Question

In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. Find the ratio of area of triangle QPO and area of the parallelogram ABCD.

• A. 7 : 9
• B. 9 : 7
• C. 8 : 9
• D. 9 : 8

Answer 1

The correct option is D 9 : 8

Joining MN
Area of $△QPO$ = Area of $△QAM+$ Area of AMONB + Area of $△PBN$ ....(1)

We know,
$△QAM\cong △CDM$ (By ASA)
$△PBN\cong △DCN$ (By ASA)

Then, Equation (1) becomes
Area of $△QPO$ = Area of $△CDM+$ Area of AMONB + Area of $△DCN$...(2)

Area of $△QPO$ = Area of AMONB + Area of $△CDM+$ Area of $△CON+$ Area of $△COD$

Area of $△QPO$ = Area of ABCD + Area of $△COD$ ....(3)

Area of $△COD$
$=\frac{1}{4}$ $\times$ Area of MNCD
$=\frac{1}{4}of\frac{1}{2}$ of Area of ABCD

Area of $△COD=\frac{1}{3}$ of Area of ABCD

Substitute Equation (4) in Equation (3),
Area of $△QPO$
$=$Area of ABCD $+\frac{1}{3}$ of Area of ABCD
$=\frac{9}{8}$ of Area of ABCD

Area of $△QPO:$ Area of ABCD = 9 : 8