Question

The area of two similar triangle are 100 $c{m}^2$ and 49 $c{m}^2$ respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other.

Answer 1

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

The area of two similar triangle are 1$00c{m}^2$ and $49c{m}^2$respectively.

The altitude of the bigger triangle is $5cm$

Let the altitude of the smaller triangle be $x$.

Therefore $\sqrt{\frac{100}{49}}=\frac{5}{x}$

$\frac{10}{7}=\frac{5}{x}$

$x=3.5cm$

The altitude of the smaller triangle is $3.5cm$.