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805+ Level|   Geometry|      
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Bunuel
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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 30 Mar 2020, 00:57
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45% (02:58) correct 55% (03:26) wrong based on 62 sessions

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Competition Mode Question



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

A. 0.5
B. 1.5
C. 2
D. 2.5
E. 3.5


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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 30 Mar 2020, 03:26
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Bunuel

Competition Mode Question



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

A. 0.5
B. 1.5
C. 2
D. 2.5
E. 3.5
Show more

Rule: For any 2 chords AB & CD of a circle intersecting at point E, \(AE*BE\) = \(CE*DE\)

From the figure, \(AE\) = \(x\), \(BE\) = \(20.5 - x\), \(CE\) = \(7\) & \(DE\) = \(15\)
--> \(AE*BE\) = \(7*15\) = \(105\)
& \(AE\) + \(BE\) = \(20.5\)

--> \((AE - BE)^2 = (AE + BE)^2 - 4*AE*BE\)
--> \((AE - BE)^2 = (20.5)^2 - 4*105 = 420.25 - 420 = 0.25\)
--> \(AE - BE = \sqrt{0.25} = 0.5\)

Option A
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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 30 Mar 2020, 04:52
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The two chords AB & CD (diameter) intersects at E, which means: AE*BE=CE*DE

Assuming AE=x and BE=20.5-x,
AE*BE = CE*DE = 7*15 = 105
x*(20.5-x) = 10*10.5, or
x*(20.5-x) = 10.5*10

The difference of the lengths of BE and AE is |AE-BE| = |20.5-2x|.
Whether x=10 or x=10.5, |BE-AE| is always 0.5cm

FINAL ANSWER IS (A)

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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 30 Mar 2020, 06:19
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Quote:
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

A. 0.5
B. 1.5
C. 2
D. 2.5
E. 3.5
Show more

CE=7, DE=22-7=15
AE=x, BE=20.5-x
CE*DE=AE*BE, 7*15=x(20.5-x)
\(x^2-20.5x+105=0…2x^2-41x+210=0\)
\(2x^2-20x-21x+210=0…2x(x-10)-21(x-10)=0\)
\((2x-21)(x-10)=0…x=(10.5,10)\)
AE=20.5-10.5=10, BE=20.5-10=10.5
|AE-BE|=0.5

Ans (A)
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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 27 Aug 2022, 18:21
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Dillesh4096
Bunuel

Competition Mode Question



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

A. 0.5
B. 1.5
C. 2
D. 2.5
E. 3.5

Rule: For any 2 chords AB & CD of a circle intersecting at point E, \(AE*BE\) = \(CE*DE\)

From the figure, \(AE\) = \(x\), \(BE\) = \(20.5 - x\), \(CE\) = \(7\) & \(DE\) = \(15\)
--> \(AE*BE\) = \(7*15\) = \(105\)
& \(AE\) + \(BE\) = \(20.5\)

--> \((AE - BE)^2 = (AE + BE)^2 - 4*AE*BE\)
--> \((AE - BE)^2 = (20.5)^2 - 4*105 = 420.25 - 420 = 0.25\)
--> \(AE - BE = \sqrt{0.25} = 0.5\)

Option A
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hi dilesh can you explain how you get to:
--> (AE−BE)2=(AE+BE)2−4∗AE∗BE(AE−BE)2=(AE+BE)2−4∗AE∗BE
--> (AE−BE)2=(20.5)2−4∗105=420.25−420=0.25(AE−BE)2=(20.5)2−4∗105=420.25−420=0.25
--> AE−BE=0.25−−−−√=0.5
here?
i dont get where the equation is coming from
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In a circle of radius 11 cm, CD is a diameter and AB is a chord of len 31 Dec 2023, 12:35
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rdrdrd1201
Dillesh4096
Bunuel

Competition Mode Question



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

A. 0.5
B. 1.5
C. 2
D. 2.5
E. 3.5

Rule: For any 2 chords AB & CD of a circle intersecting at point E, \(AE*BE\) = \(CE*DE\)

From the figure, \(AE\) = \(x\), \(BE\) = \(20.5 - x\), \(CE\) = \(7\) & \(DE\) = \(15\)
--> \(AE*BE\) = \(7*15\) = \(105\)
& \(AE\) + \(BE\) = \(20.5\)

--> \((AE - BE)^2 = (AE + BE)^2 - 4*AE*BE\)
--> \((AE - BE)^2 = (20.5)^2 - 4*105 = 420.25 - 420 = 0.25\)
--> \(AE - BE = \sqrt{0.25} = 0.5\)

Option A

hi dilesh can you explain how you get to:
--> (AE−BE)2=(AE+BE)2−4∗AE∗BE(AE−BE)2=(AE+BE)2−4∗AE∗BE
--> (AE−BE)2=(20.5)2−4∗105=420.25−420=0.25(AE−BE)2=(20.5)2−4∗105=420.25−420=0.25
--> AE−BE=0.25−−−−√=0.5
here?
i dont get where the equation is coming from
Show more

Think of AE as x and BE as y, where \(x + y = 20.5\); what we're looking for is \(|x-y|\). Now remember the properties that:

\((x - y)^2 = x^2 - 2xy + y^2\)
\((x + y)^2 = x^2 + 2xy + y^2\)

To get \((x - y)^2\) from \((x + y)^2\), we have to subtract \(4xy\) to get:

\((x + y)^2 - 4xy = x^2 + 2xy + y^2 - 4xy = x^2 - 2xy + y^2 = (x - y)^2\).

Therefore, we can solve as follows:

\((x - y)^2 = (x + y)^2 - 4xy\)
-> \((x - y)^2 = (20.5)^2 - 4(105)\) (because \(AE * BE = CE * DE = 7 * 15 = 105\) by the Intersecting Chords Theorem)
-> \((x - y)^2 = 0.25\)
-> \((x - y) = \sqrt{0.25} = 0.5\) or \(-0.5\)
-> \(|x - y| = 0.5\)

Answer choice A. A great solution by Bunuel, hope this explanation helps you, it took me a while to also understand this question!
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