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प्रश्न
In the figure, the centre of the circle is O and ∠STP = 40°.
- m (arc SP) = ? By which theorem?
- m ∠SOP = ? Give reason.
उत्तर
Given: ∠STP = 40°
i. m(arc SP) = 2∠STP
⇒ m(arc SP) = 2 × 40° = 80°
Hence, m(arc SP) is 80° by inscribed angle theorem.
ii. m∠SOP = 2 × m∠STP
= 2 × 40° = 80°
As a result, m∠SOP is 80° since the angle subtended by an arc at the centre of the circle is twice the angle subtended by the arc at any point on the circle.
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संबंधित प्रश्न
In the adjoining figure chord EF || chord GH.
Prove that chord EG ≅ chord FH.
Fill in the boxes and write the complete proof.
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In the given figure, O is centre of circle. ∠QPR = 70° and m(arc PYR) = 160°, then find the value of the following ∠QOR.
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A circle with centre P is inscribed in the ABC. Side AB, side BC and side AC touch the circle at points L, M and N respectively. Radius of the circle is r.
Prove that: `"A" (triangle "ABC") =1/2 ("AB" + "BC" + "AC") xx "r"`
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= `1/2` m(arc SQ) + `square` .....[Inscribed angle theorem]
= `1/2 [square + square]`
In figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.
Proof: Draw seg GF.
∠EFG = ∠FGH ......`square` .....(I)
∠EFG = `square` ......[inscribed angle theorem] (II)
∠FGH = `square` ......[inscribed angle theorem] (III)
∴ m(arc EG) = `square` ......[By (I), (II), and (III)]
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The angle inscribed in the semicircle is a right angle. Prove the result by completing the following activity.
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Proof: Segment AC is a diameter of the circle.
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∴ m∠ABC = `square`
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Proof:
Draw seg OD.
∠ACB = `square` ......[Angle inscribed in semicircle]
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∠DOB = `square` ......[Definition of measure of an arc](i)
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∴ seg AD ≅ seg BD
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Find
(i) m(arc MN)
(ii) m(arc LN)
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In the above figure, ∠L = 35°, find :
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Solution :
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∴ `square = 1/2` m(arc MN)
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