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In the above figure, ∠L = 35°, find : m(arc MN) m(arc MLN) Solution : ∠L = 12 m(arc MN) ............(By inscribed angle theorem)∴ □=12 m(arc MN)∴ 2 × 35 = m(arc MN) - Geometry Mathematics 2

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Question


In the above figure, ∠L = 35°, find :

  1. m(arc MN)
  2. m(arc MLN)

Solution :

  1. ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
    ∴ `square = 1/2` m(arc MN)
    ∴ 2 × 35 = m(arc MN)
    ∴ m(arc MN) = `square`
  2. m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
    = 360° – 70°
    ∴ m(arc MLN) = `square`
Fill in the Blanks
Sum

Solution

  1. ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
    35° = `1/2` m(arc MN)
    ∴ 2 × 35 = m(arc MN)
    ∴ m(arc MN) = 70°
  2. m(arc MLN) = 360° – m(arc MN) ...........[Definition of measure of arc]
    = 360° – 70°
    ∴ m(arc MLN) = 290°
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Inscribed Angle Theorem
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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.


Given: ∠ABC is inscribed angle in a semicircle with center M

To prove: ∠ABC is a right angle.

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Arc AXC is intercepted by the inscribed angle ∠ABC

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∴ m∠ABC = `square`

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Given: In a circle with center C, ∠PQR and ∠PSR are inscribed in same arc PQR. Arc PTR is intercepted by the angles.

To prove: ∠PQR ≅ ∠PSR.

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m∠`square` = m∠PSR      .....[By (i) and (ii)]

∴ ∠PQR ≅ ∠PSR


In the figure, a circle with center C has m(arc AXB) = 100° then find central ∠ACB and measure m(arc AYB).


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Proof:

Draw seg OD.

∠ACB = `square`     ......[Angle inscribed in semicircle]

∠DCB = `square`    ......[CD is the bisector of ∠C]

m(arc DB) = `square`   ......[Inscribed angle theorem]

∠DOB = `square`   ......[Definition of measure of an arc](i)

seg OA ≅ seg OB     ...... `square` (ii)

∴ Line OD is `square` of seg AB    ......[From (i) and (ii)]

∴ seg AD ≅ seg BD


In the figure, if O is the center of the circle and two chords of the circle EF and GH are parallel to each other. Show that ∠EOG ≅ ∠FOH


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∠STQ = `1/2` [m(arc PR) + m(arc SQ)]

Activity: In ΔPTS,

∠SPQ = ∠STQ – `square`  ......[∵ Exterior angle theorem]

∴ ∠SPQ = 34°

∴ m(arc QS) = 2 × `square`° = 68°   ....... ∵ `square`

Similarly, m(arc PR) = 2∠PSR = `square`°

∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58°  ......(I)

But ∠STQ = 58°  .....(II) (given)

∴  `1/2` [m(arc PR) + m(arc QS)] = ∠______  ......[From (I) and (II)]


In the figure, the centre of the circle is O and ∠STP = 40°.

  1. m (arc SP) = ? By which theorem?
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