In-a-trapezoid-ABCD-sides-AB-and-CD-are-parallel-and-side-BC-CD-5-If-DC-B-120-and-BA-D-60-Find-the-area-of-ABCD-

Question Number 111539 by Aina Samuel Temidayo last updated on 04/Sep/20

In a trapezoid ABCD, sides AB and

CD are parallel and side BC = CD = \sqrt{5} .

If D \hat{C B} = 120 ° and B \hat{A D} = 60 ° . Find

the area of ABCD .

Answered by 1549442205PVT last updated on 04/Sep/20

Commented by 1549442205PVT last updated on 04/Sep/20

Draw the line d parallel to BC cutting

AB at point E . Then BCDE is

parallelogram, so from the hypothesis

BC = CD = \sqrt{5} and \hat{DAB} = 60 ° we have

DE = BC = \sqrt{5} \hat{, DEA} = \hat{CBA} = 60 ° (since

\hat{BCD} = 60 °) It follows that Δ ADE is

equilateralu (since it has two angles

equal to 60 °) . From DF = \frac{\sqrt{5} . \sqrt{3}}{2} = \frac{\sqrt{15}}{2}

(DF is the altitude of the equilateral

triangle) AB = AE + BE = 2 \sqrt{5}

Therefore, S_{ABCD} = \frac{AB + CD}{2} \times DF

= \frac{2 \sqrt{5} + \sqrt{5}}{2} \times \frac{\sqrt{15}}{2} = \frac{3 \sqrt{75}}{4} = \frac{15 \sqrt{3}}{4} ({cm}^{2})

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

Thanks .