Question

In quadrilateral ABCD, the diagonals intersect at O and AB $\parallel$ CD. The areas of triangles AOB and DOC are equal and altitude of $△$AOB = 5 cm. The area of triangle ABC is _______.

• A. 45 $c{m}^2$
• B. 75 $c{m}^2$
• C. 90 $c{m}^2$
• D. 105 $c{m}^2$

Answer 1

The correct option is A 45 $c{m}^2$

Draw the diagonals which intersect at O.
Let ‘y’ be the altitude of triangle AOB.

As AB $\parallel$ CD, altitude of $△$DOC = x as shown.

Area ($△$AOB) = Area ($△$DOC)
$\frac{1}{2}\times 6y=\frac{1}{2}\times \left(3x\right)$

Given $y=5cm$
$\frac{1}{2}\times 6\times 5=\frac{1}{2}\times \left(3x\right)$
$x=10cm$

Area of $△ABC=\frac{1}{2}\times 6(x+y)$
Area of $△ABC=\frac{1}{2}\times 6\times 15$ $=45c{m}^2$