Course Content
Class 10th Maths
0/86
Online Class For 10th Standard Students (CBSE) (English Medium)
About Lesson

Exercise 6.4 Page: 143

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

Solution: Given, ΔABC ~ ΔDEF,

Area of ΔABC = 64 cm2

Area of ΔDEF = 121 cm2

EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,

= AC2/DF2 = 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

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,ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

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

= (2CD)2/CD2 [∴ AB = 2CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD2/CD2 = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

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:

Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.

We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 × Base × Height

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.

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

Solution:

SayΔABC and ΔPQR are two similar triangles and equal in area

Now let us prove ΔABC ≅ ΔPQR.

Since, ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2

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

⇒ BC2/QR2

⇒ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

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:

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

In ΔABC,

F is the mid-point of AB (Already given)

E is the mid-point of AC (Already given)

So, by the mid-point theorem, we have,

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

Since, opposite sides of parallelogram are equal and parallel

∴ BDEF is parallelogram.

Similarly, in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

∴ ΔFBD ≅ ΔDEF

Similarly, we can prove that

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

As we know, if triangles are congruent, then they are equal in area.

So,

Area(ΔFBD) = Area(ΔDEF) ……………………………(i)

Area(ΔAFE) = Area(ΔDEF) ……………………………….(ii)

and,

Area(ΔEDC) = Area(ΔDEF) ………………………….(iii)

Now,

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

Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF)

From equation (i)(ii) and (iii),

⇒ Area(ΔDEF) = (1/4)Area(ΔABC)

⇒ Area(ΔDEF)/Area(ΔABC) = 1/4

Hence, Area(ΔDEF): Area(ΔABC) = 1:4

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: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.

We have 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)

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

∴ ∠B = ∠E

AB/DE = BM/EN [Already Proved in equation (i)]

∴ ΔABC ~ ΔDEF [SAS similarity criterion]

⇒ 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

Hence, proved.

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 whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(ΔBQC) = ½ Area(ΔAPC)

Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

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

Since, Diagonal = √2 side = √2 BC = AC

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = 1/2area(ΔAPC)

Hence, proved.

Tick the correct answer and justify:

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:

GivenΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2a.

As, ΔABC ~ ΔBDE

∴ Area(ΔABC)/Area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1

Hence, the correct answer is (C).

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:

Given, Sides of two similar triangles are in the ratio 4 : 9.

Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)= 16/81 = 16:81

Hence, the correct answer is (D).

Exercise Files
No Attachment Found
No Attachment Found
Wisdom Academy