Question

Find the area of an isosceles triangle whose perimeter is $36$ cm and base is $16$ cm.

• A. $40$ sq. cm
• B. $42$ sq. cm
• C. $46$ sq. cm
• D. $48$ sq. cm

Answer 1

The correct option is D $48$ sq. cm

In isosceles triangle,$a=a=c$, two sides equal, base $=c$
$\therefore$ perimeter $=36=a+a+16,36-16=2a,a=10$
$a=10,b=10,c=16$

$s=\frac{10+10+16}{2}=\frac{36}{2}=18$

$\therefore$ Area of triangle=$\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{18(18-10)(18-10)(18-16)}=\sqrt{2304}=48sq.cm$