Question

The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm. find the longest side of the smaller triangle.

Answer 1

It is given that the triangles are similar.
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Let the longest side of smaller triangle be x cm.

^{$$\begin{gathered} \frac{\mathrm{ar}(\mathrm{Larger}\mathrm{triangle})}{\mathrm{ar}(\mathrm{Smaller}\mathrm{triangle})}=\frac{(\mathrm{Longest}\mathrm{side}\mathrm{of}\mathrm{larger}\mathrm{traingle}{)}^2}{(\mathrm{Longest}\mathrm{side}\mathrm{of}\mathrm{smaller}\mathrm{traingle}{)}^2} \\ \Rightarrow \frac{169}{121}=\frac{{26}^2}{x^2} \\ \Rightarrow x=\sqrt{\frac{26\times 26\times 121}{169}} \\ =22 \end{gathered}$$}

Hence, the longest side of the smaller triangle is 22 cm.