Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 6 Triangles Exercise 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} and area of △DEF = 121 cm^{2} and EF = 15.4 cm.

△ABC ~ △DEF

∴ \(\frac{a r(\Delta \mathrm{ABC})}{a r(\Delta \mathrm{DEF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{AC}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)

[∵ We know that if two triangles are similar then the ratio of their area is equal to square of the ratio of their corresponding sides.]

∴ BC = \(\frac{8 \times 15.4}{11}\) = 8 × 1.4

BC = 11.2 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 triangles AOB and COD.

Solution:

ABCD is a trapezium in which AB || DC. Its diagonals AC and BD intersect each other at point O. AB = 2 CD.

Now, in △AOB and △COD,

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angles)

∴ △AOB ~ △COD

(By AAA similarity criterion)

∴ Ratio of required ar△AOB and ar△COD

= 4:1.

Question 3.

In Fig., ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac{a r(\mathrm{ABC})}{a r(\mathrm{DBC})}\) = \(\frac{\mathbf{A O}}{\text { DO }}\).

Solution:

Given : △ABC and △DBC aretwo triangles on the same base BC. AD intersects BC at O.

To Prove : \(\frac{a r(\mathrm{ABC})}{a r(\mathrm{DBC})}\) = \(\frac{\mathrm{AO}}{\mathrm{DO}}\)

Construction : Draw AL ⊥ BC, DM ⊥ BC

Proof: In △ALO and △DMO,

∠1 = ∠2 (Vertically opposite angles)

∠L = ∠M (Each 90°)

∴ △ALO ~ △DMO (AA similarity criterion)

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

(∵ If two triangles are similar, then their corresponding sides are proportional)

Question 4.

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

Solution:

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

To prove : △ABC ~ △DEF

Proof: Since △ABC ~ △DEF,

∴ \(\frac{a r(\triangle \mathrm{ABC})}{a r(\triangle \mathrm{DEF})}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)

⇒ \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\) = 1

[∵ ar (△ABC) = ar(△DEF)]

⇒ BC^{2} = EF^{2}

⇒ BC = EF

Also, since △ABC ~ △DEF, therefore they are equiangular.

and ∠B = ∠E

and ∠C = ∠F

Now in △ABC and △DEF,

∠B = ∠E, ∠C = ∠F

and BC = EF

△ABC ≅ △DEF

(ASA congmence theorem) (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:

Given : D, E and F are respectively the mid-points of sides AB, BC and CA of △ABC.

Required : To find ar(△DEF): ar(△ABC)

Proof : It is given that in △ABC, D and E are respectively the mid-points of the sides AB and BC.

∴ DE || AC

⇒ DE || FC ……(i)

Similarly in △ABC, D and F are the mid-points of sides AB and AC respectively.

∴ DF || BC

⇒ DF || EC ……..(ii)

From (i) and (ii) it is clear that DECF is a parallelogram.

Similarly △DEF is a parallelogram.

Now in △DEF and △ABC

∠DEF = ∠A

(From parallelogram ADEF)

and ∠EDF = ∠C

(From parallelogram DECF)

∴ By AA similarity

△DEF ~ △CAB

∴ 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 and DEF are two similar triangles.

AP and DM are respectively the medians of sides BC and EF.

To prove: \(\frac{a r(\triangle \mathrm{ABC})}{a r(\Delta \mathrm{DEF})}\) = \(\frac{A P^{2}}{D M^{2}}\)

Proof: △ABC ~ △DEF (Given)

∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BC}}{\mathrm{EF}}\) = \(\frac{\mathrm{2BP}}{\mathrm{2EM}}\)

(∵ AP and DM are medians)

∴ BC = 2BP and EF = 2EM AB = BP

⇒ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BP}}{\mathrm{EM}}\) ….(1)

In △ABP and △DEM,

∠B = ∠E (∵ △ABC ~ △DEF)

\(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BP}}{\mathrm{EM}}\) [Proved in (i)]

∴ △ABP ~ △DEM (By SAS similarity criterion)

∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{AP}}{\mathrm{DM}}\) ….(ii)

Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides

∴ \(\frac{a r(\triangle \mathrm{ABC})}{a r(\triangle \mathrm{DEF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{AP}^{2}}{\mathrm{DM}^{2}}\)

(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 : ABCD is a square. Equilateral triangle has been described on side AB of the square and equilateral triangle ACF has been described on diagonals AC.

To prove : \(\frac{a r(\triangle \mathrm{ABE})}{a r(\triangle \mathrm{ACF})}\) = \(\frac{1}{2}\)

Proof: In right △ABC,

AB^{2} + BC^{2} = AC^{2}

(By Pythagoras Theorem)

or AB^{2} + AB^{2} = AC^{2}

(∵ AB = BC, sides of the same square)

2AB^{2} = AC^{2} ………(i)

Now, each of △ABE and △ACF is equilateral, therefore is equiangular and therefore similar,

or △ABE ~ △ACF

Here any side of first triangle is parallel to any side of the other triangle.

∴ \(\frac{a r(\triangle \mathrm{ABE})}{a r(\triangle \mathrm{ACF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)

[∵ The ratio of the areas of non-similar triangles is equal to the square of the ratio of their corresponding sides]

⇒ \(\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}\) = \(\frac{1}{2}\) [Using (i)]

Hence area of △ABE = \(\frac{1}{2}\)area of △ACF

(Hence Proved)

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 triangles ABC and BDE is

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Solution:

△ABC and △BDE are two equilateral triangles such that D is the mid-point of the side BC

∴ BD = DC = \(\frac{1}{2}\) BC

or BC : BD = 2 : 1

Let the side of the equilateral triangle be 2a. △ABC ~ △BDE

Since in an equilateral triangle angles are the same.

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:

△ABC ~ △DEF (Given)

∴ Correct choice = (D)

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