The areas of two similar triangles are $169\ cm^2$ and $121\ cm^2$ respectively. If the longest side of the larger triangle is $26\ cm$, find the longest side of the smaller triangle.

Given:

The areas of two similar triangles are $169\ cm^2$ and $121\ cm^2$ respectively.

The longest side of the larger triangle is $26\ cm$.

To do:

We have to find the longest side of the smaller triangle.

Solution:

Let the longer side of the smaller triangle be $x$.

We know that,

The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\frac{ar(larger\ triangle)}{ar(smaller\ triangle)} = (\frac{\ side\ of\ the\ larger\ triangle}{\ side\ of\ the\ smaller\ triangle})^2$

                                                                                 $= \frac{169}{121}$

$(\frac{\ side\ of\ the\ larger\ triangle}{\ side\ of\ the\ smaller\ triangle})=\sqrt{\frac{169}{121}}$

                $= \frac{13}{11}$

Sides of similar triangles are proportional, this implies,

$\frac{(longer\ side\ of\ the\ larger\ triangle)}{(longer\ side\ of\ the\ smaller\ triangle)}=\frac{13}{11}$

$\frac{26}{x}=\frac{13}{11}$

$x = 2\times11$

$x=22\ cm$

The longest side of the smaller triangle is $22\ cm$.

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Updated on: 10-Oct-2022

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