Question

If the perimeter of an isosceles triangle is $30$ cm and each of the equal sides is $12$ cm, then the area of the triangle is

• A. $3\sqrt{15}{\text{cm}}^2$
• B. $6\sqrt{15}{\text{cm}}^2$
• C. $9\sqrt{15}{\text{cm}}^2$
• D. $10\sqrt{15}{\text{cm}}^2$

Answer 1

The correct option is C $9\sqrt{15}{\text{cm}}^2$

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

Here $S$ is the semi- perimeter, and $a,b,c$ are sides of the triangle

The given triangle is isosceles.

In isosceles triangle, two sides are equal

so, $a=b=12$ cm

and perimeter $=30$ cm

semi perimeter = $\frac{perimeter}{2}$

$S=\frac{30}{2}$

$S=15$ cm

We find C,

perimeter $=30$ cm

$a+b+c=30$ cm

$\Rightarrow$ $12+12+c=30$ cm

$\Rightarrow$ $24+C=30$ cm

$\Rightarrow$ $C=30$ cm $-24$ cm

$\Rightarrow$ $C=6$ cm.

Area of triangle = $\sqrt{15(15-12)(15-12)(15-6)}$

$=\sqrt{\mathrm{15.3.3.9}}$

$=\sqrt{\mathrm{15.9.9}}$

$=\sqrt{9^2.15}$

$=9\sqrt{15}c{m}^2$

$\therefore$ Area of triangle = $9\sqrt{15}c{m}^2$

so, the answer is $C--9\sqrt{15}c{m}^2$