Question

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

$\frac{15\sqrt{7}}{4}cm$

Side of a triangle are 11 cm, 15 cm, 16 cm
$\therefore s=\frac{a+b+c}{2}=\frac{11+15+16}{2}=\frac{42}{2}=21$
$\therefore Area=\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{21(21-11)(21-15)(21-16)}$
= $\sqrt{21\times 10\times 6\times 5}$
= $2\times 3\times 5\sqrt{7}=30\sqrt{7}c{m}^2$
Now length of altitude to the longest side,
= $\frac{Area\times 2}{base}=\frac{30\sqrt{7}\times 2}{16}$
= $\frac{15\sqrt{7}}{4}cm$