Question

If the altitudes of triangle are $10cm,12cm$ and $15cm$ then its semi perimeter is :

• A. $\frac{45}{\sqrt{7}}cm$
• B. $\frac{7}{\sqrt{2}}cm$
• C. $\frac{15}{\sqrt{14}}cm$
• D. $\frac{60}{\sqrt{7}}cm$

Answer 1

Answer: (D)

$\frac{60}{\sqrt{7}}cm$

Let the triangle be $ABC$.
According to the question
$ar\left(△ABC\right)=A=5a=6b=\frac{15}{2}c$
$\therefore a=\frac{A}{5},b=\frac{A}{6},c=\frac{2A}{15}$
$s=\frac{A}{4}$
$A^2=s(s-a)(s-b)(s-c)$
$\Rightarrow A^2=\frac{A}{4}(\frac{A}{4}-\frac{A}{5})(\frac{A}{4}-\frac{A}{6})(\frac{A}{4}-\frac{2A}{15})$
$\Rightarrow A^2={\left(\frac{240}{\sqrt{7}}\right)}^2$
$\Rightarrow A=\frac{240}{\sqrt{7}}$
$\therefore s=\frac{240}{4\sqrt{7}}=\frac{60}{\sqrt{7}}$.