Question

The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.

Answer 1

SOLUTION :  

Given :  Area of two similar triangles is 100 cm² and 49 cm² . Altitude of bigger ∆ =  5 cm.

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

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

100/49 = (5 /altitude 2)²

On taking square root on both sides,

√100 /49 =√(5 /altitude 2)²

10/7  = 5 / altitude 2

10 × altitude 2 = 5 × 7

10 × altitude 2 = 35

Altitude 2 = 35/10

Altitude 2 = 3.5 cm

Hence, the corresponding altitude of the other ∆ is 3.5 cm.

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Answer 2

We are provided with enough details in order to find the answer :)

Two triangles are said to be similar if their sides are divided in same ratio. Few postulates are followed to prove two triangles similar.

Given,

The areas of two triangles = 100 cm^2 and 49 cm^2

Area of bigger triangle, A = 100 cm^2
Area of smaller triangle, B = 49 cm^2
Altitude of bigger triangle, C = 5 cm
Altitude of smaller triangle, D = ?

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

A / B = ( C / D ) ^ 2
100 / 49 = ( 5 / D ) ^ 2

On calculating square root of LHS & RHS,

10 / 7 = 5 / D

On cross - multiplying,

10D = 35
D = 35 / 10
D = 3.5 cm

Therefore,

Altitude of smaller triangle = 3.5 cm