Question

ABCD is a parallelogram and 'Q' is a point on AB. If area of QCD is 10 sq cm, then find the ar(QAD) + ar(QBC) in sq cms.

• A. 10

Answer 1

The correct option is A 10
$\text{Given: Area of QCD is 10}c{m}^2.$


Form the above figure :
Area of ABCD = Area of QCD + Area of QAD + Area of QBC

We know that, if a triangle and a parallelogram are on the same base and between same parallel lines, then

$\text{Area of triangle}=\frac{1}{2}\times \text{Area of parallelogram}$

$\therefore \text{Area of ABCD}=2\times \text{Area of QCD}$

$\Rightarrow 2\times \text{Area of QCD}=\text{Area of QCD + Area of QAD}\text{+ Area of QBC}$

$\Rightarrow \text{Area of QAD + Area of QBC}$$\text{= Area of QCD}$

$\Rightarrow \text{Area of QAD + Area of QBC}{\text{= 10 cm}}^2$