Page 1 Question:68 The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is a 11.5 cm b 12 cm c v 69 cm d 23 cm Solution: b 12 cm Let AB be the chord of the given circle with centre O and a radius of 13 cm. Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. ? BM = 10 2 cm = 5cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 13 2 = OM 2 + 5 2 ? 169 = OM 2 + 25 ? OM 2 = 169 25 = 144 ? OM = v 144cm = 12cm Hence, the distance of the chord from the centre is 12 cm. ( ) Page 2 Question:68 The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is a 11.5 cm b 12 cm c v 69 cm d 23 cm Solution: b 12 cm Let AB be the chord of the given circle with centre O and a radius of 13 cm. Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. ? BM = 10 2 cm = 5cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 13 2 = OM 2 + 5 2 ? 169 = OM 2 + 25 ? OM 2 = 169 25 = 144 ? OM = v 144cm = 12cm Hence, the distance of the chord from the centre is 12 cm. ( ) Question:69 A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is a 25 cm b 12.5 cm c 30 cm d 9 cm Solution: c 30 cm Let AB be the chord of the given circle with centre O and a radius of 17 cm. From O, draw OM perpendicular to AB. Then OM = 8 cm and OB = 17 cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 17 2 = 8 2 + MB 2 ? 289 = 64 + MB 2 ? MB 2 = 289 64 = 225 ? MB = v 225cm = 15cm The perpendicular from the centre of a circle to a chord bisects the chord. ? AB = 2 × MB = 2x15 cm = 30 cm Hence, the required length of the chord is 30 cm. Question:70 In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ? a 30° b 45° c 60° d 90° Solution: b 45° Since an angle in a semicircle is a right angle, ?BAC = 90° ? ?ABC + ?ACB = 90° Now, AB = AC Given Page 3 Question:68 The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is a 11.5 cm b 12 cm c v 69 cm d 23 cm Solution: b 12 cm Let AB be the chord of the given circle with centre O and a radius of 13 cm. Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. ? BM = 10 2 cm = 5cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 13 2 = OM 2 + 5 2 ? 169 = OM 2 + 25 ? OM 2 = 169 25 = 144 ? OM = v 144cm = 12cm Hence, the distance of the chord from the centre is 12 cm. ( ) Question:69 A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is a 25 cm b 12.5 cm c 30 cm d 9 cm Solution: c 30 cm Let AB be the chord of the given circle with centre O and a radius of 17 cm. From O, draw OM perpendicular to AB. Then OM = 8 cm and OB = 17 cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 17 2 = 8 2 + MB 2 ? 289 = 64 + MB 2 ? MB 2 = 289 64 = 225 ? MB = v 225cm = 15cm The perpendicular from the centre of a circle to a chord bisects the chord. ? AB = 2 × MB = 2x15 cm = 30 cm Hence, the required length of the chord is 30 cm. Question:70 In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ? a 30° b 45° c 60° d 90° Solution: b 45° Since an angle in a semicircle is a right angle, ?BAC = 90° ? ?ABC + ?ACB = 90° Now, AB = AC Given ? ?ABC = ?ACB = 45° Question:71 In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ? a 30° b 15° c 60° d 90° Figure Solution: c 60° We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference. Thus, ?AOB = 2 × ?ACB = 2 ×30° = 60° Question:72 In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ? a 40° b 50° c 80° d 100° Solution: b 50° OA = OB ? ?OBA = ?OAB = 40° Now, ?AOB = 180°  40°+40° = 100° ? ?ACB = 1 2 ?AOB = 1 2 ×100 ° = 50° Question:73 In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is a 8 cm b 15 cm ( ) Page 4 Question:68 The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is a 11.5 cm b 12 cm c v 69 cm d 23 cm Solution: b 12 cm Let AB be the chord of the given circle with centre O and a radius of 13 cm. Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. ? BM = 10 2 cm = 5cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 13 2 = OM 2 + 5 2 ? 169 = OM 2 + 25 ? OM 2 = 169 25 = 144 ? OM = v 144cm = 12cm Hence, the distance of the chord from the centre is 12 cm. ( ) Question:69 A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is a 25 cm b 12.5 cm c 30 cm d 9 cm Solution: c 30 cm Let AB be the chord of the given circle with centre O and a radius of 17 cm. From O, draw OM perpendicular to AB. Then OM = 8 cm and OB = 17 cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 17 2 = 8 2 + MB 2 ? 289 = 64 + MB 2 ? MB 2 = 289 64 = 225 ? MB = v 225cm = 15cm The perpendicular from the centre of a circle to a chord bisects the chord. ? AB = 2 × MB = 2x15 cm = 30 cm Hence, the required length of the chord is 30 cm. Question:70 In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ? a 30° b 45° c 60° d 90° Solution: b 45° Since an angle in a semicircle is a right angle, ?BAC = 90° ? ?ABC + ?ACB = 90° Now, AB = AC Given ? ?ABC = ?ACB = 45° Question:71 In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ? a 30° b 15° c 60° d 90° Figure Solution: c 60° We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference. Thus, ?AOB = 2 × ?ACB = 2 ×30° = 60° Question:72 In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ? a 40° b 50° c 80° d 100° Solution: b 50° OA = OB ? ?OBA = ?OAB = 40° Now, ?AOB = 180°  40°+40° = 100° ? ?ACB = 1 2 ?AOB = 1 2 ×100 ° = 50° Question:73 In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is a 8 cm b 15 cm ( ) c 18 cm d 6 cm Solution: a 8 cm Join OC. Then OC = radius = 17 cm CL = 1 2 CD = 1 2 ×30 cm = 15cm In right ?OLC, we have: OL 2 = OC 2  CL 2 = 17 2  15 2 = 289 225 = 64 ? OL = v 64 = 8cm ? Distance of CD from AB = 8 cm Question:74 AB and CD are two equal chords of a circle with centre O such that ?AOB = 80°, then ?COD = ? a 100° b 80° c 120° d 40° Solution: b 80° Given: AB = CD We know that equal chords of a circle subtend equal angles at the centre. ? ?COD = ?AOB = 80° Question:75 ( ) Page 5 Question:68 The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is a 11.5 cm b 12 cm c v 69 cm d 23 cm Solution: b 12 cm Let AB be the chord of the given circle with centre O and a radius of 13 cm. Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB. We know that the perpendicular from the centre of a circle to a chord bisects the chord. ? BM = 10 2 cm = 5cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 13 2 = OM 2 + 5 2 ? 169 = OM 2 + 25 ? OM 2 = 169 25 = 144 ? OM = v 144cm = 12cm Hence, the distance of the chord from the centre is 12 cm. ( ) Question:69 A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is a 25 cm b 12.5 cm c 30 cm d 9 cm Solution: c 30 cm Let AB be the chord of the given circle with centre O and a radius of 17 cm. From O, draw OM perpendicular to AB. Then OM = 8 cm and OB = 17 cm From the right ?OMB, we have: OB 2 = OM 2 + MB 2 ? 17 2 = 8 2 + MB 2 ? 289 = 64 + MB 2 ? MB 2 = 289 64 = 225 ? MB = v 225cm = 15cm The perpendicular from the centre of a circle to a chord bisects the chord. ? AB = 2 × MB = 2x15 cm = 30 cm Hence, the required length of the chord is 30 cm. Question:70 In the given figure, BOC is a diameter of a circle and AB = AC. Then, ?ABC = ? a 30° b 45° c 60° d 90° Solution: b 45° Since an angle in a semicircle is a right angle, ?BAC = 90° ? ?ABC + ?ACB = 90° Now, AB = AC Given ? ?ABC = ?ACB = 45° Question:71 In the given figure, O is the centre of a circle and ?ACB = 30°. Then, ?AOB = ? a 30° b 15° c 60° d 90° Figure Solution: c 60° We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference. Thus, ?AOB = 2 × ?ACB = 2 ×30° = 60° Question:72 In the given figure, O is the centre of a circle. If ?OAB = 40° and C is a point on the circle, then ?ACB = ? a 40° b 50° c 80° d 100° Solution: b 50° OA = OB ? ?OBA = ?OAB = 40° Now, ?AOB = 180°  40°+40° = 100° ? ?ACB = 1 2 ?AOB = 1 2 ×100 ° = 50° Question:73 In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is a 8 cm b 15 cm ( ) c 18 cm d 6 cm Solution: a 8 cm Join OC. Then OC = radius = 17 cm CL = 1 2 CD = 1 2 ×30 cm = 15cm In right ?OLC, we have: OL 2 = OC 2  CL 2 = 17 2  15 2 = 289 225 = 64 ? OL = v 64 = 8cm ? Distance of CD from AB = 8 cm Question:74 AB and CD are two equal chords of a circle with centre O such that ?AOB = 80°, then ?COD = ? a 100° b 80° c 120° d 40° Solution: b 80° Given: AB = CD We know that equal chords of a circle subtend equal angles at the centre. ? ?COD = ?AOB = 80° Question:75 ( ) In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circles is a 6 cm b 9 cm c 7.5 cm d 8 cm Solution: c 7.5 cm Let OA = OC = r cm. Then OE = (r  3) cm and AE = 1 2 AB = 6cm Now, in right ?OAE, we have: OA 2 = OE 2 +AE 2 ? (r) 2 = (r  3) 2 + 6 2 ? r 2 = r 2 + 9  6r + 36 ? 6r = 45 ? r = 45 6 = 7. 5 cm Hence, the required radius of the circle is 7.5 cm. Question:76 In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is a 10 cm b 12 cm c 6 cm d 8 cm Solution: a 10 cm Let the radius of the circle be r cm. Let OD = OB = r cm. Then OE = (r  4) cm and ED = 8 cm Now, in right ?OED, we have:Read More
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