Question

Calculate the area of the triangle whose sides are $18$ cm, $24$ cm and $30$ cm in length. Also find the length of the altitude corresponding to the smallest side.

• A. $216{\text{cm}}^2,24$ cm
• B. $220{\text{cm}}^2,25$ cm
• C. $225{\text{cm}}^2,25$ cm
• D. $168{\text{cm}}^2,19$ cm

Answer 1

The correct option is A $216{\text{cm}}^2,24$ cm

According to the Heron's formula, Area (A) of the triangle having sides $a,b,c$ units is

$A=\sqrt{s(s-a)(s-b)(s-c)}$

where

$s=\frac{a+b+c}{2}$

For the given triangle,

$a=18$ cm

$b=24$ cm

$c=30$ cm

$s=\frac{18+24+30}{2}=36$

$A=\sqrt{36(36-18)(36-24)(36-30)}$

$A=\sqrt{36\times 18\times 12\times 6}$

$A=\sqrt{216\times 216}$

$=216$ $c{m}^2$

Smallest side $=18$ cm

Area of the triangle =$\frac{1}{2}\times base\times altitude=216$

$\frac{1}{2}\times 18\times$ altitude$=216$

Altitude $=\frac{216}{9}=24$ cm