In an isosceles trapezoid ABCD, the perimeter is 42 cm, the lateral side is 10 cm. Find the area of the trapezoid if its acute angle is equal.

Let’s draw the heights of the ВK and CH trapezoid.
In a right-angled triangle СDН, the angle DСН = (180 – 90 – 60) = 30. Then the kate DН is located opposite the angle 30, and therefore, DН = СD / 2 = 10/2 = 5 cm.
Since the trapezoid is isosceles, then AK = DN = 5 cm.
CH ^ 2 = СD ^ 2 – DН ^ 2 = 100 – 25 = 75.
CH = √75 = 5 * √3 cm.
The perimeter of the trapezoid is 42 cm.
AB + BC + СD + AD = 42 cm.
BC + AD = 42 – 10 – 10 = 22 cm.
AD = (AK + KН + DН) = (KН + 10) = (BC + 10).
Sun + Sun + 10 = 22.
2 * BC = 12 cm.
BC = 12/2 = 6 cm.
Then AD = 5 + 6 + 5 = 16 cm.
The area of the trapezoid is equal to: Savsd = (BC + AD) * СН / 2 = (6 + 16) * 5 * √3 / 2 = 55 * √3 cm2.
Answer: The area of the trapezoid is 55 * √3 cm2.