Radius of the circle, r = 10 cm
Area of sector OPRQ

In ΔOPQ,
∠OPQ = ∠OQP (As OP = OQ)
∠OPQ + ∠OQP + ∠POQ = 180°
2∠OPQ = 120°
∠OPQ = 60°
ΔOPQ is an equilateral triangle.
So, area of ΔOPQ

Area of minor segment PRQ
= Area of sector OPRQ − Area of ΔOPQ
= 52.33 − 43.30
= 9.03 cm2
Area of major segment PSQ
= Area of circle − Area of minor segment PRQ
=π(10)²−9.03=314−9.03=304.97 cm²

Proof: We know that a tangent to the circle is perpendicular to the radius through the point of contact.
∴ ∠OPT = ∠OQT = 90°
In ΔOPT and ΔOQT,
OT = OT (Common)
OP = OQ (Radius of the circle)
∠OPT = ∠OQT (90°)
∴ ΔOPT ≅ ΔOQT (RHS congruence criterion)
⇒ TP = TQ (CPCT)
Hence, the lengths of the tangents drawn from an external point to a circle are equal.

$$\begin{gathered} \mathrm{Given}: \mathrm{l} \mathrm{and} \mathrm{m} \mathrm{are} \mathrm{two} \mathrm{parallel} \mathrm{tangents} \mathrm{to} \mathrm{the} \mathrm{circle} \mathrm{with} \mathrm{centre} \mathrm{O} \mathrm{touching} \\ \mathrm{the} \mathrm{circle} \mathrm{at} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{respectively}. \mathrm{DE} \mathrm{is} \mathrm{a} \mathrm{tangent} \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{C}, \mathrm{which} \mathrm{intersect} \\ \mathrm{l} \mathrm{at} \mathrm{D} \mathrm{and} \mathrm{m} \mathrm{at} \mathrm{E}. \\ \mathrm{To} \mathrm{prove}: \angle \mathrm{DOE} = 90° \\ \mathrm{Contruction}: \mathrm{Join} \mathrm{OC}. \\ \mathrm{Proof}: \end{gathered}$$

$$\begin{gathered} \mathrm{In} ∆\mathrm{ODA} \mathrm{and} ∆\mathrm{ODC}, \\ \mathrm{OA} = \mathrm{OC} ( \mathrm{Radii} \mathrm{of} \mathrm{the} \mathrm{same} \mathrm{circle} ) \\ \mathrm{AD} = \mathrm{DC} (\mathrm{Length} \mathrm{of} \mathrm{tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{to} \mathrm{a} \mathrm{circle} \\ \mathrm{are} \mathrm{equal} ) \\ \mathrm{DO} = \mathrm{OD} ( \mathrm{Commmon} \mathrm{side} ) \\ \therefore ∆\mathrm{ODA} \cong ∆\mathrm{ODC}, (\mathrm{SSS} \mathrm{congruence} \mathrm{criterion}) \\ \therefore \angle \mathrm{DOA} = \angle \mathrm{COD} ..........\left(1\right) \\ \mathrm{Similarly}, ∆\mathrm{OEB}\cong ∆\mathrm{OEC} \\ \therefore \angle \mathrm{EOB} = \angle \mathrm{COE} ...........\left(2\right) \\ \mathrm{Now}, \mathrm{AOB} \mathrm{is} \mathrm{a} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{circle}. \mathrm{Hence}, \mathrm{it} \mathrm{is} \mathrm{a} \mathrm{straight} \mathrm{line}. \\ \angle \mathrm{DOA} +\angle \mathrm{COD} + \angle \mathrm{COE} + \angle \mathrm{EOB} =180° \\ \mathrm{From} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{have}: \\ \Rightarrow 2\angle \mathrm{COD} + 2\angle \mathrm{COE} = 180° \\ \Rightarrow \angle \mathrm{COD} + \angle \mathrm{COE} = 90° \\ \Rightarrow \angle \mathrm{DOE} = 90° \\ \mathrm{Hence}, \mathrm{proved}. \end{gathered}$$

$$\begin{gathered} \mathrm{Given}: \mathrm{A} \mathrm{circle} \mathrm{with} \mathrm{centre} \mathrm{O} \mathrm{and} \mathrm{a} \mathrm{tangent} \mathrm{XY}\hspace{0.17em}\mathrm{to} \mathrm{the} \mathrm{circle} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{P} \\ \mathrm{To} \mathrm{prove}: \mathrm{OP} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{XY}. \\ \mathrm{Construction}: \\ \mathrm{Take} \mathrm{a} \mathrm{point} \mathrm{Q} \mathrm{on} \mathrm{XY} \mathrm{other} \mathrm{than} \mathrm{P} \mathrm{and} \mathrm{join} \mathrm{OQ}. \end{gathered}$$

$$\begin{gathered} \mathrm{Proof}: \\ \mathrm{Here} \mathrm{the} \mathrm{point} \mathrm{Q} \mathrm{must} \mathrm{lie} \mathrm{outside} \mathrm{the} \mathrm{circle} \mathrm{as} \\ \mathrm{if} \mathrm{it} \mathrm{lies} \mathrm{inside} \mathrm{the} \mathrm{tangent} \mathrm{XY} \mathrm{will} \mathrm{become} \mathrm{secant} \mathrm{to} \mathrm{the} \mathrm{circle}. \\ \mathrm{Therefore}, \mathrm{OQ} \mathrm{is} \mathrm{longer} \mathrm{thanthe} \mathrm{radius} \mathrm{OP}\hspace{0.17em}\mathrm{of} \mathrm{the} \mathrm{circle}, \mathrm{That} \mathrm{is}, \mathrm{OQ}> \mathrm{OP}. \\ \mathrm{This} \mathrm{happens} \mathrm{for} \mathrm{every} \mathrm{point} \mathrm{on} \mathrm{the} \mathrm{line} \mathrm{XY} \mathrm{expect} \mathrm{the} \mathrm{point} \mathrm{P}. \\ \mathrm{So} \mathrm{OP} \mathrm{is} \mathrm{the} \mathrm{shortest} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{O} \mathrm{to} \mathrm{the} \mathrm{points} \mathrm{on} \mathrm{XY}. \\ \mathrm{And} \mathrm{hence} \mathrm{OP} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{XY}. \\ \mathrm{Hence}, \mathrm{proved}. \end{gathered}$$

Given: A circle C ( 0, r ) and a tangent l at point A.

To prove: OA $\perp$ l

Construction: Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C.

Proof: We know that, among all line segment joining the point O to a point on l, the perpendicular is shortest to l.

OA = OC  (Radius of the same circle)

Now, OB = OC + BC.

$$\begin{gathered} \therefore \mathrm{OB}>\mathrm{OC} \\ \Rightarrow \mathrm{OB}>\mathrm{OA} \\ \Rightarrow \mathrm{OA}<\mathrm{OB} ( \mathrm{OC} = \mathrm{OA} = \mathrm{radius}) \end{gathered}$$

But among all the line segments, joining the point O to a point on AB, the shortest one is the perpendicular from O on AB.

Hence OA is perpendicular to l.