Question

Base of a triangle is $9$ and height is $5$. Base of another triangle is $10$ and height is $6$. Find the ratio of areas of these triangles.

Answer 1

Let $ABC$ and $PQR$ be two right triangles with $AB\perp BC$ and $PQ\perp QR$.
Given:
$BC=9,AB=5,PQ=6andQR=10$
$\therefore \frac{\text{Area}△ABC}{\text{Area}△PQR}=\frac{\frac{1}{2}\times AB\times BC}{\frac{1}{2}\times PQ\times QR}=\frac{5\times 9}{6\times 10}=\frac{3}{4}$

Hence, the required ratio is $3:4$