Question

In triangle $ABC,\angle A=\angle B={30}^{\circ },AB=18cm$ as shown in figure. The perimeter of the triangle will be to equal to

• A. $(18\sqrt{3}+12)cm$
• B. $(12\sqrt{3}+18)cm$
• C. $12\sqrt{6}+18cm$
• D. $18\sqrt{6}+12cm$

Answer 1

The correct option is B

$(12\sqrt{3}+18)cm$

Draw CD perpendicular to AB

In $\Delta ACD\&\Delta BCD$

$\angle CAD=\angle CBD\left(each{30}^{\circ }\right)$ (Given)

$\angle CDA=\angle CDB\left(each{90}^{\circ }\right)$ (CD perpendicular to AB)

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

$\Delta ACD\cong \Delta BCD$

$AD=DB=9cm$ (Corresponding sides of congruent triangle )

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

The corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ x& :& x\sqrt{3}& :& 2x\\ \downarrow & & \downarrow & & \downarrow \\ CD& & AD& & AC\\ \downarrow & & \downarrow & & \downarrow \\ 3\sqrt{3}cm& & 9cm& & 6\sqrt{3}cm\end{array}$

AC = CB (Corresponding sides of congruent triangle)

Perimeter of triangle =$AC+CB+AB$

$6\sqrt{3}cm+6\sqrt{3}cm+18$

$12\sqrt{3}+18cm$