Question

The sides of a triangle are 11 cm 15 cm and 16 cm The altitude to largest side is

• A. $30\sqrt{7}$ cm
• B. $\frac{15\sqrt{7}}{2}$ cm
• C. $\frac{15\sqrt{7}}{4}$ cm
• D. 30 cm

Answer 1

The correct option is C $\frac{15\sqrt{7}}{4}$ cm

$\Rightarrow$ Sides of triangle are $11,15$ and $16$.

$\Rightarrow$ By Heron's formula semi-perimeter of triangle = $\frac{11+15+16}{2}=\frac{42}{2}=21cm$

$\Rightarrow$ $Area=\sqrt{21(21-11)(21-15)(21-16)}$

$\Rightarrow$ $Area=\sqrt{21\times 10\times 6\times 5}$

$\Rightarrow$ $Area=\sqrt{6300}=\sqrt{900\times 7}=30\sqrt{7}$

$\Rightarrow$ We have to fine altitude on largest side of triangle i.e 16

$\therefore$ $\frac{1}{2}\times b\times h=30\sqrt{7}$

$\therefore$ $\frac{1}{2}\times 16\times h=30\sqrt{7}$

$\therefore$ $h=\frac{30\sqrt{7}}{8}$

$\therefore$ $h=\frac{15\sqrt{7}}{4}cm$