# A circle is inscribed in an isosceles trapezoid touching all its sides.

**A circle is inscribed in an isosceles trapezoid touching all its sides. Find the area of a circle and a trapezoid, if its lateral side is 4 dm, an acute angle is 30 degrees.**

Let’s apply the formula for the length of the midline of a trapezoid.

KP = (BC + AD) / 2 = 4 dm.

(BC + AD) = 2 * 4 = 8 dm.

Since a circle is inscribed in the trapezoid, the sum of the lengths of its lateral moans is equal to the sum of the lengths of its bases. Then (AB + CD) = (BC + AD) = 8 dm.

Then AB = BC = 8/2 = 4 dm.

Let’s draw the height BH of the trapezoid. In a right-angled triangle BAN, the BH leg is cut against an angle of 30, then BH = AB / 2 = 4/2 = 2 dm.

The radius of a circle inscribed in a trapezoid is half the length of its height.

R = OM = BH / 2 = 2/2 = 1 dm.

Determine the area of the circle. Scr = n * R ^ 2 = n dm2.

Determine the area of the trapezoid. Strap = KP * BH = 4 * 2 = 8 dm2.

Answer: The area of the circle is n cm2, the area of the trapezoid is 8 dm2.