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 cm²
• B. 75 cm²
• C. 90 cm²
• D. 105 cm²

Answer 1

The correct option is A

45 cm²

Draw the diagonals and they 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 $y=5cm$

$\Rightarrow$ $\frac{1}{2}\times 6\times 5=\frac{1}{2}\times \left(3x\right)$

$\Rightarrow$ $x=10cm$

Hence Area of $△ABC=\frac{1}{2}\times 6(x+y)$

$=\frac{1}{2}\times 6\times 15$
$=45c{m}^2$