# Q6.    An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

The perimeter of an isosceles triangle is 30 cm (Given).

The length of the sides which are equal is12 cm.

Let the third side length be 'a cm'.

Then, $Perimeter = a+b+c$

$\Rightarrow 30= a+12+12$

$\Rightarrow a = 6cm$

So, the semi-perimeter of the triangle is given by,

$s= \frac{1}{2}Perimeter =\frac{1}{2}\times30cm = 15cm$

Therefore, using Herons' Formula, calculating the area of the triangle

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

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

$= \sqrt{15(9)(3)(3)}$

$= 9\sqrt{15}\ cm^2$

Hence, the area of the triangle is $9\sqrt{15}cm^2.$

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