﻿ A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.  from Mathematics Circles Class 10 Uttarakhand Board

Prove that the lengths of tangents drawn from an external point to a circle are equal.

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.

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Prove that the tangent at any point of a circle is perpendicular to theradius through the point of contact.

In fig., l and m are two parallel tangents to a circle with centre O,touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that   $\angle$DOE = 900

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

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.

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.

# A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.

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

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