In an isosceles trapezoid, the bases are 8 cm and 12 cm, an acute angle of 30 degrees.

In an isosceles trapezoid, the bases are 8 cm and 12 cm, an acute angle of 30 degrees. Find the perimeter and area of the trapezoids.

Consider an isosceles trapezoid ABCD, with a smaller base BC = 8 cm, and a large AD = 12 cm. An acute angle is 30 °, and an acute angle with a larger base, <A = <D = 30 °. We omit the heights ВK and CM, they form two equal triangles ABK and CMD, AB = CD – as the lateral sides of the trapezoid, ВK = CM – as the heights of the trapezoid, AK = MD.
KM = BC = 8cm.
AD = AK + MD + KM
AK = (AD-KM) / 2 = (12-8) / 2 = 2 cm.
Consider a triangle ABK, angle <K = 90 °, <A = 30 °, AK = 2 cm.
By the property of the sine of an angle in a right-angled triangle:
cos A = AK / AB, hence:
AB = AK / cos A = 2 / cos 30 = 4 / √3 cm
sin A = BK / AB, hence:
VK = AB * sin A = (4 / √3) * sin 30 = (4 / √3) * (1/2) = 2 / √3 cm.
Let’s define the perimeter of the trapezoid:
p = 2 * AB + BC + AD = 2 * (4 / √3) + 8 + 12 = 20 + 8 / √3 = 24.6 cm.
Determine the area of ​​the trapezoid:
S = ((BC + AD) / 2) * BK = ((8 + 12) / 2) * 2 / √3 = 20 / √3 cm² = 11.36 cm².
Answer: the perimeter of the trapezoid is 24.6 cm, the area is 11.36 cm².



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