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RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9A
• RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9B
• RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9C

Question 1.
Solution:
In parallelogram ABCD.
∠ A = 72°
But ∠ A = ∠ C (opposite angle of a ||gm)
∴ ∠ C = 72°
∴ AD || BC
∴ ∠ A + ∠ B = 180° (co-interior angles)
=> 72° + ∠B = 180°
=> ∠B = 180° – 72°
=> ∠B = 108°
But ∠ B = ∠ D (opposite angles of a ||gm)
∴ ∠D = 108°
Hence ∠D = 108°, ∠ C = 72° and ∠ D = 108° Ans.

Question 2.
Solution:
In || gm ABCD, BD is its diagonal
and ∠DAB = 80° and ∠DBC = 60°
∴AB || DC
∴∠DAB + ∠ADC – 180°
(co-interior angles)
=> 80° + ∠ADC = 180°
=> ∠ ADC = 180° – 80°
=> ∠ ADC = 100°
But ∠ ADB = ∠ DBC (Alternate angles)
∴∠ ADB = 60°
But ∠ ADB + ∠ CDB = 100°
(∴∠ ADC = 100°)
60° + ∠CDB = 100°
=> ∠CDB = 100° – 60° = 40°
Hence ∠CDB = 40° and ∠ ADB = 60° Ans.

Question 3.
Solution:
Given : In ||gm ABCD,
∠ A = 60° Bisectors of ∠ A and ∠ B meet DC at P.
To Prove : (i) ∠APB = 90°
(ii) AD = DP and PB = PC = BC
(iii) DC = 2AD
Proof: ∴ AD || B (opposite sides of a ||gm)
∴∠ A + ∠ B = 180° (co-interior angles)
But AP and BP are the bisectors of ∠A and ∠B
∴\(\frac { 1 }{ 2 } \)∠A + \(\frac { 1 }{ 2 } \) ∠B = 90°

=> ∠PAB + ∠PBA = 90°
But in ∆ APB,
∠PAB + ∠PBA + ∠APB = 180° (angles of a triangle)
=> 90° + ∠APB = 180°
=> ∠APB = 180° – 90° = 90°
Hence ∠APB = 90°
(ii) ∠ A + ∠ D = 180° (co-interior angles)
and ∠ A – 60°
∴ ∠D = 180° – 60° = 120°
But ∠DAP = \(\frac { 1 }{ 2 } \) ∠A = \(\frac { 1 }{ 2 } \) x 60° = 30°
∴∠DPA = 180° – (∠DAP + ∠D)
= 180° – (30° + 120°)
= 180° – 150° = 30°
∠DAP = ∠DPA (each = 30°)
Hence AD = DP (sides opposite to equal angles)
In ∆ BCP,
∠ C = 60° (opposite to ∠ A)
∠CBP = \(\frac { 1 }{ 2 } \) ∠ B = \(\frac { 1 }{ 2 } \) x 120° = 60°
But ∠CPB + ∠CBP + ∠C = 180°
(Angles of a triangle)
=> ∠CPB + 60° + 60° = 180°
=> ∠CPB + 120° = 180°
=> ∠CPB = 180° – 120° = 60°
∆ CBP is an equilateral triangle and BC = CP = BP
=> PB – PC = BC
(iii) DC = DP + PC
= AD + BC
(∴ DP = AD and PC = BC proved)
= AD + AD (∴ AD = BC opposite sides of a ||gm)
= 2AD
Hence DC = 2AD.
Hence proved.

Question 4.
Solution:
In ||gm ABCD,
AC and BD are joined
∠BAO = 35°, ∠ DAO = 40°
∠COD = 105°
∴ ∠AOB = ∠COD
(vertically opposite angles)
∴∠AOB = 105°
(i) Now in ∆ AOB,
∠ABO + ∠AOB + ∠OAB = 180°
(angles of a triangle)
=> ∠ABO + 105° + 35° = 180°
=> ∠ABO + 140° = 180°
=> ∠ABO = 180° – 140°
∠ ABO = 40°
(ii) ∴ AB || DC
∴ ∠ ABO = ∠ ODC (alternate angles)
∴ ∠ ODC = 40°
(iii) ∴ AD || BC
∴ ∠ ACB = ∠ DAO or ∠ DAC
(alternate angles)
= 40°
(iv) ∴ ∠ A + ∠ B = 180° (co-interior angles)
=> (40° + 35°) + ∠B = 180°
=> ∠B = 180° – 75° = 105°
=> ∠ CBD + ∠ABO = 105°
=> ∠CBD + 40° = 105°
=> ∠CBD = 105° – 40° = 65°
Hence ∠ CBD = 65° Ans.

Question 5.
Solution:
In ||gm ABCD
( ∠A = (2x + 25)° and ∠ B = (3x – 5)°
∴AD || BC (opposite sides of parallelogram)
∴∠ A + ∠B = 180° (co-interior angles)
=> 2x + 25° + 3x – 5° = 180°
=> 5x + 20° = 180°
=> 5x = 180° – 20°

=> 5x = 160° => x = \(\frac { { 160 }^{ o } }{ 5 } \) = 32°
∴x = 32°
Now ∠A = 2x + 25° = 2 x 32° + 25°
= 64° + 25° = 89°
∠B = 3x – 5 = 3 x 32° – 5°
= 96° – 5° = 91°
∠ C = ∠ A (∴ opposite angles of ||gm)
= 89°
Similarly ∠B = ∠D
∠D = 91°
Hence ∠ A = 89°, ∠ B = 91°, ∠ C = 89°, ∠D = 91° Ans.

Question 6.
Solution:
Let ∠A and ∠B of a ||gm ABCD are adjacent angles.
∠A + ∠B = 180°
Let ∠B = x

Then ∠ A = \(\frac { 4 }{ 5 } \) x
∴ x + \(\frac { 4 }{ 5 } \) x = 180°
\(\frac { 9 }{ 5 } \) x = 180°
=>\(\frac { { 180 }^{ o }\quad X\quad 5 }{ 9 } \) = 100°
∴ ∠A = \(\frac { 4 }{ 5 } \) x 100° = 80°
and ∠B = 100°
But ∠ C = ∠ A and ∠ D = ∠ B
(opposite angles of a || gm)
∴∠C = 80°, and ∠D = 100°
Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠ D = 100° Ans.

Question 7.
Solution:
Let the smallest angle ∠ A and the other angle ∠ B
Let ∠ A = x
Then ∠ B = 2x – 30°
But ∠ A + ∠ B = 180° (co-interior angles)
∴x + 2x – 30° = 180°
=> 3x = 180° + 30° = 210°
=> x = \(\frac { { 210 }^{ o } }{ 3 } \) = 70°
∴ ∠ A = 70°

and ∠ B = 2x – 30° = 2 x 70° – 30°
= 140° – 30° = 110°
But ∠C = ∠ A and ∠D = ∠B
(opposite angles of a ||gm)
∠C = 70° and ∠D = 110°
Hence ∠A = 70°, ∠B = 110°, ∠C = 70° and ∠D = 110° Ans.

Question 8.
Solution:
In ||gm ABCD,
AB = 9.5 cm and perimeter = 30 cm
=> AB + BC + CD + DA = 30cm
=> AB + BC + AB + BC = 30 cm
( ∴ AB = CD and BC – DA opposite sides)

=> 2(AB + BC) = 30cm
=> AB + BC = 15cm
=> 9 5cm + BC = 15cm
∴BC = 15cm – 9.5cm = 5.5cm
Hence AB = 9.5cm, BC = 5.5cm,
CD = 9.5cm and DA = 5.5cm Ans.

Question 9.
Solution:
ABCD is a rhombus
AB = BC = CD = DA
(i)∴ AB || DC
∴ ∠ B + ∠ C = 180° (co-interior angles)
=> 110° + ∠C = 180°
=> ∠C = 180° – 110° = 70°

Question 10.
Solution:
In a rhombus,
Diagonals bisect each other at right angles
∴ AC and BC bisect each other at O at right angles.
But AC = 24 cm and BD = 18 cm

Question 11.
Solution:
Let ABCD he the rhombus whose diagonal are AC and BD which bisect each other at right angles at O.

Question 12.
Solution:
ABCD is a rectangle whose diagonals AC and BD bisect each other at O.

Question 13.
Solution:
ABCD is a square. A line CX cuts AB at X and diagonal BD at O such that
∠ COD = 80° and ∠ OXA = x°
∴∠ BOX = ∠ COD
(vertically opposite angles)
∴∠BOX = 80°
∴Diagonal BD bisects ∠ B and ∠ D
∴ ∠ABO or ∠ABD = ∠ ADO or ∠ ADB
∴ ∠OBA or ∠OBX = 45°
Now in ∆ OBX,
Ext. ∠ OXA = ∠ BOX + ∠ OBX
=>x° = 80° + 45° = 125° Ans.

Question 14.
Solution:
Given : In ||gm ABCD, AC is joined. AL ⊥ BD and CM ⊥ BD
To prove :
(i) ∆ ALD ≅ ∆ CMB
(ii) AL = CM
Proof : In ∆ ALD and ∆ BMC
AD = BC (opposite sides of ||gm)
∠L = ∠M (each 90°)
∠ ADL = ∠ CBM (Alternate angles)
∴ ∆ ALD ≅ ∆ BMC. (AAS axiom)
∴ or A ALD ≅ A CMB.
AL = CM (c.p.c.t.) Hence proved.

Question 15.
Solution:
Given : In ∆ ABCD, bisectors of ∠A and ∠B meet each other at P.
To prove : ∠APB = 90°
Proof : AD || BC
∠A + ∠B = .180° (co-interior angles)
PA and PB are the bisectors of ∠ A and ∠B

Question 16.
Solution:
In ||gm ABCD,
P and Q are the points on AD and BC respectively such that AP = \(\frac { 1 }{ 3 } \) AD and CQ = \(\frac { 1 }{ 3 } \) BC

Question 17.
Solution:
Given : In ||gm ABCD, diagonals AC and BD bisect each other at O.
A line segment EOF is drawn, which meet AB at E and DC at F.
To -prove : OE = OF

Question 18.
Solution:
Given : ABCD is a ||gm.
AB is produced to E. Such that AB = BE
DE is joined which intersects BC in O.
To prove : ED bisects BC i.e. BO = OC

Question 19.
Solution:
Given : In ||gm ABCD, E is the midpoint BC
DE is joined and produced to meet AB on producing at F.

Question 20.
Solution:
Given : ∆ ABC and lines are drawn through A, B and C parallel to respectively BC, CA and AB forming ∆ PQR.
To prove : BC = \(\frac { 1 }{ 2 } \) QR

Question 21.
Solution:
Given : In ∆ ABC, parallel lines are drawn through A, B and C respectively to the sides BC, CA and AB intersecting each other at P, Q and R.

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.

Other Exercises

• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
• RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.

Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.

Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.

Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : \(\frac { upper\quad limit+lower\quad limit }{ 2 } \) is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.

Question 5.
Solution:
The given data can be represent in form of frequency table as given below:

Question 6.
Solution:
The frequency distribution table of the given data is given below :

Question 7.
Solution:
The frequency distribution table of the

Question 8.
Solution:
The frequency table is given below :

Question 9.
Solution:
The frequency table of given data is given below :

Question 10.
Solution:
The frequency distribution table of the given data in given below :

Question 11.
Solution:
The frequency table of the given data:

Question 12.
Solution:
The cumulative frequency of the given table is given below:

Question 13.
Solution:
The given table can be represented in a group frequency table in given below :

Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :

Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :

 

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

MCQ
Question 1.
Solution:
(b) ∆ABC ~ ∆DEF

Question 2.
Solution:
(a) In the given figure,

Question 3.
Solution:
(b) Length of pole AB = 6 m
and CD = 11 m
and distance between the BD = 12 m
Draw AE || BD, then
ED = AB = 6 m, AE = BD = 12 m
CE = CD – ED = 11 – 6 = 5 m

Now, in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = 13²
AC = 13
Distance between their tops = 13 m

Question 4.
Solution:
(c) Area of two similar triangles ABC and DEF are 25 cm² and 36 cm².
Altitude of first ∆ABC is AL = 3.5 m.
Let DM be the altitude of second triangle.

Short-Answer Questions
Question 5.
Solution:
∆ABC ~ ∆DEF
and 2AB = DE

Question 6.
Solution:
In the given figure,

DE || BC, AL ⊥ BC
AD = x cm,
DB = (3x + 4) cm AE = (x + 3) cm, EC = (3x + 19) cm
In ∆ABC, DE || BC

Question 7.
Solution:
Let AB be the ladder, and A is the window.
Length of ladder AB = 10 m and AC = 8 m
Distance between the foot of the ladder from the house = x m

In right ∆ABC,
AB² = AC² + BC²
⇒ (10)² = (8)² + (x)²
⇒x² = (10)² – (8)²
⇒ x² = 100 – 64 = 36
⇒ x² = (6)²
⇒ x = 6
Distance between the foot of ladder and foot of the house = 6m.

Question 8.
Solution:
In ∆ABC, AB = BC = CA = 2a cm
AD ⊥ BC which bisects the base BC at D

BD = DC = \(\frac { 2a }{ 2 }\) = a cm
Now, in right ∆ABD
AB² = AD² + BD²
(2a)² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
AD = √3a² = √3 a
Height of altitude = √3 a cm

Question 9.
Solution:
Let EF = x cm
∆ABC ~ ∆DEF

Question 10.
Solution:
ABCD is a trape∠ium in which AB || DC
AB = 2CD
Diagonals AC and BD intersect each other at O.

Question 11.
Solution:
Let corresponding sides of two similar triangles ABC and DEF are in the ratio 2 : 3.

Question 12.
Solution:
In the given figure,

Question 13.
Solution:
Given : In ∆ABC, AD is the bisector of ∠A which meets BC at D.

Construction : Produce BA and draw CE || DA meeting BA produced at E.
DA || CE
∠3 = ∠1 (corresponding angles)
∠4 = ∠2 (alternate angles)
But ∠3 = ∠4 (AD is the bisector of ∠A)
∠1 = ∠2
AC = AE (Sides opposite to equal angles)
Now, in ∆AEC,
AD || EC

Question 14.
Solution:
Given : In ∆ABC,
AB = BC = CA = a cm

Question 15.
Solution:
In rhombus ABCD, diagonals AC = 24 cm and BD = 10 cm.

The diagonals of a rhombus bisect each other at right angles.
AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² = 12² + 5² = 144 + 25 = 169 = (13)²
AB = 13
Each side of a rhombus = 13 cm

Question 16.
Solution:
Given : ∆ABC ~ ∆DEF

Long-Answer Questions
Question 17.
Solution:
Given: In the given figure, ∆ABC and ∆DBC are on the same base BC but in opposite sides.
AD and BC intersect at O.

Question 18.
Solution:
In the given figure,
XY || AC and XY divides
∆ABC into two regions equal in area

Question 19.
Solution:
In the given figure, ∆ABC is an obtuse triangle, obtuse angle at B.
AD ⊥ CB produced.

To prove : AC² = AB² + BC² + 2BC x BD
Proof: In ∆ADB, ∠D = 90°
AB² = AD² + DB² (Pythagoras Theorem)
⇒ AD² = AB² – DB² ……(i)
Similarly, in right ∆ADC,
AC² = AD² + DC²
= AB² – DB² + (DB + BC)² [From (i)]
= AB² – DB² + DB² + BC² + 2BC x BD
= AB² + BC² + 2 BC x BD

Question 20.
Solution:
In the given figure,
PA, QB and RC is perpendicular to AC.
AP = x, QB = z, RC =y, AB = a and BC = b.

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself 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 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.

Question 1.
Solution:
Co-ordinates of
A are ( – 6, 5)
B are (5, 4)
C are ( – 3, 2)
D are (2, – 2)
E are ( – 1, – 4)
Ans.

Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:

Question 3.
Solution:
(i) (7, 0) lies on x-axis as its ordinate is (0)
(ii) (0, – 5) lies on y-axis as its abscissa is (0)
(iii) (0, 1) lies on y-axis as its abscissa is (0)
(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)
Ans.

Question 4.
Solution:
(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)
(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)
(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).

Question 5.
Solution:
In the given equation, .
y = x+1
Put x = 0, then y = 0 + 1 = 1
x = 1, then, y = 1 + 1=2
x = 2, then, y = 2 + 1 = 3
Now, plot the points as given in the table given below on the graph, and join them as shown.

Question 6.
Solution:
In the given equation
y = 3x + 2
Put x = 0,
then y = 3 x 0 + 2 = 0 + 2 = 2
x = 1, then
y = 3 x 1 + 2 = 3 + 2 = 5
and x = – 2, then
y = 3 x ( – 2) + 2 = – 6 + 2 = – 4

Question 7.
Solution:
In the given equation,
y = 5x – 3
Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2
x = 0 then,
y = 5 x 0 – 3 = 0 – 3 = – 3
and x = 2, then
y = 5 x 2 – 3 = 10 – 3 = 7

Question 8.
Solution:
In the given equation,

y = 3x,
Put x = 0,
then y = 3 x 0 = 0
Put x = 1, then
y = 3 x 1 = 3
Put x = – 1, then
y = 3 ( – 1) = – 3
Now, plot the points as given in the table below

and join them as shown.

Question 9.
Solution:
In the given equation, y = – x,
Put x = 1,
then y = – 1
Put x = 2, then
y = – 2
Put x = – 2, then
y = – ( – 2) = 2

Hope given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4A
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4B
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles MCQS
• RS Aggarwal Solutions Class 10 Chapter 4 Triangles Test Yourself

Question 1.
Solution:
Given : Area of ∆ABC = 64 cm²
and area of ∆DEF =121 cm²
EF = 15.4 cm

Question 2.
Solution:

Question 3.
Solution:
∆ABC ~ ∆PQR
ar (∆ABC) = 4ar (∆PQR),

Question 4.
Solution:
Areas of two similar triangles are 169 cm² and 121 cm²
Longest side of largest triangle = 26 cm
Let longest side of smallest triangle = x
∆s are similar

Question 5.
Solution:
Area of ∆ABC = 100 cm²
and area of ∆DEF = 49 cm²

Question 6.
Solution:
Given : Corresponding altitudes of two similar triangles are 6 cm and 9 cm
We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.
Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9 (Dividing by 9)

Question 7.
Solution:
The areas of two similar triangles are 81 cm² and 49 cm²
Altitude of the first triangle = 6.3 cm
Let altitude of second triangle = x cm
The areas of two similar triangles are in the ratio of the squares on their corresponding altitude,
Let area of ∆ABC = 81 cm²
and area of ∆DEF = 49cm²
Altitude AL = 6 – 3 cm
Let altitude DM = x cm

Question 8.
Solution:
Areas of two similar triangles are 100 cm² and 64 cm²
Let area of ∆ABC = 100 cm²
and area of ∆DEF = 64 cm²
Median DM of ∆DEF = 5.6 cm
Let median AL of ∆ABC = x
The areas of two similar triangles is proportional to the squares of their corresponding median.

Question 9.
Solution:
Given : In ∆ABC, PQ is a line which meets AB in P and AC in Q.
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm QC = 4.5 cm

Question 10.
Solution:
In ∆ABC,
DE || BC
DE = 3 cm, BC = 6 cm
area (∆ADE) = 15 cm²

Question 11.
Solution:
Given : In right ∆ABC, ∠A = 90°
AD ⊥ BC
BC = 13 cm, AC = 5 cm
To find : Ratio in area of ∆ABC and ∆ADC
In ∆ABC and ∆ADC
∠C = ∠C (common)
∠BAC = ∠ADC (each 90°)
∆ABC ~ ∆ADC (AA axiom)

Question 12.
Solution:
In the given figure ∆ABC,
DE || BC and DE : BC = 3 : 5.
In ∆ABC and ∆ADE,
DE || BC
∆ABC ~ ∆ADE

Question 13.
Solution:
In ∆ABC, D and E are the midpoints of sides AB and AC respectively.
DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
∆ADE ~ ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C are helpful to complete your math homework.

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