Question

One side of a triangle is 15 in., and the area of the triangle is 90 sq. in. Find the area of a smaller triangle in which the corresponding side is 9 in.

Answer 1

Area (triangle) = 90 sq. in 
Side = 15 in.

Let the given side be AB

{Only if the smaller triangle is similar to the larger triangle, this is possible)

Let the given side of the smaller triangle be PQ.
PQ = 9 in. (Given)

$\frac{area (large-triangle)}{area (small-triangle)}$ = $\frac{AB^2}{PQ^2}$ (Theorem)

$\frac{90}{area (small-triangle)} = \frac{15^2}{9^2}$

$\frac{90}{area (small-triangle)} = \frac{225}{81}$

$Area (Small-triangle) = \frac{90 * 81}{225}$

= 32.4 sq. in