Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Answer 1

SOLUTION :  

Given:

Two isosceles ∆s have equal vertical angles and their areas are in the ratio of 36: 25.

Let the two isosceles triangles be  ΔABC and ΔPQR with ∠A=∠P.

Therefore,

AB/AC = PQ/PR

In ΔABC and ΔPQR,

∠A = ∠P   (given)

AB/AC = PQ/ PR  (sides of a isosceles∆)

∴ ΔABC – ΔPQR    (By SAS similarity)

Let AD and PS be the altitudes of ΔABC and ΔPQR.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

arΔABC/arΔPQR = (AD/PS)²

36/25 = (AD/PS)²

√36/25 = √(AD/PS)²

[On taking square root both sides]

AD/ PS = 6/5

Hence, the ratio of their corresponding heights is 6: 5.

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Answer 2

$<b>$ Solution

Let in ABC

AB = AC

B = C = 180- Angle A/ 2

And in PQR

PQ = QR

Q= R = 180- Angle P / 2

Given verticle angle of two angle are equal

A= P

B= C = Q= R

ABC congruent to PQR ( by AAA criteria)


We know that area of two similar triangles in the ratio is same

as the ratio between the squares of their corresponding altitudes and corresponding height of two given triangle are AD and PS


Area of ( ABC) / Area of ( PQR) = Ad²/ ps²

36² / 25²= Ad² / Ps²

= 6/5

AD : PS =6:5


corresponding height is 6:5