The side of an isosceles triangle is 20 m, and the area is 98 m2. Find the height of the triangle drawn to the base.

The area of an isosceles triangle is:

S = (AC ^ 2 * sin (ACB)) / 2.

From here we derive sin (ACB) = 2 * S / (AC ^ 2).

sin (ACB) = 2 * 98/20 ^ 2 = 0.49.

Therefore, the ACB angle is 29.3 °.

Now you can find the ACD angle:

29.3 / 2 = 14.65 °.

So the height of the CD is:

CD = AC * cos (14.65) = 20 * 0.97 = 19.3 (m).

Answer: the height is 19.3 m.