Question

Two isosceles triangle have equal Vertical angle and their area are in the ratio 36:25. find the distance of their corresponding height.​

Answer 1

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

= 36²/25² = AD²/PS²

AD : PS = 6 : 5 ( ratio of their corresponding heights).