Question

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as to form a trapezoid inscribed within a semicircle. What is the value of the area enclosed by the trapezoid?

(a) \(\bigg(\frac{157\times\sqrt3}{4}\bigg)\) cm²

(b) (49 x √3) cm²

(c) \(\bigg(\frac{147\times\sqrt3}{4}\bigg)\) cm²

(d) \(\bigg(\frac{100}{\sqrt3}\bigg)\) cm²

Answer 1

(c)\(\bigg(\frac{147\times\sqrt3}{4}\bigg)\) cm²

Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. 

Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF is a regular hexagon. 

⇒ ∠DAO = 60º                  (ΔDAO is an equilateral triangle)

⇒ DA = AO = 7 cm.          (∵ AB = 14 cm)

∴ Area of regular hexagon = \(\frac{3\sqrt3}{2}\) x (side)²

= \(\frac{3\sqrt3}{2}\) x 49 cm²

⇒ Area of trapezoid = \(\frac{1}{2}\) x Area of hexagon

= \(\frac{1}{2}\)x \(\frac{3\sqrt3}{2}\) x 49 = \(\frac{147\times\sqrt3}{4}\) cm².