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A farmer has a semicircular field as shown in the figure given below :
In segment ACPA, he grows rice. In the triangular region APB, he grows vegetables. He allows his cows to graze in the segment BDPB. The diameter of the semicircular field with centre O is $120$ m, $\angle PAO = 30^\circ$ and $\angle POB = 60^\circ$.
Based on the above information, answer the following questions :
(i) Find the radius of the field and the measure of $\angle ABP$.
(ii) Find the area of sector AOP in terms of $\pi$.
(iii) (a) Find the area of the region in which cows are allowed to graze. (Use $\pi = 3\cdot14$)
OR
(iii) (b) Find the area of the region in which vegetables are grown.
In segment ACPA, he grows rice. In the triangular region APB, he grows vegetables. He allows his cows to graze in the segment BDPB. The diameter of the semicircular field with centre O is $120$ m, $\angle PAO = 30^\circ$ and $\angle POB = 60^\circ$.
Based on the above information, answer the following questions :
(i) Find the radius of the field and the measure of $\angle ABP$.
(ii) Find the area of sector AOP in terms of $\pi$.
(iii) (a) Find the area of the region in which cows are allowed to graze. (Use $\pi = 3\cdot14$)
OR
(iii) (b) Find the area of the region in which vegetables are grown.
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Solution
(i) Radius of the field = $60$ m (1/2 Mark)
$\angle ABP = 60^\circ$ (1/2 Mark)
(ii) Area of sector AOP $= \frac{120}{360} \times \pi \times 60 \times 60$ (1/2 Mark)
$= 1200 \pi$ m$^2$ (1/2 Mark)
(iii) (a) Area of the region in which cows are allowed to graze
$= \text{Area of segment BDPB}$
$= \frac{60}{360} \times 3.14 \times (60)^2 - \frac{\sqrt{3}}{4} \times (60)^2$ (1 Mark)
$= (1884 – 900\sqrt{3})$ m$^2$ (1 Mark)
OR
(iii) (b) As $\Delta OPB$ is an equilateral $\Delta$
$\therefore PB = 60$ m (1 Mark)
In right $\Delta APB$, $AP = \sqrt{(120)^2 - (60)^2} = 60\sqrt{3}$ m
Area of the region in which vegetables are grown
$= \text{Area of } \Delta APB = \frac{1}{2} \times 60 \times 60\sqrt{3} = 1800\sqrt{3}$ m$^2$ (1 Mark)
(i) Radius of the field = $60$ m (1/2 Mark)
$\angle ABP = 60^\circ$ (1/2 Mark)
(ii) Area of sector AOP $= \frac{120}{360} \times \pi \times 60 \times 60$ (1/2 Mark)
$= 1200 \pi$ m$^2$ (1/2 Mark)
(iii) (a) Area of the region in which cows are allowed to graze
$= \text{Area of segment BDPB}$
$= \frac{60}{360} \times 3.14 \times (60)^2 - \frac{\sqrt{3}}{4} \times (60)^2$ (1 Mark)
$= (1884 – 900\sqrt{3})$ m$^2$ (1 Mark)
OR
(iii) (b) As $\Delta OPB$ is an equilateral $\Delta$
$\therefore PB = 60$ m (1 Mark)
In right $\Delta APB$, $AP = \sqrt{(120)^2 - (60)^2} = 60\sqrt{3}$ m
Area of the region in which vegetables are grown
$= \text{Area of } \Delta APB = \frac{1}{2} \times 60 \times 60\sqrt{3} = 1800\sqrt{3}$ m$^2$ (1 Mark)