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AB and CD are parallel chords of a circle 3 cm apart. If AB = 4 cm, CD = 10 cm, then what is the radius of the circle?
- a)7 cm
- b)√19 cm
- c)√29 cm
- d)14 cm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
AB and CD are parallel chords of a circle 3 cm apart. If AB = 4 cm, CD...
Let O be the centre of the circle and EF be a perpendicular drawn to the lines from O
Let the radius of the circle be r cm
Perpendicular drawn from the center bisects the chord
∴ AF = FB = AB/2 = 2 cm
DE = EC = CD/2 = 5 cm
In triangle AFO
⇒ OF = √(AO2 - AF2) [∵ Pythagoras theorem]
⇒ OF = √(r2 - 4)
In triangle OEC
⇒ OE = √(CO2 - CE2)
⇒ OE = √(r2 - 25)
⇒ EF = OE + OF
⇒ 3 = √(r2 - 4) + √(r2 - 25)
⇒ 3 - √(r2 - 25) = √(r2 - 4)
Squaring on both the sides
⇒ 9 + r2 - 25 - 6√(r2 - 25) = r2 - 4
⇒ - 12 = - 6√(r2 - 25)
⇒ 2 = √(r2 - 25)
⇒ 4 = r2 - 25
⇒ r2 = 29
∴ r = √29 cm
∴ AF = FB = AB/2 = 2 cm
DE = EC = CD/2 = 5 cm
In triangle AFO
⇒ OF = √(AO2 - AF2) [∵ Pythagoras theorem]
⇒ OF = √(r2 - 4)
In triangle OEC
⇒ OE = √(CO2 - CE2)
⇒ OE = √(r2 - 25)
⇒ EF = OE + OF
⇒ 3 = √(r2 - 4) + √(r2 - 25)
⇒ 3 - √(r2 - 25) = √(r2 - 4)
Squaring on both the sides
⇒ 9 + r2 - 25 - 6√(r2 - 25) = r2 - 4
⇒ - 12 = - 6√(r2 - 25)
⇒ 2 = √(r2 - 25)
⇒ 4 = r2 - 25
⇒ r2 = 29
∴ r = √29 cm
Most Upvoted Answer
AB and CD are parallel chords of a circle 3 cm apart. If AB = 4 cm, CD...
We can draw a diagram to visualize the problem:
Let O be the center of the circle and let E be the midpoint of AB and CD. Since AB and CD are parallel, OE is perpendicular to both AB and CD.
Since OE is perpendicular to AB, it bisects AB into two equal parts. Therefore, AE = BE = 4/2 = 2 cm.
Similarly, OE is perpendicular to CD, so it bisects CD into two equal parts. Therefore, CE = DE = 10/2 = 5 cm.
We can draw radii OA, OB, OC, and OD. Since OE is perpendicular to AB and CD, it is also perpendicular to OA, OB, OC, and OD.
Since AE = BE = 2 cm, and CE = DE = 5 cm, we can see that triangles AEO and CEO are right triangles with sides in the ratio of 2:5.
Let r be the radius of the circle. Then, AO = BO = r, and CO = DO = r + 3 cm.
By the Pythagorean theorem, we have:
AE^2 + EO^2 = AO^2
2^2 + (r + 3)^2 = r^2
4 + r^2 + 6r + 9 = r^2
6r + 13 = 0
6r = -13
r = -13/6
Since the radius of a circle cannot be negative, we discard this solution.
Therefore, there is no real solution for the radius of the circle given the given information.
Let O be the center of the circle and let E be the midpoint of AB and CD. Since AB and CD are parallel, OE is perpendicular to both AB and CD.
Since OE is perpendicular to AB, it bisects AB into two equal parts. Therefore, AE = BE = 4/2 = 2 cm.
Similarly, OE is perpendicular to CD, so it bisects CD into two equal parts. Therefore, CE = DE = 10/2 = 5 cm.
We can draw radii OA, OB, OC, and OD. Since OE is perpendicular to AB and CD, it is also perpendicular to OA, OB, OC, and OD.
Since AE = BE = 2 cm, and CE = DE = 5 cm, we can see that triangles AEO and CEO are right triangles with sides in the ratio of 2:5.
Let r be the radius of the circle. Then, AO = BO = r, and CO = DO = r + 3 cm.
By the Pythagorean theorem, we have:
AE^2 + EO^2 = AO^2
2^2 + (r + 3)^2 = r^2
4 + r^2 + 6r + 9 = r^2
6r + 13 = 0
6r = -13
r = -13/6
Since the radius of a circle cannot be negative, we discard this solution.
Therefore, there is no real solution for the radius of the circle given the given information.
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AB and CD are parallel chords of a circle 3 cm apart. If AB = 4 cm, CD = 10 cm, then what is the radius of the circle?a)7 cmb)√19 cmc)√29 cmd)14 cmCorrect answer is option 'C'. Can you explain this answer?
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