The areas of two similar triangles are $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. If the longest side of the larger triangle is $26 \mathrm{~cm}$, find the longest side of the smaller triangle.
Given: The area of two similar triangles is $169 \mathrm{~cm}^{2}$ and $121 \mathrm{~cm}^{2}$ respectively. The longest side of the larger triangle is $26 \mathrm{~cm}$.
To find: Longest side of the smaller triangle
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
$\frac{a r(\text { larger triangle })}{a r(\text { smaller triangle })}=\left(\frac{\text { side of the larger triangle }}{\text { side of the smaller triangle }}\right)^{2}$
$\frac{169}{121}=\left(\frac{\text { side of the larger triangle }}{\text { side of the smaller triangle }}\right)^{2}$
Taking square root on both sides, we get
$\frac{13}{11}=\frac{\text { side of the larger triangle }}{\text { side of the smaller triangle }}$
$\frac{13}{11}=\frac{26}{\text { side of the smaller triangle }}$
side of the smaller triangle $=\frac{11 \times 26}{13}=22 \mathrm{~cm}$
Hence, the longest side of the smaller triangle is $22 \mathrm{~cm}$.