Find the area of an isosceles trapezoid in which the bases are 42cm and 54cm, and the angle for a large base is 45 degrees.
1. The tops of the trapezoid – A, B, C, D. AD = 54 cm. BC = 42 cm. BH – height (drawn to AD).
S is the area of the trapezoid. ∠А = 45 °.
2. The length of the segment АН is calculated by the formula: АН = (АD – ВС) / 2 (according to the properties isosceles trapezoid).
AH = (54 – 42) / 2 = 6 cm.
3. We calculate the length of the height BH through the tangent ∠A, which is equal to the quotient of dividing the leg BH by the leg AH.
BH: AH = tangent ∠A = tangent 45 ° = 1.
BH = AH x 1 = 6 x 1 = 6 cm.
4. S = (АD + ВС) / 2 х ВН = (54 + 42) / 2 х 6 = 288 cm².
Answer: S is equal to 288 cm².
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