# The areas of two similar triangles are 100 cm2 and 49 cm2 respectively.

Question:

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

Solution:

Given: The area of two similar triangles is $100 \mathrm{~cm}^{2}$ and $49 \mathrm{~cm}^{2}$ respectively. If the altitude of bigger triangle is $5 \mathrm{~cm}$

To find: their corresponding altitude of other triangle

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

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

$\frac{100}{49}=\left(\frac{5}{\text { altitude2 }}\right)^{2}$

Taking square root on both side

$\frac{10}{7}=\frac{5}{\text { altitude } 2}$

altitude $2=3.5 \mathrm{~cm}$