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In the figure above, O is the center of the circle. If AB has a length

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In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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New post 10 Jan 2019, 02:21
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In the figure above, O is the center of the circle. If AB has a length of 16 and OB has a length of 10, what is the length of CD ?


A. 2

B. 4

C. \(2\sqrt{3}\)

D. \(8-\sqrt{35}\)

E. \(8-\sqrt{39}\)


Attachment:
2019-01-10_1419.png
2019-01-10_1419.png [ 16.83 KiB | Viewed 273 times ]

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Re: In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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New post 10 Jan 2019, 03:49
Bunuel wrote:
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In the figure above, O is the center of the circle. If AB has a length of 16 and OB has a length of 10, what is the length of CD ?


A. 2

B. 4

C. \(2\sqrt{3}\)

D. \(8-\sqrt{35}\)

E. \(8-\sqrt{39}\)


In right triangle OBD, BD = (1/2)*AB = 8, AB = 10

Attachment:
2019-01-10_1419.png


i.e. OD = 6 (Using pythagorean triplet 6-8-10)

CD = CO - OD = 10-6 = 4

Answer: Option B
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Re: In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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New post 10 Jan 2019, 04:42
my answer is option B ie 4
OBD being a right angled triangle with OB =10 and BD=8, we get OD=6.
Therefore, CD=10-6=4
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Re: In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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New post 10 Jan 2019, 06:08

Solution


Given:
    • O is the center of the circle
    • AB = 16
    • OB = 10

To find:
    • The length of CD

Approach and Working:
OD bisects AB
    • Thus, \(DB = \frac{AB}{2} = \frac{16}{2} = 8\)

In triangle ODB,
    • \(OB^2 = OD^2 + DB^2\)
    • Implies, \(100 = OD^2 + 64\)
    • OD = 6

OC = radius = OD + CD
    • CD = 10 – 6 = 4

Hence, the correct answer is Option B

Answer: B

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Re: In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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New post 10 Jan 2019, 06:42
[quote="Bunuel"]Image
In the figure above, O is the center of the circle. If AB has a length of 16 and OB has a length of 10, what is the length of CD ?


A. 2

B. 4

C. \(2\sqrt{3}\)

D. \(8-\sqrt{35}\)

E. \(8-\sqrt{39}\)

oc=ob=oa=10
and od=6 triangle ado = 3:4:5 = 6:8:10
so dc= oc-od = 10-6 = 4
IMO B
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Re: In the figure above, O is the center of the circle. If AB has a length &nbs [#permalink] 10 Jan 2019, 06:42
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