Question

The areas of two similar triangles are $12c{m}^2$ and $48c{m}^2$. If the height of the smaller one is $2.1cm,$ then the corresponding height of the bigger one is

• A. $4.41cm$
• B. $8.4cm$
• C. $4.2cm$
• D. $0.525cm$

Answer 1

The correct option is C

$4.2cm$

Given, $\frac{\text{Area of smaller triangle}}{\text{Area of bigger triangle}}=\frac{12}{48}$

Let height of the bigger triangle be x.

We know that

$\frac{\text{Area of smaller triangle}}{\text{Area of bigger triangle}}=\frac{{\text{(Height of smaller triangle)}}^2}{{\text{(Height of bigger triangle)}}^2}$

$\Rightarrow \frac{{\text{(Height of smaller triangle)}}^2}{{\text{(Height of bigger triangle)}}^2}=\frac{12}{48}=\frac{1}{4}$

$\Rightarrow \frac{(2.1{)}^2}{x^2}=\frac{1}{4}$

$\Rightarrow x=\sqrt{4\times (2.1{)}^2}=2\times 2.1=4.2cm$

Therefore, height of the bigger triangle is 4.2 cm.