Question

ABCD is a rectangle such that $\angle CAB={60}^{\circ }$ and $AC=aunits$.
The area of rectangle ABCD is ______.

• A. $\frac{\sqrt{3}}{2}{a}^2$
• B. $\frac{\sqrt{2}}{3}{a}^2$
• C. $\frac{\sqrt{3}}{4}{a}^2$
• D. $\frac{\sqrt{3}}{8}{a}^2$

Answer 1

The correct option is C $\frac{\sqrt{3}}{4}{a}^2$

ABCD is a rectangle. Hence$\angle ABC={90}^0$.



So, in $\Delta ABC$, right-angled at B,
$Sin\theta =\frac{opposite}{hypotenuse}Sin{60}^0=\frac{BC}{AC}$
On substituiting the value of $Sin{60}^0=\frac{\sqrt{3}}{2}$ and AC = a units, we get

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{BC}{a}\Rightarrow BC=\frac{\sqrt{3}}{2}a$---------(i)

$Cos\theta =\frac{adjacent}{hypotenuse}Cos{60}^{\circ }=\frac{AB}{AC}$
On substituiting the value of $Cos{60}^0=\frac{1}{2}$ and AC = a units, we get
$\Rightarrow \frac{1}{2}=\frac{AB}{a}\Rightarrow AB=\frac{a}{2}$---------(ii)

From(i) and (ii),
So, area of the rectangle = length $\times$ breadth
$=AB\times BC=\frac{a}{2}\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}{a}^2}{4}$