Question

ABCD is a rectangle such that $\angle CAB={60}^{\circ }$ and AC=a units. The area of rectangle ABCD is $\frac{\sqrt{3}}{2}{a}^2$

• A. True
• B. False

Answer 1

The correct option is B False


In $\Delta ABC$, right-angled at B,
$sin{60}^{\circ }=\frac{BC}{AC}\Rightarrow \frac{\sqrt{3}}{2}=\frac{BC}{a}\Rightarrow BC=\frac{\sqrt{3}}{2}a$
$andcos{60}^{\circ }=\frac{AB}{AC}\Rightarrow \frac{1}{2}=\frac{AB}{a}\Rightarrow AB=\frac{a}{2}$
Area of rectangle
$=AB\times BC=\frac{a}{2}\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}{a}^2}{4}$