Question

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

• A. $3cm$
• B. $2.5cm$
• C. $4cm$
• D. $3.5cm$

Answer 1

The correct option is D $3.5cm$

Given: Area of $two$ similar triangle $81{cm}^2$ and $49{cm}^2$

Altitude of bigger triangle $=4.5cm$

For similar triangle,

${\text{Ratio on sides}}^2=\text{Ratio of their Area}$

$\therefore$ $\frac{{4.5}^2}{x^2}=\frac{81}{49}$

$\frac{45\times 45}{x^2\times 100}=\frac{81}{49}$

$100x^2=25\times 49$

$x^2=\frac{25\times 49}{100}$

$x^2=\frac{49}{4}$

$x=\frac{7}{2}$

$x=3.5cm$