In an isosceles triangle ABC, side AB = 5.6 cm and AC = 4 cm. Calculate its area.

In an isosceles triangle ABC it is known:

Side AB = 5.6 cm;
Base AC = 4 cm.
Let’s calculate the area of the triangle.

Solution.

1) The area is equal to the product of the base and the height and divided by half.

To do this, find the height of the triangle VM. In an isosceles triangle, the height bisects the base.

So we get:

AM = AC = AC / 2 = 4 cm / 2 = 4/2 cm = 2 cm;

2) Triangle ABM is rectangular with angle M.

BM = √ (AB ^ 2 – AM ^ 2) = √ (5.6 ^ 2 – 2 ^ 2) = √ ((5.6 – 2) * (5.6 + 2)) = √ (3.6 * 7.6) = √27.36;

3) The area is equal to:

S = 1/2 * 4 * √27.36 = 4/2 * √27.36 = 2 * √27.36 = 2√27.36.

Answer: S = 2√27.36 cm ^ 2.