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In a circular museum hall of radius $14 \text{ m}$, some statues are displayed. Statues are kept inside the inner concentric circle of radius $7 \text{ m}$. One such statue lying in sector $\text{OAB}$, is fenced along line segments $\text{OA, AP, PB}$ and $\text{BO}$ where $\text{P}$ is a point on outer circle.
Based on above information, answer the following questions :
(i) Find $m\angle \text{AOP}$.
(ii) Prove that $\triangle \text{OAP} \cong \triangle \text{OBP}$.
(iii) (a) Find the length of fencing required to protect the statue.
(Take $\sqrt{3} = 1.73$)
OR
(b) Find area of quadrilateral $\text{OAPB}$. (Take $\sqrt{3} = 1.73$)
Based on above information, answer the following questions :
(i) Find $m\angle \text{AOP}$.
(ii) Prove that $\triangle \text{OAP} \cong \triangle \text{OBP}$.
(iii) (a) Find the length of fencing required to protect the statue.
(Take $\sqrt{3} = 1.73$)
OR
(b) Find area of quadrilateral $\text{OAPB}$. (Take $\sqrt{3} = 1.73$)
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Solution:
(i) $\text{OA = 7 m}$, $\text{OP = 14 m}$
If $\angle \text{AOP} = \theta$, $\cos \theta = \frac{\text{OA}}{\text{OP}} = \frac{7}{14} = \frac{1}{2}$ (1 Mark)
$\Rightarrow m\angle \text{AOP} = 60^\circ$
(ii) In $\triangle \text{OAP}$ and $\triangle \text{OBP}$
$\angle \text{A} = \angle \text{B} = 90^\circ$ (radius $\perp$ to tangent at the point of contact) (1 Mark)
$\text{OP = OP}$ (common)
$\text{OA = OB}$ (radii of the same circle)
Thus, $\triangle \text{OAP} \cong \triangle \text{OBP}$ (by RHS congruency) (1 Mark)
(iii) (a) $\sin 60^\circ = \frac{\text{AP}}{\text{OP}} = \frac{\text{AP}}{14}$ (1 Mark)
$\Rightarrow \text{AP} = 14 \sin 60^\circ = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \text{ m}$ or $12.11 \text{ m}$
Total fencing required = $\text{OA + AP + PB + BO} = 2(\text{OA + AP}) = 2(7 + 7\sqrt{3}) = 14(1 + \sqrt{3}) = 14(1 + 1.73) = 14 \times 2.73 = 38.22 \text{ m}$ (1 Mark)
OR
(b) $\text{Area OAPB} = \text{Area } \triangle \text{OAP} + \text{Area } \triangle \text{OBP} = 2 \times \text{Area } \triangle \text{OAP}$ (1 Mark)
$= 2 \times \frac{1}{2} \times \text{OA} \times \text{AP} = \text{OA} \times \text{AP} = 7 \times 7\sqrt{3} = 49\sqrt{3} = 49 \times 1.73 = 84.77 \text{ m}^2$ (1 Mark)
(i) $\text{OA = 7 m}$, $\text{OP = 14 m}$
If $\angle \text{AOP} = \theta$, $\cos \theta = \frac{\text{OA}}{\text{OP}} = \frac{7}{14} = \frac{1}{2}$ (1 Mark)
$\Rightarrow m\angle \text{AOP} = 60^\circ$
(ii) In $\triangle \text{OAP}$ and $\triangle \text{OBP}$
$\angle \text{A} = \angle \text{B} = 90^\circ$ (radius $\perp$ to tangent at the point of contact) (1 Mark)
$\text{OP = OP}$ (common)
$\text{OA = OB}$ (radii of the same circle)
Thus, $\triangle \text{OAP} \cong \triangle \text{OBP}$ (by RHS congruency) (1 Mark)
(iii) (a) $\sin 60^\circ = \frac{\text{AP}}{\text{OP}} = \frac{\text{AP}}{14}$ (1 Mark)
$\Rightarrow \text{AP} = 14 \sin 60^\circ = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \text{ m}$ or $12.11 \text{ m}$
Total fencing required = $\text{OA + AP + PB + BO} = 2(\text{OA + AP}) = 2(7 + 7\sqrt{3}) = 14(1 + \sqrt{3}) = 14(1 + 1.73) = 14 \times 2.73 = 38.22 \text{ m}$ (1 Mark)
OR
(b) $\text{Area OAPB} = \text{Area } \triangle \text{OAP} + \text{Area } \triangle \text{OBP} = 2 \times \text{Area } \triangle \text{OAP}$ (1 Mark)
$= 2 \times \frac{1}{2} \times \text{OA} \times \text{AP} = \text{OA} \times \text{AP} = 7 \times 7\sqrt{3} = 49\sqrt{3} = 49 \times 1.73 = 84.77 \text{ m}^2$ (1 Mark)