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If a 32-inch chord is drawn in a circle of radius 20 inches, how far

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Bunuel
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If a 32-inch chord is drawn in a circle of radius 20 inches, how far [#permalink] New post   31 Jan 2016, 07:45
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E

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73% (01:29) correct 27% (01:51) wrong based on 110 sessions
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If a 32-inch chord is drawn in a circle of radius 20 inches, how far is the chord from the center of the circle?

A. 4 inches
B. 6 inches
C. 8 inches
D. 10 inches
E. 12 inches
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E
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Re: If a 32-inch chord is drawn in a circle of radius 20 inches, how far [#permalink] New post   31 Jan 2016, 13:14
OA=OC = 20 = Radius
AC= Length of chord = 32
OB = Distance of chord from center of circle.
In a circle, the perpendicular bisector of a chord passes through the center of the circle.
AB=BC = 16
OB^2 = OA^2 - AB^2 (Using pythagoras theorem )
= 20^2 - 16^2
= 400 - 256
= 144
=> OB = 12
Answer E
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Re: If a 32-inch chord is drawn in a circle of radius 20 inches, how far [#permalink] New post   17 Jul 2019, 02:51
Cord with center (O) will form a triangle. Draw a perpendicular from center & we will get 2 right angle triangle
OAB & ABC
OA= 20 (radius)
AB = 16 ( AC/2)
OB =12 ( avoid calculation by simply identifying 3-4-5 to 12-16-20)
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Re: If a 32-inch chord is drawn in a circle of radius 20 inches, how far [#permalink] New post   18 Aug 2022, 23:46
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Solution:

We have a circle with center O, radius 20 and a chord AB of length 32

Attachment:
circle2.png
circle2.png [ 10.28 KiB | Viewed 847 times ]


Let us draw a line OC perpendicular to chord AB from center O

We know that a perpendicular from the center to a chord will bisect the chord. This means \(AC=CB=\frac{32}{2}=16\)

Thus, we need the length \(OC\)

Attachment:
circle3.png
circle3.png [ 17.71 KiB | Viewed 829 times ]


Applying Pythagoras theorem on triangle \(OCB\), we can write \(OB^2=OC^2+CB^2\)

\(⇒ 20^2=OC^2+16^2 \)
\(⇒ 400=OC^2+256 \)
\(⇒ OC^2=400-256\)
\(⇒ OC^2=144\)
\(⇒ OC=12\)

Hence the right answer is Option E
GMAT Club Bot
gmatclubot
Re: If a 32-inch chord is drawn in a circle of radius 20 inches, how far [#permalink]
18 Aug 2022, 23:46

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