm²
and area of rectangle = l x b = 12 x 8 = 96 cm²
Difference in areas = 100 – 96 = 4 cm²
Square has 4 cm² more area

Question 16.
Solution:
Length of go down (l) = 50 m
Breadth (b) = 40 m
and height (h) = 10 m
Area of 4 walls = 2 (l + b) x h
= 2 (50 + 40) x 10 m
= 2 x 90 x 10 = 1800 m²
and area of ceiling = l x b = 50 x 40 = 2000 m²
Total area of walls and ceiling = 1800 + 2000 = 3800 m²
Rate of whitewashing = Rs. 20 per m²
Total cost = Rs. 20 x 3800 = Rs. 76000

Question 17.
Solution:
Area of 4 walls of a room= 168m²
Breadth of the room (b) = 10m
and height (h) = 4m.
Let l be the length of room
2 (l + b) h = 168
⇒ 2 (l + 10) x 4 = 168
⇒ l + 10 = \(\frac { 168 }{ 2 x 4 }\) = 21
⇒ l = 21 – 10 = 11m
Length of the room = 11 m

Question 18.
Solution:
Area of 4 walls of a room = 77 m²
Length of room (l) = 7.5 m
and breadth (b) = 3.5 m
Let h be the height,
then area of four walls = 2 (l + b) h
= 2 (7.5 + 3.5) h = 77
⇒ 2 x 11 x h = 77
⇒ h = \(\frac { 77 }{ 2 x 11 }\)
Height of room = 3.5 m

Question 19.
Solution:
Area of 4 walls = 120 m²
Height (h) = 4m.
Let breadth (b) = x
and length (l) = 2x
Area of 4 walls = 2 (l + h) x h = 2(2x + x) x 4 = 8 x 3x = 24x
24x = 120
x = \(\frac { 120 }{ 24 }\) = 5
Length of room = 2x = 2 x 5 = 10m
and breadth = x = 5m
Area of floor = l x b = 10 x 5 = 50 m²

Question 20.
Solution:
Length of a room (l) = 8.5 m
Breadth (b) = 6.5 m
and height (h) = 3.4 m
Area of four walls = 2 (l + b) x h
= 2 (8.5 +6.5) x 3.4 m²
= 2 x 15 x 3.4 m²
= 30 x 3.4= 102.0 m²
Area of two doors of size 1.5 m x 1 m = 2 x 1.5 x 1 m = 3 m²
and area of two windows of size 2 m x 1 m = 2 x 2 x 1 = 4 m²
Area of remaining portion = 102 – (3 + 4) = 102 – 7 m2 = 95 m²
Rate of painting = Rs. 160 per m²
Total cost = Rs. 160 x 95 = Rs. 15200

 

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RS Aggarwal Class 7 Solutions Chapter 19 Three-Dimensional Shapes Ex 19

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 19 Three-Dimensional Shapes Ex 19.

Question 1.
Solution:
(i) A cuboid has six rectangular faces 12 edges and 8 vertices.
(ii) A cylinder has one curved face and two flat faces.
(iii) A cone has one curved face and one flat face..
(iv) A sphere has a curved surface.

Question 2.
Solution:
(i) True : A cylinder has no vertex
(ii) True : A cube has 6 faces, 12 edges and 8 vertices.
(iii) True : A cone has one vertex.
(iv) False : A sphere has no edge.
(v) True : A sphere has one curved surface

Question 3.
Solution:
(i) Examples of cone : Ice cream cone, conical tent, conical cap, conical vessal
(ii) Examples of a sphere : A ball, a football, a volleyball, a basket ball, a hand ball.
(iii) Examples of a cuboid : A tin, a cardboard box, a book, a room, a matchbox.
(iv) Examples of a cylinder : Circular pipe, a pencil, road roller, a gas cylinder, a jar.

 

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RS Aggarwal Class 7 Solutions Chapter 18 Reflection and Rotational Symmetry Ex 18B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A
• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B

Question 1.
Solution:
(a) An equilateral triangle has three lines of symmetry which are the angle bisectors.
(b) It has three order of rotational symmetry.

Question 2.
Solution:
The rectangle should be rotated through 180° and 360° to be in symmetrical position with the original position as given below :

Question 3.
Solution:
A square has four orders of rotational symmetry and angles through which the rotational symmetry are 90°, 180°, 270° and 360° as given below:

Question 4.
Solution:
(i) A rhombus has two lines of symmetry which are its diagonal.
(ii) Order of rotational symmetry of a rhombus is not possible. Therefore it has no rotational symmetry.

Question 5.
Solution:
Three letters of the English Alphabet which have two lines of symmetry and rotational symmetry of order 2 are H, I and N.

Question 6.
Solution:
The figure which has only on line of symmetry but no rotational symmetry order is an isosceles triangle.

Question 7.
Solution:
No, only isosceles trapezium has a line of symmetry but not every trapezium.

Question 8.
Solution:
The line of symmetry of a semicircle is the perpendicular bisector of the diameter No, it has not any rotational symmetry.

Question 9.
Solution:
A scalene triangle has neither any line of symmetry nor a rotational symmetry.

Question 10.
Solution:
In the given figure, the line of symmetry has been drawn which is one. There is no rotational symmetry of this figure.

Question 11.
Solution:
(i) The given figure has two lines of symmetry which has been drawn.

(ii) It has three orders of the rotational symmetry which are 90°, 270° and 360°.

Question 12.
Solution:
There is one letter of the English Alphabet Z which has no line of symmetry but it has order two of rotational symmetry of 180° and 360°.

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RS Aggarwal Class 7 Solutions Chapter 18 Reflection and Rotational Symmetry Ex 18A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18A
• RS Aggarwal Solutions Class 7 Chapter 18 Reflection and Rotational Symmetry Ex 18B

Tick (✓) the correct answer in each of Q.1 to Q.9
Question 1.
Solution:
(a) A scalene triangle has no line of symmetry

Question 2.
Solution:
(c) A rectangle has two lines of symmetry which are the lines joining the mid points of opposite sides.

Question 3.
Solution:
(d) A square has four lines of symmetry which are two diagonal and two the lines joining the mid points of opposite sides.

Question 4.
Solution:
(b) A rhombus has two lines of symmetry which are the diagonals.

Question 5.
Solution:
(d) A circle has an unlimited number of lines of symmetry as its lines of symmetry is its diameter which are infinite in number.

Question 6.
Solution:
(a) ∆ABC in which AB = AC, is an isosceles triangle and AD ⊥ BC.

AD is the line of symmetry

Question 7.
Solution:
(a) ABCD is a kite in which AB = AD and BC = DC

Its line of symmetry will be one diagonal AC.

Question 8.
Solution:
(c) The letter O of the English Alphabet has two lines of symmetry as shown here in the figure.

Question 9.
Solution:
(a) The letter Z of the English Alphabet has no line of symmetry.

Question 10.
Solution:
The line/lines of symmetry have been drawn as given below :

Question 11.
Solution:
(i) True
(ii) True
(iii) True : The bisectors of angles are its lines of symmetry.
(iv) False : A rhombus has two lines of symmetry which are its diagonals.
(v) True : The two diagonals and two perpendicular bisector of its opposite sides are the lines of symmetry.
(vi) True : The perpendicular bisectors of opposite sides are the two lines of symmetry of the rectangle.
(vii) True : Each of the English Alphabet H, I, O and X has two lines of symmetry.

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RS Aggarwal Class 7 Solutions Chapter 17 Constructions CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
• RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

Question 1.
Solution:
Given, ∠ABO = 60°
∠CDO = 40°
⇒ ∠ABO = ∠BOC = 60° [alternate angles]

Question 2.
Solution:
Here, AB || EC
∠BAC = ∠ACE = 70° (alternate angles)
⇒ ∠BCA = 180° – ∠BAC
⇒ ∠BCA = 180°- 120°
⇒ ∠BCA = 60°

Question 3.
Solution:
(i) ∠AOC = ∠BOD = 50° [vertically opposite angles]
(ii) ∠BOC = 180° – 50° (linear pair)
= 130°

Question 4.
Solution:
Here, 3x + 20 + 2x – 10 = 180
⇒ 5x + 10 = 180
⇒ 5x = 170
⇒ x = 34
∠AOC = (3 x 34 + 20)° = (102 + 20)° = 122°
∠BOC = (2 x 34 – 10)° = (68 – 10)° = 58°

Question 5.
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 65° + 45° + ∠C= 180°
⇒ ∠C = 180° – 110° = 90°

Question 6.
Solution:
Let x = 2k and y = 3k
2k + 3k = 120° [Exterior angle property]
⇒ 5k = 120°
⇒ k = 24°
x = 2 x 24° = 48° and y = 3 x 24° = 72°
In ∆ABC :
∠A + ∠B + ∠C = 180°
⇒ 48° + 72° + ∠C = 180°
⇒ ∠C = 180°- 120°
⇒ ∠C = 60°
z = 60°

Question 7.
Solution:
Since it is a right triangle, by using the Pythagoras theorem:
Length of the hypotenuse = √(8² + 15²) = √(64 + 225) = √289 = ± 17 cm
The length of the side can not be negative.

Question 8.
Solution:
Given:
∠BAD = ∠DAC …..(i)
To show that ∆ABC is isosceles, we should show that ∠B = ∠C
AD ⊥ BC, ∠ADB = ∠ADC = 90°
∠ADC = ∠ADB
∠BAD + ∠ABD = ∠DAC + ∠ACD (exterior angle property)
∠DAC + ∠ABD = ∠DAC + ∠ ACD [from equation (i)]
∠ABD = ∠ACD
This is because opposite angles of a triangle ∆ABC are equal.
Hence, ∆ABC is an isosceles triangle.

Mark (✓) against the correct answer in each of the following :
Question 9.
Solution:
(c) 145°
The supplement of 35° = 180° – 35° = 145°

Question 10.
Solution:
(d) 124
x° + 56° = 180° (linear pair)
⇒ x = 180° – 56°
⇒ x = 124
x = 124°

Question 11.
Solution:
(c) 65°
∠ACD = 125°
∠ACD = ∠CAB + ∠ABC (the exterior angles are equal to the sum of its interior opposite angles)
∠ABC = 125° – 60° = 65°

Question 12.
Solution:
(c) 105°
∠A + ∠B + ∠C = 180°
⇒ ∠A = 180° – (40° + 35°)
⇒ ∠A = 105°

Question 13.
Solution:
(c) 60°
Given:
2∠A = 3∠B

Question 14.
Solution:
(b) 55°
In ∆ABC :
A + B + C = 180° …(i)
Given, A – B = 33°
A = 33° + B …(ii)
B – C = 18°
C = B + 18° …(iii)
Putting the values of A and B in equation (i):
⇒ B + 33° + B + B – 18° = 180°
⇒ 3B = 180°
⇒ B = 55°

Question 15.
Solution:
(b) 3√2 cm
Here, AB = AC
In right angled isosceles triangle:
BC² = AB² + AC²
⇒ BC² = AB² + AB²
⇒ BC² = 2AB²
⇒ 36 = 2AB²
⇒ AB² = 18
⇒ AB = √18
⇒ AB = 3√2

Question 16.
Solution:
(i) The sum of the angles of a triangle is 180°.
(ii) The sum of any two sides of a triangle is always greater than the third side.
(iii) In ∆ABC, if ∠A = 90°, then BC² = (AB²) + (BC²)
(iv) In ∆ABC :
AB = AC
AD ⊥ BC
Then, BD = DC
This is because in an isosceles triangle, the perpendicular dropped from the vertex joining the equal sides, bisects the base.
(v) In the given figure, side BC of ∆ABC has produced to D and CE || BA.
If ∠ABC = 50°, then ∠ACE = 50°
AB || CE
∠BAC = ∠ACE = 50° (alternate angles)

Question 17.
Solution:
(i) True
(ii) True
(iii) False. Each acute angle of an isosceles right triangle measures 45°.
(iv) True.

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Objective Questions
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(c)
Length of rectangle AB = 16 cm
and diagonal BD = 20 cm

But, in right ∆ABD
BD² = AB² + AD²
⇒ (20)² = (16)² + AD²
⇒ 400 = 256 + AD²
⇒ AD² = 400 – 256 = 144 = (12)²
⇒ AD = 12 cm
Area = l x b = 16 x 12 = 192 cm²

Question 2.
Solution:
(b)
Diagonal of square = 12 cm
Let side = 9
diagonal = √2 a

Question 3.
Solution:
(b)
Area = 200 cm²
side = √200 =√2 x 10
and diagonal = √2 a = √2 x √2 x 10 = 20

Question 4.
Solution:
(a)
Area of square = 0.5 hectare = 0.5 x 10000 = 5000 m²
= √10000 = 100 m

Question 5.
Solution:
(c)
Perimeter of rectangle = 240m
l + b = \(\frac { 240 }{ 2 }\) = 120 m
Let breadth = x, then length = 3x .
3x + x = 120
⇒ 4x = 120
⇒ x = 30
Length = 3x = 3 x 30 = 90 m

Question 6.
Solution:
Answer = (d)
Let original side of square = x
area = x²

Question 7.
Solution:
(b)
Let side of square = a
Then its diagonal = √2 a
Now, area of square = a²
and area of square on diagonal = (√2 a)² = 2a²
Ratio = a² : 2a² = 1 : 2

Question 8.
Solution:
(c)
If perimeters of a square and a rectangle are equal Then the area of the square will be greater than that of a rectangle
A > B

Question 9.
Solution:
(b)
Perimeter of rectangle = 480m

Question 10.
Solution:
(a)
Total cost of carpet = Rs. 6000
Rate per metre = Rs. 50

Question 11.
Solution:
(a)
Sides are 13 cm, 14 cm, 15 cm

Question 12.
Solution:
(b)
Base of triangle = 12 m
and height = 8m
Area= \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 12 x 8 = 48 m²

Question 13.
Solution:
(b)
Let side = a
then area = \(\frac { \surd 3 }{ 4 }\) a²

Question 14.
Solution:
(c)
Side of an equilateral triangle = 8cm

Question 15.
Solution:
(b)
Let a be the side of an equilateral triangle

Question 16.
Solution:
(b)
One side (Base) of parallelogram = 16 cm
and altitude = 4.5 cm
Area = base x altitude = 16 x 4.5 = 72 cm²

Question 17.
Solution:
(b)
Length of diagonals of a rhombus are 24 cm and 18 cm

Question 18.
Solution:
(c)
Let r be the radius of the circle Then
c = 2πr
2πr – r = 37

Question 19.
Solution:
(c) Perimeter of room = 18 m
and height = 3 m
Area of 4 walls = Perimeter x height = 18 x 3 = 54 m²

Question 20.
Solution:
(a) Area of floor = l x b = 14 x 9 = 126 m²
Area of carpet = 126 m²

Question 21.
Solution:
(c)
Perimeter = 46 cm

Question 22.
Solution:
(b)
Ratio in area of two squares = 9 : 1
Let area of bigger square = 9x²
and of smaller square = x²
Side of bigger square = √9x² = 3x
and perimeter = 4 x side = 4 x 3x = 12x
Side of smaller square = √x² = x
Perimeter = 4x
Now ratio in their perimeter = 12x : 4x = 3 : 1

Question 23.
Solution:
(d)
Let the diagonals of two square be 2d and d
Area of bigger square 2 (2d)² = 8d²
and of smaller = 2 (d)² = 2d²
Ratio in their area = \(\frac { 8{ d }^{ 2 } }{ 2{ d }^{ 2 } } =\frac { 4 }{ 1 }\)
= 4 : 1

Question 24.
Solution:
(c)
Side of square = 84 m
Area of square = (84)² = 7056 m²
Area of rectangle = 7056 m²
Length of rectangle = 144 m

Question 25.
Solution:
(d)
Side of a square = a
Area = a²
Side of equilateral triangle = a

Question 26.
Solution:
(a)
Let a be the side of a square
Area = a²
Then area of circle = a²
Let r be the radius

Question 27.
Solution:
(b)
Let each side of an equilateral triangle = a
Then area = \(\frac {\surd 3 }{ 4 }\) a²
Now radius of the circle = a
Then area = πr² = πa²

Question 28.
Solution:
(c)
Area of rhombus = 36 cm²
Length of one diagonal = 6 cm
Length of second diagonal

Question 29.
Solution:
(d)
Area of a rhombus =144 cm²
Let one diagonal (d1) = a
then Second diagonal (d2) = 2a

Larger diagonal = 2a = 2 x 12 = 24 cm

Question 30.
Solution:
(c)
Area of a circle = 24.64 m²

Question 31.
Solution:
(c)
Let original radius = r
Then its area = πr²
Radius of increased circle = r + 1

2r = 7 – 1 = 6
⇒ r = \(\frac { 6 }{ 2 }\) = 3
Radius of original circle = 3 cm

Question 32.
Solution:
(c)
Radius of a circular wheel (r) = 1.75 m

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
(i) Radius of the circle (r) = 21 cm

Question 2.
Solution:
(i) Diameter of the circle = 28 cm

Question 3.
Solution:
Circumference of a circle = 264 cm

Question 4.
Solution:
Circumference of the circle (c) = 35.2 m

Question 5.
Solution:
Area of the circle = 616 cm²

Question 6.
Solution:
Area of a circle = 1386 m²

Question 7.
Solution:
Ratio in the radii of two circles = 4 : 5
Let radius of first circle (r1) = 4x
and radius of the second circle (r2) = 5x

Question 8.
Solution:
Length of rope (r) = 21 m
Area of the circle = πr² = \(\frac { 22 }{ 7 }\) x 21 x 21 m² = 1386 m²
The horse will graze on 1386 m² area

Question 9.
Solution:
Area of a square made of a wire = 121 cm²
Side = √Area = √121 = 11 m
Perimeter of wire = 4 x side = 4 x 11 = 44 cm
Circumference of circular wire = 44 cm

Question 10.
Solution:
Radius of circular wire = 28 cm
Circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 28 cm = 176 cm
Perimeter of the square formed by this wire = 176 cm
Side (a) = \(\frac { 176 }{ 4 }\) = 44 cm .
Area of square so formed = a² = (44)² cm² = 1936 cm²

Question 11.
Solution:
Length of rectangular sheet (l) = 34 cm
and breadth (b) = 24 cm
Area = l x b = 34 x 24 cm² = 816 cm²
Diameter of one button = 3.5 cm
Radius (r) = \(\frac { 3.5 }{ 2 }\) cm
and area of one button = πr²

Area of 64 buttons = 9.625 x 64 cm² =616 cm²
Area of remaining sheet = 816 – 616 = 200 cm²

Question 12.
Solution:
Length of ground (l) = 90 m
and breadth (b) = 32 m
Area = l x b = 90 x 32 m² = 2880 m²
Radius of circular tank (r) = 14 m

= 616 m²
Area of remaining portion = 2880 – 616 = 2264 m²
Rate of turfing = Rs. 50 per sq.m.
Total cost = Rs. 50 x 2264 = Rs. 113200

Question 13.
Solution:
Each side of square = 14 cm
Area of square = a² = 14 x 14 = 196 cm²
Radius of each circle at each corner of square = \(\frac { 14 }{ 2 }\) = 7 cm

Area of shaded portion = Area of square – area of 4 quadrants
= 196 – 154 = 42 cm²

Question 14.
Solution:
Length of field = 60 m
and breadth = 40 m
Length of rope = 14 m
Area covered by the horse

Question 15.
Solution:
Diameter of largest circle (outer circle) = 21 cm


Area of shaded portion = 346.5 – (154 + 38.5) = (346.5 – 192.5) = 154 cm²

Question 16.
Solution:
Length of plot (l) = 8m
and breadth (b) = 6m
Area of plot = l x b = 8 x 6 = 48 m²
Radius of each quadrant at the corner = 2 m

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F are helpful to complete your math homework.

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