Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

## Haryana Board 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 :

We have, ∆ABC ~ ∆DEF

Hence, BC = 11.2 cm.

Question 2.

Diagonals of trapezium ABCD with AB || CD intersects each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:

We have, AB || CD and AB = 2CD

In triangles AOB and COD,

∠ABO = ∠CDO (alternate interior ∠s)

∠AOB = ∠COD (vertically opposite ∠s)

∆AOB ~ ∆COD (By AA similarity criterion)

⇒ \(\frac{\ {ar}(\Delta \mathrm{AOB})}{\ {ar}(\Delta \mathrm{COD})}=\frac{\mathrm{AB}^2}{\mathrm{CD}^2}\)

⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{(2 \mathrm{CD})^2}{\mathrm{CD}^2}\)

⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4 \mathrm{CD}^2}{\mathrm{CD}^2}\)

⇒ \(\frac{\ {ar}(\triangle \mathrm{AOB})}{\ {ar}(\triangle \mathrm{COD})}=\frac{4}{1}\)

Hence, ar(∆AOB) : ar(∆COD) = 4 : 1.

Question 3.

In given figure ABC and DBC are two triangles on the same base BC.. If AD intersects BC at O, show that \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\).

Solution :

Given : ∆ABC and ∆DBC are two triangles on the same base BC

To Prove: \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}\)

Construction : Draw AL ⊥ BC and DM ⊥ BC

Proof : In ∆ALO and ∆DMO

∠ALO = ∠DMO = 90° (By construction)

∠AOL = ∠DOM (vertically opposite ∠s)

∆ALO ~ ∆DMO (By AA similarity criterion)

⇒ \(\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)

[corresponding sides of similar triangles are proportional]

Hence proved.

Question 4.

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

Solution:

Given : Let two triangles ABC and PQR such that ar(ABC) = ar(APQR)

To Prove: ∆ABC ∆PQR

Proof: ar(∆ABC) ≅ ar(∆PQR)

⇒ ∆ABC ~ ∆PQR

⇒ ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

and \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)

⇒ AB = PQ, BC = QR, CA = RP

∴ ∆ABC ≅ ∆PQR (By SSS congruence criteriah)

Hence Proved.

Question 5.

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

Solution:

We have,

∵ D, E and F, are the mid points of sides AB, BC and CA respectively.

∴ DF = \(\frac{1}{2}\)BC, DE = \(\frac{1}{2}\)AC and EF = \(\frac{1}{2}\)AB

[Mid point Theorem]

Hence, ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 6.

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

Solution:

Given: Two triangles ABC and PQR such that AABC – APQR and AD, PM are their medians respectively.

Hence proved.

Question 7.

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution :

Given : A square ABCD and two equilateral ∆s BCE and ACF have been described on side BC and diagonal AC respectively.

To Prove: Area (∆BCE) = \(\frac{1}{2}\) Area (∆ACF)

Proof: We have ∆BCE and ∆ACF are equilateral triangles.

∠B = ∠A (each = 60°)

∠C = ∠C (each = 60°)

∠E = ∠F (each = 60°)

∆ BCE ~ ∆ ACF (By AAA similarity criterion)

⇒ \(\frac{\ {ar}(\triangle \mathrm{BCE})}{\ {ar}(\triangle \mathrm{ACF})}=\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\)

[By theorem 6.6] ………………..(1)

But AB = BC (sides squares)

and AC^{2} = AB^{2} + BC^{2} [By Pythagoras theorem]

⇒ AC^{2} = 2BC^{2} ……………..(2)

From (1) and (2), we get

\(\frac{\ {ar}(\Delta B C E)}{\ {ar}(\Delta A C F)}=\frac{B^2}{2 B C^2}\)

⇒ \(\frac{\ {ar}(\triangle B C E)}{\ {ar}(\triangle A C F)}=\frac{1}{2}\)

⇒ ar(∆BCE) = \(\frac{1}{2}\) ar(∆ACF)

Hence Proved.

In question 8 and 9, tick the correct answer and justify:

Question 8.

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

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Solution :

Correct answer is (C).

Justification : ∆ABC and ∆BDE are two equilateral triangles.

Let AB = BC = CA = a unit

Then BD = \(\frac{a}{2}\) unit

[∵ D is the mid point of BC]

∵ ∠CAB = ∠DBE, ∠ABC = ∠BDE, ∠ACB = ∠BED [each is 60°]

∆ABC ~ ∆BDE (By AAA similarity criterion)

⇒ \(\frac{\ {ar}(\triangle \mathrm{ABC})}{\ {ar}(\Delta \mathrm{BDE})}=\frac{\mathrm{AB}^2}{\mathrm{BD}^2}\) [By theorem (6.6)]

⇒ ar(∆ABC) : ar(∆BDE) = 4 : 1.

Question 9.

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.

(A) 2 : 3

(B) 4 : 9

(C) 81 : 16

(D) 16 : 81

Solution :

Correct answer is (D).

Justification : We have,

\(\frac{\text { Side of } 1^{\text {st }} \text { triangle }}{\text { Side of } 2^{\text {nd }} \text { triangle }}=\frac{4}{9}\)

We know that areas of two similar triangles are in the ratio of the squares of their corresponding sides.

∴ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{(\text { side of } 1 \text { st } \Delta)^2}{(\text { side of } 2 \text { nd } \Delta)^2}\)

⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{4^2}{9^2}\)

⇒ \(\frac{\text { Area of Ist triangle }}{\text { Area of IInd triangle }}=\frac{16}{81}\)

⇒ Area of 1st triangle : Area of IInd triangle = 16 : 81.