The areas of two similar triangles are 25 cm2 and 36 cm2 respectively.

Question:

The areas of two similar triangles are $25 \mathrm{~cm}^{2}$ and $36 \mathrm{~cm}^{2}$ respectively. If the altitude of the first triangle is $2.4 \mathrm{~cm}$, find the corresponding altitude of the other.

 

Solution:

Given: The area of two similar triangles is 25cm2 and 36cm2 respectively. If the altitude of first triangle is 2.4cm

To find: The altitude of the other triangle 

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

$\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\text { altitude } 1}{\text { altitude } 2}\right)^{2}$

$\frac{25}{36}=\left(\frac{2.4}{\text { altitude2 }}\right)^{2}$

Taking square root on both sides, we get

$56=2.4$ altitude $2 \Rightarrow$ altitude $2=2.88 \mathrm{~cm}$

Hence, the corresponding altitude of the other is 2.88 cm.

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