Question

What is the area of the quadrilateral AMND in the given figure (in $c{m}^2$)?

• A. $42\sqrt{3}$
• B. $48\sqrt{3}$
• C. $36\sqrt{3}$
• D. $24\sqrt{3}$

Answer 1

The correct option is C $36\sqrt{3}$


Construction: From point M draw a perpendicular on side CD. The perpendicular touches CD at O.
$\angle MNO=\angle NMB=60°$ (alternate angles)
In $\Delta MON$$\cos 60°=\frac{ON}{MN}$
$\Rightarrow \frac{1}{2}=\frac{ON}{12}$
$\Rightarrow ON=6cm$

$\sin 60°=\frac{OM}{MN}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{OM}{12}$
$\Rightarrow OM=6\sqrt{3}cm$

Quadrilateral AMND is a trapezium
Area of AMND $=\frac{1}{2}\times (AM+DN)\times OM$
$=\frac{1}{2}\times (AM+OD+ON)\times OM$$=\frac{1}{2}\times (3+3+6)\times 6\sqrt{3}$ [$\because OD=AM$]
$=6\times 6\sqrt{3}$
$=36\sqrt{3}c{m}^2$

Therefore, ar(AMND)$=36\sqrt{3}c{m}^2$