Question

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. The area of the trapezium is:

(a) √p + √q

(b) \(\frac13\) (√p + √q)³

(c) \(\frac12\) (√p + √q)²

(d) (√p + √q)²

Answer 1

(d) (√p + √q)²

∠BDC = ∠DBA            (alternate angles) 

∠ACD = ∠CAB            (alternate angles)

∠DOC = ∠AOB           (vert. opp. ∠s) 

⇒ ΔDCO ~ ΔBAO

⇒ \(\frac{\text{Area of ΔABO}}{\text{Area of ΔDCO}}\) = \(\frac{AB^2}{DC^2}\)  (By property of similar triangles)

⇒ \(\frac{p}{q}=\frac{AB^2}{DC^2}\) ⇒ \(\frac{AB}{DC}\) = \(\frac{\sqrt{p}}{\sqrt{q}}\)

⇒ AB = √P . k and DC = √q . k    for some constant k.

⇒ q = \(\frac{1}{2}\) x √q . k x OL ⇒ OL = \(\frac{2q}{\sqrt{q}.k}\) = \(\frac{2\sqrt{q}}{k}\)

Area of ΔAOB = \(\frac{1}{2}\) x AB x OM

⇒ p = \(\frac{1}{2}\) x  √p . k x OM ⇒ OM = \(\frac{2q}{\sqrt{p}.k}\) = \(\frac{2\sqrt{p}}{k}\)

Area of a trapezium = \(\frac{1}{2}\) x height of trapezium × sum of parallel sides

\(\frac{1}{2}\) x (OL + OM) x (DC + AB)

= \(\bigg(\frac{\frac{2\sqrt{q}}{k}+\frac{2\sqrt{p}}{k}}{2}\bigg)\) (√q k + √ p k)

= (√q k + √ p k)²