Question

In triangle $ABC,\angle A={60}^{\circ },\angle B={45}^{\circ }$ and AB=20cm. Then the area of the triangle is equal to

• A. $300-100\sqrt{3}c{m}^2$
• B. $300+100\sqrt{3}c{m}^2$
• C. $300\sqrt{3}-100c{m}^2$
• D. $300\sqrt{3}+100c{m}^2$

Answer 1

The correct option is A

$300-100\sqrt{3}c{m}^2$

Draw CD perpendicular to AB

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

In right-angled triangle BDC

$Tan{45}^{\circ }=\frac{CD}{DB}=1$

Let $CD=DB=x$

In right-angled triangle ADC

$Tan{60}^{\circ }=\frac{CD}{AD}$

$\sqrt{3}=\frac{x}{AD}\Rightarrow AD=\frac{x}{\sqrt{3}}$

$AB=AD+DB$

$20=\frac{x}{\sqrt{3}}+x$

$20=x(\frac{1}{\sqrt{3}}+1)$

$20=x(\frac{1+\sqrt{3}}{\sqrt{3}}$)

$x=\frac{20\sqrt{3}}{\sqrt{3}+1}=\frac{20\sqrt{3}}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{20\sqrt{3}(\sqrt{3}-1)}{2}=10\sqrt{3}(\sqrt{3}-1)$

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

=$\frac{1}{2}\times 20\times x$

$=\frac{1}{2}\times 20\times 10\sqrt{3}(\sqrt{3}-1)=100\sqrt{3}(\sqrt{3}-1)=300-100\sqrt{3}c{m}^2$