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In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is
(2019)
- a)1.5
- b)2.5
- c)3.5
- d)0.5
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a circle of radius 11 cm, CD is a diameter and AB is a chord of len...
In the figure, AB and CD are two chords of a circle, intersects each other at point E.
∴ AE × BE = CE × DE
⇒ AE × BE = 7 × 15 = 105
(AE – BE)2 = (AE + BE)2 – 4 × AE × BE = (AB)2 – 4
× 105
⇒ (AE – BE)2 = (20.5)2 – 420
⇒ (AE – BE)2 = 420.25 – 420 = 0.25
⇒ AE – BE = 0.5
Most Upvoted Answer
In a circle of radius 11 cm, CD is a diameter and AB is a chord of len...
To solve this problem, let's start by drawing a diagram:
1. Draw a circle with a radius of 11 cm.
2. Mark the center of the circle as O.
3. Draw a diameter CD.
4. Draw a chord AB with a length of 20.5 cm.
5. Draw the perpendicular bisector of AB, which intersects CD at point E.
6. Mark the point where CE intersects AB as F.
Finding the length of BE and AE:
We can use the property of perpendicular bisectors to find the lengths of BE and AE.
1. Since the perpendicular bisector of AB passes through the center of the circle, it bisects the chord AB. Therefore, AF = FB = 20.5/2 = 10.25 cm.
2. As CE is perpendicular to AB, triangle CEF is a right triangle. We can use the Pythagorean theorem to find the length of EF.
CE^2 = EF^2 + CF^2
7^2 = EF^2 + 10.25^2
49 = EF^2 + 105.0625
EF^2 = 49 - 105.0625
EF^2 = -56.0625 (which is not possible)
Since we obtained a negative value for EF^2, it means that the triangle CEF is not possible. This implies that point E does not lie inside the circle.
Hence, there is no solution to this problem. The answer is not D.
1. Draw a circle with a radius of 11 cm.
2. Mark the center of the circle as O.
3. Draw a diameter CD.
4. Draw a chord AB with a length of 20.5 cm.
5. Draw the perpendicular bisector of AB, which intersects CD at point E.
6. Mark the point where CE intersects AB as F.
Finding the length of BE and AE:
We can use the property of perpendicular bisectors to find the lengths of BE and AE.
1. Since the perpendicular bisector of AB passes through the center of the circle, it bisects the chord AB. Therefore, AF = FB = 20.5/2 = 10.25 cm.
2. As CE is perpendicular to AB, triangle CEF is a right triangle. We can use the Pythagorean theorem to find the length of EF.
CE^2 = EF^2 + CF^2
7^2 = EF^2 + 10.25^2
49 = EF^2 + 105.0625
EF^2 = 49 - 105.0625
EF^2 = -56.0625 (which is not possible)
Since we obtained a negative value for EF^2, it means that the triangle CEF is not possible. This implies that point E does not lie inside the circle.
Hence, there is no solution to this problem. The answer is not D.
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In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is(2019)a)1.5b)2.5c)3.5d)0.5Correct answer is option 'D'. Can you explain this answer?
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