CBSE2026Class Class 10 · Mathematics4 Marks · Long✅ Verified
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 the outer circle.
(i) Find m∠AOP.
(ii) Prove that ΔOAP ≅ ΔOBP.
(iii)(a) Find the length of fencing required to protect the statue. (Take √3 = 1.73)
OR
(iii)(b) Find the area of quadrilateral OAPB. (Take √3 = 1.73)
✅ Answer & Solution
Here $OA=OB=7$ m (inner radius), $OP=14$ m (outer radius), and AP is tangent to the inner circle at A (so $OA\perp AP$).
(i) In right triangle OAP: $$\cos(\angle AOP)=\frac{OA}{OP}=\frac{7}{14}=\frac12 \Rightarrow \angle AOP=60°$$
(ii) In ΔOAP and ΔOBP: $OA=OB$ (radii), $OP=OP$ (common), $\angle OAP=\angle OBP=90°$ (tangent ⊥ radius). By RHS congruence rule, $$\triangle OAP\cong\triangle OBP$$
(iii)(a) $$AP=\sqrt{OP^2-OA^2}=\sqrt{196-49}=\sqrt{147}=7\sqrt3\ m$$ By congruence, $BP=AP=7\sqrt3$ m. $$\text{Fencing length}=OA+AP+PB+BO=7+7\sqrt3+7\sqrt3+7=14+14\sqrt3$$ Using $\sqrt3=1.73$: $$14+14(1.73)=14+24.22=38.22\ m$$
OR
(iii)(b) $$\text{Area of }OAPB=2\times\text{Area of }\triangle OAP=2\times\frac12\times OA\times AP=OA\times AP=7\times7\sqrt3=49\sqrt3$$ Using $\sqrt3=1.73$: $$49\times1.73=84.77\ m^2$$
