Question

Perimeter of an isosceles triangle is 30 cm and equal sides are of 12cm,then the area of triangle in sq.cm is​

Answer 1

$P = a + + c = 30\\a = b = 12\\c = 30 - 24 = 6$

Now, Area,

$A = \sqrt{s(s-a)(s-b)(s-c)} \\S = (a + b + c)/2 = 30/2 = 15\\\\A=\sqrt{15(15-12)(15-1@)(15-6)} \\\\A = \sqrt{15 * 3 * 3 * 9}\\ \\A = 9\sqrt{15}$

Therefore, Area of the Triangle is  $9\sqrt{15}$...

Answer 2

First,let the third side be x.

It is given that the length of the equal sides us 12 cm and it's perimeter is 30 cm.

So,$30=12+12+x$

$⇒ 30 = 24 + x$

$⇒24 + x = 30$

$⇒ x= 30 - 24$

$⇒ x = 6$

So,the length of the third side is 6 cm.

Thus,the semi perimeter of the isosceles triangle (s) = 30/2 cm =15 cm

By using Heron's Formula,

Area of the triangle,

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

$= \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \: {cm}^{2}$

$= \sqrt{15 \times 3 \times 3 \times 9} \: {cm}^{2}$

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