Question

The areas of two similar triangles are $81c{m}^2$and $49c{m}^2$ respectively. If the altitude of the bigger triangle is $4.5cm$, find the corresponding altitude of the smaller triangle.

Answer 1

Step 1: Note the given data and draw the diagram

Let the bigger and smaller triangles are $△ABC,△DEF$ respectively and $AP\perp BC,DQ\perp EF$ .

The area of two similar triangles is $81c{m}^2$and $49c{m}^2$.

The altitude of the bigger triangle is $4.5cm$.

Step 2: Finding the altitude of the smaller triangle

Let the altitude of the smaller triangle, $DQ=xcm$

Relation between areas and altitude of similar triangles: The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Since both the triangles are similar then

$\frac{Area∆ABC}{Area∆DEF}=\frac{A{P}^2}{D{Q}^2}$

$$\begin{gathered} \Rightarrow \frac{81}{49}=\frac{{\left(4.5\right)}^2}{x^2} \\ \Rightarrow 81\times x^2={\left(4.5\right)}^2\times 49 \\ \Rightarrow x^2=\frac{{\left(4.5\right)}^2\times 49}{81} \\ \Rightarrow x=\sqrt{\frac{{\left(4.5\right)}^2\times 49}{81}} \\ \Rightarrow x=\frac{4.5\times 7}{9} \\ \Rightarrow x=3.5cm \end{gathered}$$

Hence, the altitude of the smaller triangle is $3.5cm$.