Question

The areas of two similar triangles are $121c{m}^2$ and $81c{m}^2$ 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 $\frac{11}{9}$

Given the areas of two similar triangles are $121$ sq. cm and $81$ sq. cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding heights.

The ratio of area of triangles $=\frac{121}{81}=\frac{(11{)}^2}{(9{)}^2}$ $=$ $\frac{(h_1{)}^2}{(h_2{)}^2}$

Then ratio of heights of triangle $=\sqrt{{\left[\frac{11}{9}\right]}^2}=\frac{11}{9}$