The altitude drawn to the base of an isosceles triangle is 12 cm and the perimeter is 36 cm. Find the area of the triangle.

Correct option is B

Given:

The altitude (height) of the isosceles triangle = 12 cm

Perimeter of the triangle = 36 cm

Formula Used:

Area of a triangle $$= \frac{1}{2} \times \text{base} \times \text{height}$$​​

Solution:
Let the base of the isosceles triangle be b, and the equal sides of the triangle be s. The perimeter of the triangle is the sum of all three sides, so:

b + 2s = 36

2s = 36 - b

s $$= \frac{36 - b}{2} \quad \text{(1)}$$​

The altitude divides the triangle into two right triangles, with each having a base of $$\frac{b}{2}$$​ and a height of 12 cm. Applying the Pythagorean theorem:

​$$s^2 = \left(\frac{b}{2}\right)^2 + 12^2$$​

​$$s^2 = \frac{b^2}{4} + 144 \quad \text{(2)}$$​

Substitute s from equation (1) into equation (2):

​$$\left(\frac{36 - b}{2}\right)^2 = \frac{b^2}{4} + 144$$​

​$$\frac{(36 - b)^2}{4} = \frac{b^2}{4} + 144$$​

​$$(36 - b)^2 = b^2 + 576$$​​

​$$1296 - 72b + b^2 = b^2 + 576$$​

1296 - 72b = 576

72b = 720

b = 10

Now, substitute b = 10 into equation (1) to find s:

​$$s = \frac{36 - 10}{2} = \frac{26}{2} = 13$$​

Area =$$\frac{1}{2} \times b \times \text{height} = \frac{1}{2} \times 10 \times 12 = 60 \, \text{cm}^2$$ 
Exam Hall Method: