Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio 81:25. find the ratio of their corresponding heights.

Answer 1

Solution:-
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding heights.
Ratio of the areas of two similar triangles = 81 : 25
or ratio of the squares of the corresponding heights of these two triangles = 81 : 25                                                           ______
So, ratio the squares of their corresponding heights = √(81/25)
= 9/5
The ratio = 9 : 5  Answer

Answer 2

Let in Δ ABC with height as AD,

AB = AC

⇒ ∠B = ∠C = 180-∠A/2...... Equation-1

And in Δ PQR with height as PS

PQ = PR

⇒ ∠Q = ∠R = 180-∠P/2...... Equation- 2

Given vertical angles of the two triangles are equal.

i.e., ∠A = ∠P

and ∠B = ∠C = ∠Q = ∠R

By AAA similarity criterion,

We know ratio between the areas of two similar triangle is same as the ratio between the square of their corresponding altitudes.

And corresponding heights of two given triangles are AD and PS.

Area of Δ ABC/Area of Δ PQR = AD²/PS²

= 81²/25² = AD²/PS²

AD : PS = 9 : 5 ( ratio of their corresponding heights).