Question

In $\Delta ABC,\angle A=\angle B={30}^{\circ },AB=12cm$, the area of the triangle will be equal to

• A. $24\sqrt{3}c{m}^2$
• B. $12\sqrt{3}c{m}^2$
• C. $6\sqrt{3}c{m}^2$
• D. $18\sqrt{3}c{m}^2$

Answer 1

The correct option is B

$12\sqrt{3}c{m}^2$

Draw CD perpendicular to AB

In $\Delta ACD\&\Delta BCD$

$\angle A=\angle B,\left(Both{30}^{\circ }\right)$

$CD=CD\left(common\right)$

$\angle CDA=\angle CDB(Each{90}^{\circ }$)

Area, $\Delta ACDD\cong BCD$

Hence,$AD=DC=6cm$ (corresponding sides of congruent triangles )

In right-angled triangle ACD, the angles are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

Hence, the corresponding sides can be calculated as

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\Rightarrow 1:\sqrt{3}:2$

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& \sqrt{2}\\ \downarrow & & \downarrow & & \downarrow \\ CD& & AD& & AC\\ \downarrow & & \downarrow & & \downarrow \\ \frac{6}{\sqrt{3}}cm& & 6cm& & \frac{12}{\sqrt{13}}cm\end{array}$

Area of triangle $ABC=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}\times 12\times \frac{6}{\sqrt{3}}$

$=\frac{36}{\sqrt{3}}=\frac{36}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=12\sqrt{3}c{m}^2$