A circular grass plot 4 m in diameter is cut by a straight path | m wide, one edge of which passes through the centre of the plot. Area of the remaining portion is

Radius of circle = OC = OB = 2 m and OA = 1 m

Area of path = $$2\times$$[Area of $$\triangle$$ OAB+Area of sector BOC] ----------(i)

In right $$\triangle$$ AOB,

=> $$(AB)^2=(OB)^2-(OA)^2$$

=> $$(AB)^2=4-1=3$$

=> $$(AB)=\sqrt3$$ m ------------(ii)

Also, $$tan(\angle AOB)=\frac{AB}{OA}=\sqrt3$$

=> $$\angle OAB=60^\circ$$

Thus, $$\angle BOC=90^\circ-60^\circ=30^\circ$$ --------------(iii)

Now, in equation (i), area of path = $$2\times[(\frac{1}{2}\times OA \times AB)+(\frac{30^\circ}{360^\circ}\pi r^2)]$$

= $$2\times[(\frac{1}{2}\times1\times\sqrt3)+(\frac{1}{12}\times \pi (2)^2)]$$

= $$2\times[\frac{\sqrt3}{2}+\frac{\pi}{3}]$$

= $$\sqrt3+\frac{2\pi}{3}$$ -----------(iv)

$$\therefore$$ Area of plot = Area of shaded region = Area of circle - Area of path

= $$\pi(2)^2-(\sqrt3+\frac{2\pi}{3})$$

= $$4\pi-\sqrt3-\frac{2\pi}{3}$$

= $$\frac{10\pi}{3}-\sqrt3$$ $$m^2$$

=> Ans - (D)