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# Additional Questions Solutions- Circles Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## Class 9 : Additional Questions Solutions- Circles Class 9 Notes | EduRev

The document Additional Questions Solutions- Circles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Question 1. In the figure, OD is perpendicular to chord AB of a circle whose centre is O. If BC is a diameter, prove that CA = 2OD.
Solution:
We have a circle whose centre is O. BC is a diameter and AB a chord such that OD âŠ¥ AB. Let us join AC.
âˆµ The perpendicular from centre of a circle to a chord bisects the chord

âˆ´ D is the mid-point of AB
âˆµ O is the mid-point of the diameter BC
âˆ´ OD is the line segment joining the mid-points of two sides of Î”ABC.
âˆ´ OD is half of the third side of Î”ABC.

i.e. OD =(1/2)AC
or 2OD = AC

Question 2. l is a line intersecting two concentric circles having common centre O, at A, B, C and D. Prove that AB = CD.
Solution:
We have two concentric circles with common centre O.
Line l intersects these circles at A, B, C and D.
Let us draw OP âŠ¥ â„“ For the inner circle

âˆµ OP âŠ¥ BC                 [construction]
âˆ´ P is the mid-point of BC                 [âˆµ Perpendicular from the centre to a chord bisects the chord]
âˆ´ PB = PC                 â€¦(1)
For the outer circle.

âˆ´ PA = PD                ..(2)
Subtracting (1) from (2), we have PA - PB = PD - PC
â‡’ AB = CD

Question 3. AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED.

Solution: We have a circle with centre O. Chord AB = chord CD and on production, AB and CD meet at E.

Let us join OE and draw OP âŠ¥ AB and OQ âŠ¥ CD.
âˆµ In a circle, equal chords are equidistant from the centre.
âˆ´ OP = OQ.                [âˆµ AB = CD]
Now, in right Î”OPE and right Î”OQE, we have OP = OQ                 [Proved],
OE = OE                [Common]
âˆµ Using RHS criterion, Î”OPE â‰Œ Î”OQE
â‡’ P E = QE                 [c.p.c.t.]           â€¦(1)
But AB = CD                [Given]
â‡’ (1/2) AB = (1/2)CD or PB = QD                 [âˆµ OP âŠ¥ AB and OQ âŠ¥ CD] â€¦(2)
Subtracting (2) from (1), we have PE âˆ  PB = QE âˆ  QD
â‡’ EB = ED

Question 4. If O be the centre of the circle, find the value of â€˜xâ€™ in each of the following figures.

Solution: (i) âˆµ OA = OB                 [Radii of the same circle]
âˆ´ âˆ A = âˆ B                 [Angles opposite to equal side in a triangle are equal]
In Î”ABC, âˆ A + âˆ B + âˆ O = 180Âº
âˆ´ x + x + 70Âº = 180Âº                 [âˆµ âˆ O = 70Âº (given) and âˆ A = âˆ B]
â‡’ 2x + 70Âº = 180Âº
â‡’ 2x = 180Âº - 70Âº = 110Âº
â‡’ x= (1100/2)= 55Âº
Thus, x = 55Â°

(ii) In Î”AOC, âˆ A + âˆ ACO + âˆ AOC = 180Âº
â‡’ 40Âº + âˆ ACO + 90Âº = 180Âº
â‡’ âˆ ACO = 180Âº - 40Âº - 90Âº = 50Âº
âˆµ AB is a diameter.
âˆ´ âˆ ACB = 90Âº                 [Angle in a semicircle]
âˆ´ 50Âº + x = 90Âº
â‡’ xÂº = 90Âº - 50Âº = 40Âº
Thus, x = 40Âº

(iii) âˆµ âˆ AOC + âˆ COB = 180Âº                 [Linear pairs]
âˆ´ 120Âº + âˆ COB = 180Âº
â‡’ âˆ COB = 180Âº - 120Âº = 60Âº
âˆµ The arc CB is subtending âˆ COB at the centre and âˆ CDB at the remaining part.
âˆ´ âˆ CDB = (1/2)âˆ COB
â‡’ x= (1/2)(60Âº) = 30Âº
â‡’ x = 30Âº

(iv) In Î”AOC,
âˆµ AO = OC                 [Radii of the same circle]
âˆ´ âˆ OAC = âˆ OCA                [Angles opposite to equal sides are equal]
â‡’ âˆ OAC = 50Âº
âˆ´ Exterior âˆ COB = 50Âº + 50Âº = 100Âº.
Now, the arc BC is subtending âˆ BOC at the centre and âˆ BDC at the remaining part of the circle.
âˆ´ âˆ BDC = (1/2)âˆ BOC
â‡’ x= (1/2)(100Âº) = 50Âº
Thus, x = 50Âº

(v) In Î”OAC, OA = OC                 [Radii of the same circle]
âˆ´ âˆ AOC = âˆ ACO                [âˆµ Angles opposite to equal sides are equal]
Now, âˆ AOC + âˆ ACO + âˆ OAC = 180Âº
â‡’ âˆ AOC + âˆ ACO + 50Âº = 180Âº
â‡’ âˆ AOC + âˆ ACO = 180Âº - 50Âº = 130Âº
â‡’ âˆ AOC = âˆ ACO = (1300/2) = 65Âº
Now, âˆ AOB + âˆ AOC = 180Âº                [Linear pairs]
âˆ´ âˆ AOB + 65Âº = 180Âº
â‡’ âˆ AOB = 180Âº - 65Âº = 125Âº
âˆµ The arc AB subtends âˆ AOB at the centre and âˆ ADB at the remaining part of the circle.
âˆ´ âˆ ADB = (1/2) âˆ AOB
â‡’ x= (1/2)(125Âº) = 62 (1/2)
âˆ´ x= 62(1/2)

(vi) âˆµ âˆ BDC = âˆ BAC                [Angles in the same segment]
âˆ´ âˆ BDC = 40Âº
Now, in Î”BDC, we have âˆ BDC + âˆ CBD + âˆ BCD = 180Âº
âˆ´ 40Âº + 80Âº + x = 180Âº
â‡’ 120Âº + x = 180Âº
â‡’ x = 180Âº - 120Âº = 60Âº
Thus, x = 60Âº

Question 5. In the adjoining figure, O is the centre of the circle. Prove that âˆ  XOZ = 2(âˆ  XZY + âˆ  YXZ).
Solution:
Let us join OY.
âˆµ The arc XY subtends âˆ XOY at the centre and âˆ XZY at a point Z on the remaining part of the circle.

âˆ´ âˆ XOY = 2âˆ XZY                â€¦(1)
Similarly, âˆ YOZ = 2âˆ YXZ                â€¦(2)
Adding (1) and (2), we have âˆ XOY + âˆ YOZ = 2âˆ XZY + 2âˆ YXZ
â‡’ âˆ XOZ = 2[âˆ XZY + âˆ YXZ]

Question 6. Show that the sum of the opposite angles of a cyclic quadrilateral is 180Âº.
Solution:
We have a cyclic quadrilateral. Let us join AC and BD. Since, angles in the same segment are equal.
âˆ´ âˆ ACB = âˆ ADB                â€¦(1)
and âˆ BAC = âˆ BDC                â€¦(2)
Adding (1) and (2), we have

âˆ ACB + âˆ BAC = âˆ ADB + âˆ BDC
â‡’ âˆ ACB + âˆ BAC = âˆ ADC
Adding âˆ ABC to both sides, we have âˆ ACB + âˆ BAC + âˆ ABC = âˆ ADC + âˆ ABC
But, âˆ ACB + âˆ BAC + âˆ ABC = 180Âº                 [Sum of the angles of Î”ABC = 180Âº]
âˆ´ âˆ ADC + âˆ ABC = 180Âº
â‡’ âˆ B + âˆ D = 180Âº
Since, âˆ A + âˆ B + âˆ C + âˆ D = 360Âº
â‡’ âˆ A + âˆ C = 360Âº âˆ  180Âº = 180Âº

Question 7. Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
Solution:
We have a cyclic quadrilateral ABCD in which the bisectors âˆ A, âˆ B, âˆ C and âˆ D for a quadrilateral PQRS.
From Î”ABP, we have âˆ PAB + âˆ PBA + âˆ P = 180Âº                [Sum of the three angles of Î”ABP]

â‡’ (1/2) âˆ A + (1/2)âˆ B + âˆ P = 180Âº                â€¦(1)
From Î”CDR, we have
âˆ RCD + âˆ RDC + âˆ R = 180Âº                [Sum of the three angles of Î”CDR.]
â‡’ (1/2)âˆ C +(1/2)âˆ D + âˆ R = 180Â°                 â€¦(2)
Adding (1) and (2), we have (1/2)âˆ A + (1/2)âˆ B + (1/2)âˆ C +(1/2)âˆ D + âˆ P + âˆ R = 360Âº
â‡’ (1/2) (âˆ A + âˆ B + âˆ C + âˆ D) + âˆ P + âˆ R = 360Âº
â‡’ (1/2)(360Âº) + âˆ P + âˆ R = 360Âº                 [âˆµ âˆ A + âˆ B + âˆ C + âˆ D = 360Â°]
â‡’ âˆ P + âˆ R = 360Âº -(1/2) (360Âº) = 180Âº
Similarly, âˆ Q + âˆ S = 180Âº
Thus, the pairs of opposite angles of quadrilateral PQRS are supplementary.
Hence, PQRS is cyclic.

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