Question

In a triangle $ABC,AB=8cm,\angle B={30}^{\circ },\angle A={45}^{\circ }$. The area of the triangle will be equal to.

• A. $8(\sqrt{3}-1)c{m}^2$
• B. $16(\sqrt{3}-1)c{m}^2$
• C. $16(\sqrt{3}+1)c{m}^2$
• D. $8(\sqrt{3}-1)c{m}^2$

Answer 1

The correct option is B

$16(\sqrt{3}-1)c{m}^2$

Draw CD perpendicular to AB

In triangle ACD

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

So,$AD=DC$ , Let it be equal to $x$

In triangle ADC, the angles are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$

The corresponding sides can be calculated as

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

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

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

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

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

The corresponding sides can be calculated as

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

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

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

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

$AB=AD+DB$

$8=x+x\sqrt{3}$

$x(1+\sqrt{3})=8$

$x=\frac{8}{(1+\sqrt{3})}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}$

$x=\frac{8(\sqrt{3}-1)}{(3-1)}=4(\sqrt{3}-1)cm$

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

$=\frac{1}{2}\times 8\times x$

$=\frac{1}{2}\times 8\times 4(\sqrt{3}-1)$

$16(\sqrt{3}-1)c{m}^2$