Question

In a triangle $\Delta ABC,AB=12cm.\angle A={45}^{\circ },\angle B={60}^{\circ }$. The area of triangle ABC will be equal to.

• A. $108-72\sqrt{3}c{m}^2$
• B. $54-72\sqrt{3}c{m}^2$
• C. $108-36\sqrt{3}c{m}^2$
• D. $72-36\sqrt{3}c{m}^2$

Answer 1

The correct option is C

$108-36\sqrt{3}c{m}^2$

Draw CD perpendicular to AB

In $\Delta ACD$

$\angle ACD=180-({90}^{\circ }+{45}^{\circ })={45}^{\circ }$

Hence, $AD=DC,$ Let it be $x$

In Triangle ACD, the angles are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$

$\Rightarrow \sin \left(45\right):\sin \left(45\right):\sin \left(90\right)$

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

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

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ \downarrow & & \downarrow & & \downarrow \\ AD& & CD& & AC\\ \downarrow & & \downarrow & & \downarrow \\ x& & x& & \sqrt{2x}\end{array}$

In triangle BCD, the angles are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

Hence, the corresponding sides can be calculated as

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

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

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

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ \downarrow & & \downarrow & & \downarrow \\ DB& & CD& & CB\\ \downarrow & & \downarrow & & \downarrow \\ \frac{x}{\sqrt{3}}& & x& & \frac{2x}{\sqrt{3}}\end{array}$

Hence , $AB=AD+DB$

$AB=x+\frac{x}{\sqrt{3}}$

$12=x(1+\frac{1}{\sqrt{3}})$

$\frac{x(\sqrt{3}+1)}{\sqrt{3}}=12$

$x=\frac{12\sqrt{3}}{(\sqrt{3}+1)}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}$

$x=\frac{12.\sqrt{3}(\sqrt{3}-1)}{(3-1)}=6(3-\sqrt{3})=18-6\sqrt{3}$

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

$=\frac{1}{2}\times 12\times (18-6\sqrt{3})$

$=6(18-6\sqrt{3})$

=$108-36\sqrt{3}c{m}^2$