Rectangle Area?

Aaah! $\Delta ADC$ will be a $3$ - $4$ - $5$ right triangle. Let's see how:

As, $ABCD$ is a rectangle, $\Delta ADC$ will be a right triangle. So, $\text{Hypotenuse of } \Delta ADC =AC=\text{Diameter of the circle}$

Now, $\text{Diameter} =\frac{\text{circumference}} {\pi}$ $\Rightarrow AC=\frac{5\pi}{\pi}=5$ Thus, $DC=\sqrt{5^2-4^2}=3\, \color{teal} {[\text{Applying Pythagorian}]}$

Therefore, area of $ABCD$ will be, $AD\times DC=4\times 3=\color{#0C6AC7} {\boxed {12}}$

2.pi.r = 5.pi

d = 5

applying pythagoras theorm in ABD,

AB = 3

ar(ABCD) = 4 * 3 = 12