Class 10 Mathematics

# Triangle

## Exercise 6.4 (NCERT)

Question 1: Let Δ ABC ∼ Δ DEF and their areas be, respectively, 64 sq cm and 121 sq m. If EF = 15.4 cm, find BC.

Solution: In Δ ABC ∼ Δ DEF;

text(ar ABC)/text(ar DEF)=(BC^2)/(EF^2)

Or, (BC^2)/(15.4^2)=(64)/(121)

Or, (BC)/(15.4)=(8)/(11)

Or, BC=11.2 cm

Question 2: Diagonals of a trapezium ABCD with AB || CD intersect each other at point O. if AB = 2 CD, find the ratio of the areas of triangles AOB and COD. Solution: In triangles AOB and COD;

∠ AOB = ∠ COD (Opposite angles)

∠ OAC = ∠ OCD (Alternate angles)

Hence; Δ AOB ∼ Δ COD

Hence;

text(ar AOB)/text(ar COD)=(AB^2)/(CD^2)

=((2CD)^2)/(CD^2)=(4CD^2)/(CD)^2=4:1

Question 3: In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that text(ar ABC)/text(ar DBC)=(AO)/(DO) Solution: Let us draw altitudes AM and DN on BC; respectively from A and D

text(ar ABC)/text(ar DBC)=(1/2xxBCxxAM)/(1/2xxBCxxDN)

=(AM)/(DN)

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ∼ ΔDNO

Hence;

(AM)/(DN)=(AO)/(DO)

Or, text(ar ABC)/text(ar DBC)=(AO)/(DO)

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

Solution: Let us take two triangles ABC and PQR with equal areas.

Then, we have;

text(ar ABC)/text(ar PQR)=1/1

In this case;

(AB^2)/(PQ^2)=(AC^2)/(PR^2)=1/1

Or, (AB)/(PQ)=(AC)/(PQ)=1

Hence; the triangles are congruent.

Question 5: D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC. Solution: Since D, E and F are mid points of AB, BC and AC

Hence; ΔBAC ∼ΔDFE

So,

(DF)/(BC)=(EF)/(AB)=(DE)/(AC)=1/2

So,

text(ar DEF)/text(ar ABC)=1^2/2^2=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: In case of two similar triangles ABC and PQR;

text(ar ABC)/text(ar PQR)=(AB^2)/(PQ^2)=(AC^2)/(PR^2)

Let us assume AD and PM are the medians of these two triangles.

Then;

(AB^2)/(PQ^2)=(AC^2)/(PR^2)=(AD^2)/(PM^2)

Or, text(ar ABC)/text(ar PQR)=(AD^2)/(PM^2)

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 us take a square with side ‘a’

Then the diagonal of square will be a\sqrt2

Area of equilateral triangle with side ‘a’

=sqrt3/4a^2

Area of equilateral triangle with side a\sqrt2

=sqrt3/4(a\sqrt2)^2

Ratio of two areas can be given as follows:

(sqrt3/4xx\a^2)/(sqrt3/4xx2a^2)=1/2

Proved

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

• 2 : 1
• 1 : 2
• 4 : 1
• 1: 4

Solution: (c) 4 : 1

For explanation; refer to question 5

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

• 2 : 3
• 4 : 9
• 81 : 16
• 16: 81

Solution: (d) 16 : 91

Note: Areas are in duplicate ratio of corresponding sides in case of similar figures.

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Theorem (Part 1)

Theorem (Part 2)

Exercise 6.1

Exercise 6.2 (Part 1)

Exercise 6.2 (Part 2)

Exercise 6.3 (Part 1)

Exercise 6.3 (Part 2)

Exercise 6.3 (Part 3)

Exercise 6.5 (Part 1)

Exercise 6.5 (Part 2)

Exercise 6.6 (Part 1)

Exercise 6.6 (Part 2)