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The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.

Question:

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.

 

Solution:

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 $22 \mathrm{~cm}$.

 

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