The document Short Answer Type Questions- Circles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Question 1. The diameter of circle is 3.8 cm. Find the length of its radius. Solution: **Since, the diameter of circle is double its radius.

âˆ´ Diameter = 2 x Radius

â‡’ (1/2)x Diameter = Radius

â‡’ Radius = (1/2) x 3.8 cm

= 1/2 x 38/10 cm

= 19/10 = 1.9 cm

**Question 2. In the adjoining figure, O is the centre of the circle. The chord AB = 10 cm is such that OP âŠ¥ AB. Find the length of AP. Solution:** âˆµ OP âŠ¥ AB

âˆ´ P is the mid-point of AB.

â‡’ AP =(1/2)AB

â‡’ AP = (1/2)x 10 cm = 5 cm

**Question 3. In the figure AOC is a diameter of the circle and arc AXB = (1/2)**** arc BYC. Find âˆ BOC. Solution: **âˆµ arc AXB = (1/2)arc BYC

âˆ´ âˆ AOB = (1/2) âˆ BOC

Also âˆ AOB + âˆ BOC = 180Â°

â‡’ (1/2)âˆ BOC + âˆ BOC = 180Â°

â‡’ (3/2) âˆ BOC = 180Â°

â‡’ âˆ BOC = (1/2) Ã— 180Â° = 120Â°

**Question 4. In the figure âˆ ABC = 45Â°. Prove that OA âŠ¥ OC. Solution: **Since, the angle subtended at the centre by an arc is double the angle subtended by it at any other point on the remaining part of the circle.

âˆ´ âˆ ABC =(1/2)âˆ AOC

â‡’ âˆ AOC = 2 âˆ ABC = 2 Ã— 45Â° = 90Â° [âˆµ âˆ ABC = 45Â°]

Thus, OA âŠ¥ OC.

**Question 5. Look at the adjoining figure, in which O is the centre of the circle. If AB = 8 cm and OP = 3 cm, then find the radius of the circle. Solution: **âˆµ OP âŠ¥ AB

âˆ´ P is the mid-point of AB.

â‡’ AP = (1/2)AB = (1/2)x 8 cm = 4 cm

Now, in right Î”OPA, Radius, OA = = 5 cm

**Question 6. In the adjoining figure, O is the centre of the circle. Find the length of AB. Solution:** Since chord AB and chord CD subtend equal angles at the centre,

i.e. âˆ AOB = âˆ COD [Each = 60Â°]

âˆ´ Chord AB = Chord CD

â‡’ Chord AB = 5 cm [âˆµ Chord CD = 5 cm]

Thus, the required lenth of chord AB is 5 cm.

**Question 1. In the adjoining figure, O is the centre of the circle and OP = OQ. If AP = 4 cm, then find the length of CD. Solution:** âˆµ OP = OQ

âˆ´ Chord AB and chord CD are equidistant from the centre.

Thus, the required length of CD is 8 cm.

**Question 2. AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that AB = 24 cm and CD = 10 cm and the distance between AB and CD is 17 cm. Find the radius of the circle. Solution: **âˆµ Perpendicular from the centre to a chord bisects the chord.

âˆ´ AP = (1/2)AB = (1/2)x 24 cm

= 12 cm

Similarly, CQ = (1/2)CD =(1/2)x 10 cm = 5 cm

Let OP = x cm

â‡’ OQ = (17 - x) cm

Now, in right Î”APO, x^{2 }+ 12^{2} = OA^{2} ...(1)

Again, in right Î”COQ, OC^{2 }= CQ^{2} + (17 - x)^{2}

= 5^{2 }+ 17^{2} + x^{2} - 34x

= 25 + 289 + x^{2} - 34x = 314 + x^{2} - 34x ...(2)

From (1) and (2),

we have x^{2} + 314 - 34x = x^{2} + 12^{2}

â‡’ x^{2} - x^{2} - 34x = 144 - 314

â‡’ - 34x = -170

â‡’ x= (-170/34) = 5

Now from (1), we have OA^{2 }= 5^{2} + 12^{2} = 25 + 144

â‡’ OA^{2 }= 169

â‡’ OA = 13 cm

Thus, the required radius of the circle = 13 cm.

**Question 3. If O is the centre of the circle, then find the value of x. Solution:** âˆµ AOB is a diameter.

âˆ´ âˆ AOC + âˆ COB = 180Â° [Linear pairs]

â‡’ 130Â° + âˆ COB = 180Â°

â‡’ âˆ COB = 180Â° - 130Â° = 50Â°

Now, 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‡’ âˆ CDB = (1/2)x 50Â° = 25Â°

Thus, the measure of x = 25Â°

**Question 4. The radius of a circle is 17 cm. A chord of length 30 cm is drawn. Find the distance of the chord from the centre. Solution:** Length of chord AB = 30 cm.

Since, OP âŠ¥ AB

âˆ´ P is the mid-point of AB.

AP = (1/2)AB =(1/2) x 30 cm = 15 cm

Now, in right Î”APO, AO^{2} = AP^{2} + OP^{2}

â‡’ 17^{2} = 15^{2 }+ OP^{2}

âˆ´ OP^{2} = 17^{2} - 15^{2} = (17 - 15)(17 + 15)

= 2 x 32 = 64

â‡’ OP = 64 = 8 cm

âˆ´ The distance of the chord AB from the centre O is 8 cm.

**Question 5. Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 5 cm. Solution:** âˆµ The perpendicular distance, OP = 4 cm

âˆ´ In right Î”APO, AO^{2} = AP^{2} + OP^{2}

â‡’ 5^{2} = AP^{2 }+ 4^{2}

â‡’ AP^{2} = 5^{2 }- 4^{2 }= (5 - 4)(5 + 4)

= 1 x 9 = 9

â‡’ AP = âˆš9= 3 cm

Since, the perpendicular from the centre to a chord of a circle divides the chord into two equal parts.

âˆ´ AP = (1/2) AB

â‡’ AB = 2AP

â‡’ AB = 2 x 3 cm = 6 cm

Thus, the required length of chord AB is 6 cm.

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