Question

solutions of this question

Answer 1

The answer is in photograph

Answer 2

Given:

The perimeter of an isosceles triangle = 30 cm and

Each of the equal sides of an  isosceles triangle (a) = 12 cm

Let the unequal side = b

We have to find, the area of the triangle is:

Solution:

We know that,

The perimeter of an isosceles triangle = 2a + b

⇒ 2(12) + b = 30

⇒ b = 30 - 24 = 6 cm

The area of an isosceles triangle = $\dfrac{1}{4} b\sqrt{4a^2-b^2}$

= $\dfrac{1}{4} \times 6\sqrt{4(12)^2-6^2}$ $cm^{2}$

= $\dfrac{3}{2} \times \sqrt{576-36}$ $cm^{2}$

= $\dfrac{3}{2} \times \sqrt{540}$ $cm^{2}$

= $\dfrac{3}{2} \times 6\sqrt{15}$ $cm^{2}$

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

∴ The area of an isosceles triangle = $9\sqrt{15}$ $cm^{2}$

Thus, the area of an isosceles triangle is "$9\sqrt{15}$ $cm^{2}$".