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In a circular museum hall of radius $14$ m, some statues are displayed. Statues are kept inside the inner concentric circle of radius $7$ m. One such statue lying in sector $OAB$, is fenced along line segments $OA, AP, PB$ and $BO$ where $P$ is a point on outer circle.
Based on above information, answer the following questions :
(i) Find $m\angle AOP$. (1 Mark)
(ii) Prove that $\triangle OAP \cong \triangle OBP$. (1 Mark)
(iii) (a) Find the length of fencing required to protect the statue. (Take $\sqrt{3} = 1.73$) (2 Marks)
OR
(b) Find area of quadrilateral $OAPB$. (Take $\sqrt{3} = 1.73$)
Based on above information, answer the following questions :
(i) Find $m\angle AOP$. (1 Mark)
(ii) Prove that $\triangle OAP \cong \triangle OBP$. (1 Mark)
(iii) (a) Find the length of fencing required to protect the statue. (Take $\sqrt{3} = 1.73$) (2 Marks)
OR
(b) Find area of quadrilateral $OAPB$. (Take $\sqrt{3} = 1.73$)
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Solution:
(i) $OA = 7$ m, $OP = 14$ m
If $\angle AOP = \theta \therefore \cos \theta = \frac{7}{14} = \frac{1}{2}$
$\Rightarrow m\angle AOP = 60\degree$ (1 Mark)
(ii) In $\triangle OAP$ and $\triangle OBP$
$\angle A = \angle B = 90\degree$ (radius $\perp$ to tangent at the point of contact)
$OP = OP$ (common)
$OA = OB$ (radii of the same circle)
Thus, $\triangle OAP \cong \triangle OBP$ (by RHS congruency) (1 Mark)
(iii) (a) $\sin 60\degree = \frac{AP}{14}$
$\Rightarrow AP = 7\sqrt{3}$ m or $12.11$ m (1 Mark)
Total fencing required = $2(OA + AP) = 2(7 + 12.11) = 2(19.11) = 38.22$ m (1 Mark)
OR
(b) $\sin 60\degree = \frac{AP}{14}$
$AP = 7\sqrt{3}$ m or $12.11$ m (1 Mark)
Area $OAPB = 2 \times \frac{1}{2} \times OA \times AP = 7 \times 12.11 = 84.77$ m$^2$ (1 Mark)
(i) $OA = 7$ m, $OP = 14$ m
If $\angle AOP = \theta \therefore \cos \theta = \frac{7}{14} = \frac{1}{2}$
$\Rightarrow m\angle AOP = 60\degree$ (1 Mark)
(ii) In $\triangle OAP$ and $\triangle OBP$
$\angle A = \angle B = 90\degree$ (radius $\perp$ to tangent at the point of contact)
$OP = OP$ (common)
$OA = OB$ (radii of the same circle)
Thus, $\triangle OAP \cong \triangle OBP$ (by RHS congruency) (1 Mark)
(iii) (a) $\sin 60\degree = \frac{AP}{14}$
$\Rightarrow AP = 7\sqrt{3}$ m or $12.11$ m (1 Mark)
Total fencing required = $2(OA + AP) = 2(7 + 12.11) = 2(19.11) = 38.22$ m (1 Mark)
OR
(b) $\sin 60\degree = \frac{AP}{14}$
$AP = 7\sqrt{3}$ m or $12.11$ m (1 Mark)
Area $OAPB = 2 \times \frac{1}{2} \times OA \times AP = 7 \times 12.11 = 84.77$ m$^2$ (1 Mark)