Find the length of the midline of a 60-degree acute-angled rectangular trapezoid with a larger side

Find the length of the midline of a 60-degree acute-angled rectangular trapezoid with a larger side and a larger base equal to 10?

Given: trapezoid, where ∠BAD = 90 °, ∠ADC = 60 °, AD = CD = 10. Find the length of the midline EF. Let’s omit the height of the trapezoid CG. By construction, CG ⊥ AD. Therefore, ∠CGD = 90 °. Hence, ΔCGD is a right-angled triangle with hypotenuse CD and legs CG and GD, with ∠GDC = 60 °. Then ∠GCD = 90 ° – 60 ° = 30 °. The leg of a right-angled triangle, lying opposite an angle of 30º, is equal to half of the hypotenuse. Hence, GD = 0.5 * CD = 0.5 * 10 = 5. Naturally, AG = AD – GD = 10 – 5 = 5. There is no doubt that the ABCG figure is a rectangle. Therefore, BC = AG = 5. Finally, using the fact “The middle line of the trapezoid is parallel to the bases and equal to their half-sum”, we have EF = (AD + BC) / 2 = (10 + 5) / 2 = 7.5.