Question

Vertical angles of two isosceles Triangles are equal. If there areas in the ratio 16: 25, then find the ratio of their altitudes drawn from vertex to the opposite side.
(Class 10 Maths Sample Question Paper)

Answer 1

FIGURE IS IN THE ATTACHMENT

Given :

∠A = ∠P

∠B = ∠C, ∠Q = ∠R

To Prove: let ∠A = ∠P = x

Proof:

In ∆ ABC
∠A + ∠B + ∠C = 180°
x° + ∠B +∠B = 180°
x° + 2∠B= 180°
2∠B= 180° - x
∠B= (180° - x)/2…….(1)

In ∆ PQR

∠P + ∠Q + ∠R = 180°
x° + ∠Q +∠Q = 180°
x° + 2∠Q= 180°
2∠Q = 180° - x
∠Q= (180° - x)/2…….(2)

In ∆ ABC & ∆PQR
∠A = ∠P    ( GIVEN)
∠B = ∠Q  ( From eq 1 & 2)
∆ABC ~ ∆ PQR  ( AA similarity)
ar(∆ABC)/ar(∆PQR) = AD²/ PE²

[ By theorem ratio of the areas of two similar triangles]

16/25 = AD²/PE²
√ 16/25 = AD /PE
⅘ = AD/PE
AD / PE = ⅘

Hence, the ratio of their altitudes drawn from vertex to the opposite side is ⅘.

HOPE THIS WILL HELP YOU.....

Answer 2

Answer:

Step-by-step explanation:

Given :

∠A = ∠P

∠B = ∠C, ∠Q = ∠R

To Prove: let ∠A = ∠P = x

Proof:

In ∆ ABC

∠A + ∠B + ∠C = 180°

x° + ∠B +∠B = 180°

x° + 2∠B= 180°

2∠B= 180° - x

∠B= (180° - x)/2…….(1)

In ∆ PQR

∠P + ∠Q + ∠R = 180°

x° + ∠Q +∠Q = 180°

x° + 2∠Q= 180°

2∠Q = 180° - x

∠Q= (180° - x)/2…….(2)

In ∆ ABC & ∆PQR

∠A = ∠P    ( GIVEN)

∠B = ∠Q  ( From eq 1 & 2)

∆ABC ~ ∆ PQR  ( AA similarity)

ar(∆ABC)/ar(∆PQR) = AD²/ PE²

[ By theorem ratio of the areas of two similar triangles]

16/25 = AD²/PE²

√ 16/25 = AD /PE

⅘ = AD/PE

AD / PE = ⅘

Hence, the ratio of their altitudes drawn from vertex to the opposite side is ⅘.