Question

Find the area of the quadrilateral $ABCD$. Here, $AC=22cm$, $BM=3cm$, $DN=3cm$ and $BM\perp AC$ and $DN\perp AC$.

Answer 1

Solve for area of the given quadrilateral

Given,

1. Length of the diagonal $AC$$=22cm$
2. Offsets i.e. perpendiculars distances from opposite vertices to the diagonal, $h_1$$=3cm$and $h_2$$=3cm$.

We know that, the area of quadrilateral $\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathit{d}\mathbf{\times }\left(h_1+h_2\right)$

$\therefore$Area of the given quadrilateral $=\frac{1}{2}\times 22\times \left(3+3\right)$ [Where $d=AC=22cm$]

$$\begin{gathered} =\frac{1}{2}\times 22\times 6 \\ =66c{m}^2 \end{gathered}$$

Hence, area of the given quadrilateral is $66c{m}^2$