The areas of two similar triangles are 100 cm2 and 49 cm2 respectively

1.	The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude of the bigger triangles is 5 cm, find the corresponding altitude of the other.
Answer» Given: The area of the two similar triangles is 100 cm² and 49 cm². And the altitude of the bigger triangle is 5 cm. Required to find: The corresponding altitude of the other triangle We know that, The ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding altitudes. ar(bigger triangle)/ ar(smaller triangle) = (altitude of the bigger triangle/ altitude of the smaller triangle)² (100/ 49) = (5/ altitude of the smaller triangle)² Taking square root on LHS and RHS, we get (10/ 7) = (5/ altitude of the smaller triangle) = 7/2 Therefore, altitude of the smaller triangle = 3.5 cm

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

Answer»

Given: The area of the two similar triangles is 100 cm² and 49 cm². And the altitude of the bigger triangle is 5 cm.

Required to find: The corresponding altitude of the other triangle

We know that,

The ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding altitudes.

ar(bigger triangle)/ ar(smaller triangle) = (altitude of the bigger triangle/ altitude of the smaller triangle)²

(100/ 49) = (5/ altitude of the smaller triangle)²

Taking square root on LHS and RHS, we get

(10/ 7) = (5/ altitude of the smaller triangle) = 7/2

Therefore, altitude of the smaller triangle = 3.5 cm