# PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

## PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.

Let △ABC ~ △DEF and their areas be respectively 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.

Solution:

△ABC ~ △DEF ;

area of △ABC = 64 cm^{2};

area of △DEF = 121 cm^{2};

EF= 15.4 cm

Question 2.

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD. Find the ratio of the areas of traingles AOB and COD.

Solution:

ABCD is trapezium AB || DC. Diagonals AC and BD intersects each other at the point O. AB = 2 CD

∴ Required ratio of ar △AOB and △COD = 4 : 1.

Question 3.

In the fig., △ABC and △DBC are two triangles on the same base BC. If AD intersects BC at O show that

Hence proved.

Question 4.

If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Given: Two ∆s ABC and DEF are similar and equal in area.

To Prove : ∆ABC ≅ ∆DEF

Question 5.

D, E and F are respectively the mid points of the sides BC, CA and AB of ∆ABC. Determine the ratio of the areas of triangles DEF and ABC.

Solution:

Given. D, E, F are the mid-point of the sides BC, CA and AB respectively of a tABC.

To find : ar (∆DEE) : ar (∆ABC)

Proof: In ∆ABC,

F is the mid-point of AB …(given)

E is the mid-point of AC …(given)

So, by the Mid-Foint Theorem

FE || BC and FE = 1/2 BC

⇒ FE || BD and FE = BD [∵ BD = 1/2 BC]

∴ BDEF is a || gm.

(∵ Opp. sides are || and equal)

In △s FBD and DEF,

FB = DE (opp. sides of || gm BDEF)

FD = FD .. .(common)

BD = FE

. ..(opp. sides of || gm BDEF)

∴ △FBD ≅ △DEF … (SSS Congruency Theorem)

Similary we can prove that:

△AFE ≅ △DEF

and △EDC ≅ △DEF

if △s are , then they are equal in area.

∴ ar (∆FBD) = ar. (∆DEF) ……………(1)

ar (∆AFE) = ar (∆DEF) ……………(2)

ar (∆EDC) = ar (∆DEF) ……………(3)

Now ar ∆ (ABC)

= ar (∆FBD) + ar (∆DEF) + ar (∆AFE) + ar (∆EDC)

= ar.(∆DEF) + ar (∆DEF) + ar (∆DEF) + ar. (∆DEF) [Using (1), (2) and (3)]

= 4 ar (∆DEF)

⇒ (∆DEF) = 1/4 ar(∆ABC)

⇒ ar.(ΔDEF)/ar.(ΔABC)=1/4

∴ ar (∆DEF) : ar (∆ABC) = 1 : 4.

Question 6.

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Given: ∆ABC ~ ∆DEF.

AX and DY are the medians to the side BC and EF respectively.

Question 7.

Prove that the areas of the equilateral triangle described on the side of a square Is half the area of the equilateral triangle described on its diagonal.

Solution:

Given: ABCD is a square. Equilateral ∆ABE is described on the side AB of the square and equilateral ∆ACF is desribed on the diagonal AC.

Question 8.

Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4.

Solution:

Question 9.

Tick the correct answer and justify: Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are ¡n the ratio

(A) 2 : 3

(B) 4 : 9

(C) 81 : 16

(D) 16 : 81.

Solution:

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