Question

The areas of two similar triangles are 121 $c{m}^2$ and 81 $c{m}^2$ respectively. Find the ratio of their corresponding heights.

• A. $\frac{11}{9}$
• B. $\frac{10}{9}$
• C. $\frac{9}{11}$
• D. $\frac{9}{10}$

Answer 1

The correct option is A $\frac{11}{9}$

Since triangles are similar, so the ratio of square root of areas will be ratio of base $=\frac{\sqrt{121}}{\sqrt{81}}=\frac{11}{9}=\frac{b_1}{b_2}\left(Let\right)$

Let ratio of heights be $\frac{h_1}{h_2}$.

Now, $\frac{A_1}{A_2}=\frac{\frac{1}{2}{b}_1{h}_1}{\frac{1}{2}{b}_2{h}_2}=\frac{b_1}{b_2}\times \frac{h_1}{h_2}=\frac{h_1}{h_2}=\frac{11}{9}$