Question

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the longest side is.

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

Answer 1

The correct option is D

$\frac{15\sqrt{7}}{4}cm$
Semi-Perimeter , $s=\frac{11+15+16}{2}=21cm\text{Using heron's formula}Area=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21(21-11)(21-15)(21-16)}=\sqrt{21\times 10\times 6\times 5}=\sqrt{3\times 7\times 2\times 5\times 2\times 3\times 5}=\sqrt{7\times 3^2\times 2^2\times 5^2}=3\times 2\times 5\times \sqrt{7}\therefore Area=30\sqrt{7}c{m}^2$

$\text{Altitude}=\frac{2\times \text{Area of triangle}}{\text{base}}$
$\Rightarrow h=\frac{2\times 30\sqrt{7}}{16}=\frac{15\sqrt{7}}{4}cm$


So, option d is correct.