I draw a quadrilateral ABCD of which AD and BC are perpendicular on side AB and AB = 5 cm, AD = 7 cm and BC = 4 cm. I draw a triangle with equal area of that quadrilateral of which one angle is 30°.

A. Draw the given quadrilateral ABCD.

B. Draw the diagonal DB of quadrilateral ABCD.

C. Draw a parallel line through point C to diagonal DB of quadrilateral ABCD which intersects at Q produced AD.

D. CQ||DB, ΔABQ is the required triangle.

Proof:

∆DCB = ∆DBQ (on same base DB and between same parallels DB and CQ)

∴ ∆DCB = ∆DBQ

∆ABD + ∆DCB = ∆DBQ + ∆ABD (adding area of ∆ABD on both sides)

∴ quadrilateral ABCD = ∆ABQ

Taking BQ as a base, we draw another ∆BCE with one angle 30⁰ and between the parallels DQ and BC.

∴ ∆BCE = quadrilateral ABCD