Question

Find the area of an isosceles triangle each of whose equal sides measures 9 cm and whose base is 12 cm.
18√3
18√5
18√6
18√2​

Answer 1

★Area of traingle is $\sqrt{s( s-a)(s-b)(s-c)}$

↝a = 9

↝b = 9

↝C = 12

↝s = a + b + c / 2

= 9 + 9 + 12/2

= 15

★Now :Area is --

↝$\sqrt{s( s-a)(s-b)(s-c)}$

↝ $\sqrt{15( 15-9)(15-9)(15-12)}$

↝ $\sqrt{15( 6)(6)(3)}$

↝ $\sqrt{15×6×6×3}$

↝ $\sqrt{ 3×5×6×6×3}$

↝ 6×3√5

↝ 18√5

★Area of traingle is : $\sf\red{18√5}$

Your answer is : option ( B )