The areas of two similar triangles are $100\ cm^2$ and $49\ cm^2$ respectively. If the altitude of the bigger triangles is $5\ cm$, find the corresponding altitude of the other.

Given:

The areas of two similar triangles are $100\ cm^2$ and $49\ cm^2$ respectively.

The altitude of the bigger triangles is $5\ cm$.

To do:

We have to find the corresponding altitude of the other triangle.

Solution:

We know that,

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

Therefore,

$\frac{ar(bigger\ triangle)}{ar(smaller\ triangle)} = (\frac{altitude\ of\ the\ bigger\ triangle}{altitude\ of\ the\ smaller\ triangle})^2$

$\frac{100}{49} = (\frac{5}{altitude\ of\ the\ smaller\ triangle})^2$

$\frac{altitude\ of\ the\ smaller\ triangle}{5} = \sqrt{\frac{49}{100}}$

$altitude\ of\ the\ smaller\ triangle=\frac{7\times5}{10}$

$altitude\ of\ the\ smaller\ triangle=3.5$

The altitude of the other triangle is $3.5\ cm$.

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