# Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.4

### Question 1. Let âˆ† ABC ~ âˆ† DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.

**Solution:**

According to the theorem 1, we get

BC = Ã— 15.4

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:**

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

In â–³AOB and â–³COD,

âˆ AOB = âˆ COD (

Opposite angles)âˆ 1 = âˆ 2 (

Alternate angles of parallel lines)

â–³AOB ~ â–³COD by AA property.According to the theorem 1, we get

As, AB = 2CD

=

=

=

ar(AOB) : ar(COD) = 4 : 1

### Question 3. In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that .

**Solution:**

Let’s draw two perpendiculars AP and DM on line BC.

Area of triangle = Â½ Ã— Base Ã— Height……………………………(1)

In Î”APO and Î”DMO,

âˆ APO = âˆ DMO (Each 90Â°)

âˆ AOP = âˆ DOM (Vertically opposite angles)

Î”APO ~ Î”DMO by AA similarity……………………………(2)

From (1) and (2), we can conclude that

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

**Solution:**

As it is given, Î”ABC ~ Î”DEF

According to the theorem 1, we have

=1 [Since, Area(Î”ABC) = Area(Î”DEF)

BC

^{2}= EF^{2}BC = EF

Similarly, we can prove that

AB = DE and AC = DF

Thus,

Î”ABC â‰… Î”PQR[SSS criterion of congruence]

### Question 5. D, E and F are respectively the mid-points of sides AB, BC, and CA of âˆ† ABC. Find the ratio of the areas of âˆ† DEF and âˆ† ABC.

**Solution:**

As, it is given here

DF = Â½ BC

DE = Â½ AC

EF = Â½ AB

So,

Hence, Î”ABC ~ Î”DEF

According to theorem 1,

ar(Î”DEF) : ar(Î”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: AM and DN are the medians of triangles ABC and DEF respectively.

Î”ABC ~ Î”DEF

According to theorem 1,

So,

……………………….(1)

âˆ B = âˆ E (because Î”ABC ~ Î”DEF)

Hence,

Î”ABP ~ Î”DEQ [SAS similarity criterion]……………………….(2)

From (1) and (2), we conclude that

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:**

Let’s take side of square = a

Diagonal of square AC = aâˆš2

As, Î”BCF and Î”ACE are equilateral, so they are similar

Î”BCF ~ Î”ACEAccording to theorem 1,

=

= 2

Hence,

Area of (Î”BCF) = Â½ Area of (Î”ACE)

### 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:**

Here,

AB = BC = AC = a

and, BE = BD = ED = Â½a

Î”ABC ~ Î”EBD (

Equilateral triangle)According to theorem 1,

Area of (Î”ABC) : Area of (Î”EBD) = 4 : 1

Hence, OPTION (C) is correct.

### 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:**

Î”ABC ~ Î”DEF

According to theorem 1,

Area of (Î”ABC) : Area of (Î”DEF) = 16 : 81

Hence, OPTION (D) is correct.

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