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

Given the sides of triangle is 11 cm ,15 cm and 16 cm

Then $s=\frac{11+15+16}{2}=21$

Then area of triangle= $\sqrt{21(21-11)(21-15)(21-16)}=\sqrt{21\times 10\times 6\times 5}=\sqrt{6300}=30\sqrt{7}$ sqcm

But area of triangle =$\frac{1}{2}\times base\times height$

$\frac{1}{2}\times 16\times height\left(altitude\right)=30\sqrt{7}\Rightarrow altitude=\frac{60}{16}\sqrt{7}=\frac{15}{4}\sqrt{7}$ cm