Question

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Answer 1

SOLUTION :  

Given :  Area of two similar triangles is 25cm² and 36cm² . Altitude of first ∆ =  2.4cm.

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

ar(∆1)ar(∆2) = (altitude 1/ altitude 2)²

25/36 = (2.4/altitude 2)²

On taking square root on both sides,

√25/36 =√(2.4/altitude 2)²

⅚ = 2.4/ altitude 2

5 × altitude 2 = 6 × 2.4

Altitude 2 = (6 × 2.4)/5

Altitude 2 = 14.4 /5 cm

Altitude 2 = 2.88 cm

Hence, the corresponding altitude of the other ∆ is 2.88 cm.

HOPE THIS ANSWER WILL HELP YOU....

Answer 2

We know that when any two triangles are proved to be similar to each other then the ratio of their areas become equal to the ratio of squares of their altitudes.

Given,

Area of 1st triangle, A = 25 cm^2
Area of 2nd triangle, B = 36 cm^2

Also,

Altitude of 1st triangle, C = 2.4 cm
Altitude of 2nd triangle, D = ?

Since, the area and the altitude are already in similar units so we do not need to perform any conversions.

Now since the triangles are given similar,

A / B = C / D
25 / 36 = ( 2.4 / D ) ^ 2

On square rooting LHS and RHS,

5 / 6 = 2.4 / D

On cross - multiplying,

5D = 14.4
D = 14.4 / 5

On solving we get,

D = 2.88 cm

Altitude of 2nd triangle = 2.88 cm