# NCERT solutions for class 10 Maths chapter 6 Triangles (Exercise 6.4)

## Ex 6.4 Question 1.

Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Solution:

Given that, ΔABC ~ ΔDEF,

ar.(ΔABC)/ar.(ΔDEF) = AB²/DE² = BC²/EF² = AC²/DF²

Area of ΔABC = 64 cm2

Area of ΔDEF = 121 cm2

EF = 15.4 cm

∴ ar.(ΔABC)/ar.(ΔDEF) = BC²/EF²

∴ 64/121 = BC2/EF2

⇒ (8/11)2 = (BC/15.4)2

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

## Ex 6.4 Question 2.

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

Solution:

Given, AB || DC.

∠OAB = ∠OCD & ∠OBA = ∠OCD    [Alternate angles]

In ΔAOB and ΔCOD,

∠AOB = ∠COD        [Vertically opposite angles]

∠OAB = ∠OCD        [Alternate angles]

∠OBA = ∠ODC        [Alternate angles]

∴ ΔAOB ~ ΔCOD     [AAA similarity criterion]

∴ Area of (ΔAOB)/Area of (ΔCOD) = AB2/CD2

= (2CD)2/CD2 [∴ AB = 2CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD2/CD2 = 4/1 = 4:1

## Ex 6.4 Question 3.

In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Solution:

Let AP and DM are the perpendiculars on line BC.

Area of a triangle = 1/2 × Base × Height

∴ ar(ΔABC)/ar(ΔDBC) = ½BC x AP/ ½BC x DM  = AP/DM  ……(i)

In ΔAPO and ΔDMO,

∠APO = ∠DMO [Each 90°]

∠AOP = ∠DOM [Vertically opposite angles]

∴ ΔAPO ~ ΔDMO [AA similarity criterion]

∴ AP/DM = AO/DO

⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.   [From equation (i)]

## Ex 6.4 Question 4.

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

Solution:

Let, ΔABC ~ ΔPQR

∴ ar(ΔABC)/ar(ΔPQR) = BC2/QR2

⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)]

⇒ BC2/QR2

⇒ BC = QR

Similarly,

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS congruencecy]

## Ex 6.4 Question 5.

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Solution:

In ΔABC,

F is the mid-point of AB [Given]

E is the mid-point of AC [Given]

By mid-point theorem,

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

[Opposite sides of the parallelogram are equal and parallel]

∴ BDEF is parallelogram.

Similarly, in ΔFBD and ΔDEF,

FB = DE [Opposite sides of parallelogram BDEF]

FD = FD [Common]

BD = FE [Opposite sides of parallelogram BDEF]

∴ ΔFBD ≅ ΔDEF

Similarly,

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

[when triangles are congruent, they are equal in the area.]

ar(ΔFBD) = ar(ΔDEF) ……………(i)

ar(ΔAFE) = ar(ΔDEF) ………….(ii)

and,

ar(ΔEDC) = ar(ΔDEF) ………….(iii)

Now,

ar(ΔABC) = ar(ΔFBD) + ar(ΔDEF) + ar(ΔAFE) + ar(ΔEDC) ………(iv)

ar(ΔABC) = ar(ΔDEF) + ar(ΔDEF) + ar(ΔDEF) + ar(ΔDEF)

From equation (i), (ii) and (iii),

⇒ ar(ΔDEF) = (1/4)ar(ΔABC)

⇒ ar(ΔDEF)/ar(ΔABC) = 1/4

Hence, ar(ΔDEF): ar(ΔABC) = 1:4

## Ex 6.4 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:  ΔABC ~ ΔDEF.

To prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2

Since, ΔABC ~ ΔDEF (Given)

∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) …………(i)

and, AB/DE = BC/EF = CA/FD ………………(ii)

⇒AB/DE = ½BC/½EF = CD/FD

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

∴ ∠B = ∠E

AB/DE = BM/EN [Proved above]

∴ ΔABC ~ ΔDEF [SAS similarity]

⇒ AB/DE = AM/DN …………..(iii)

∴ ΔABM ~ ΔDEN

[As the areas of two similar triangles are proportional to the squares of the corresponding sides.]

∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2

## Ex 6.4 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:

Area(ΔBQC) = ½ Area(ΔAPC)

ΔAPC and ΔBQC are both equilateral triangles,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2

Diagonal = √2 side = √2 BC = AC

(√2 BC/BC)2 =2

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = 1/2area(ΔAPC)

## Ex 6.4 Question 8.

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4

Solution:

Both the triangles are equilateral and each angle of both the triangles are 60°.

By AAA similarity, ΔABC ~ ΔBDE

The side of ΔBDE = x/2

∴ ar(ΔABC)/ar(ΔBDE) = AB2/BD2 = (x/x/2)2= 4/1 = 4:1

Hence, (C) is the correct answer.

## Ex 6.4 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:-

The ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.

The ratio of the area of two triangles = (4/9)= 16/81 = 16:81

Hence, (D) is the correct answer.

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