An isosceles trapezoid with an acute angle of 60 is inscribed with a circle of radius 3√3 cm. Find the perimeter of the trapezoid.
If a circle is inscribed in a trapezoid, then the sum of the opposite sides of the trapezoid is equal to each other. AD + BC = AB + CD. Since the trapezoid is isosceles, then AD + BC = 2 * AB.
Let’s draw a segment AB, the length of which will be equal to two radii of the circle. AB = 2 * 3 * √3 = 6 * √3.
Consider a right-angled triangle ABE, and which angle BAE = 60, and leg BE = 6 * √3. Then AB = BE / Sin 60 = 6 * √3 / (√3 / 2) = 12 cm.
Then СD = BE = 12 cm.
CD + BE = 24 cm.
AD + BC = CD + BE = 24 cm.
P = 24 + 24 = 48 cm.
Answer: The perimeter of the trapezoid is 48 cm.
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