Question

The areas of two similar triangle are 81 cm and 49 cm respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer 1

$\frac{A_1}{A_2}=\frac{\frac{1}{2}\times b_1\times h_1}{\frac{1}{2}\times b_2\times h_2}=\frac{b_1{h}_1}{b_2{h}_2}$

Since, the triangles are similar their bases & heights will be proportional

$\frac{b_1}{b_2}=\frac{h_1}{h_2}=k$

$\frac{A_1}{A_2}=\frac{81}{49}=\left(k\right)\left(k\right)$

$k^2=\frac{81}{49}$

$\Rightarrow k=\frac{9}{7}$

$\therefore \frac{h_1}{h_2}=\frac{9}{7}$

$\Rightarrow \frac{4.5cm}{h_2}=\frac{9}{7}$ ($h_1\to$ altitude of bigger triangle, $h_2\to$ altitude of smaller triangle)

$\Rightarrow h_2=\frac{7\left(4.5\right)}{9}=7\left(0.5\right)=3.5cm$