Given a rectangular trapezoid, the larger base of which is 12 cm, and the radius of the inscribed circle is 3 cm. Find the area of the trapezoid.
From the center of the O circle, draw the radii OK, OH and OM, to the points of tangency.
By the property of a tangent drawn from one point, BK = BM = AH = AM = OM = 3 cm.
Then the length of the segment DH = AD – AH = 12 – 3 = 9 cm.
From a right-angled triangle DOH tgODH = OH / DH = 3/9 = 1/3.
If a circle is inscribed in a trapezoid, then the triangles formed by the side and the center of the circle are rectangular. Then the triangle COD is rectangular, and then the angle is ODH = COK.
In a right-angled triangle SOC tgO = KC / KO.
1/3 = KC / 3.
KC = 1 cm.
Then BC = BK + KC = 3 + 1 = 4 cm.
Determine the area of the trapezoid.
Savsd = (ВС + АD) * КН / 2 = (4 + 12) * 6/2 = 48 cm2.
Answer: The area of the trapezoid is 48 cm2.
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