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

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Joined: 02 Sep 2009
Posts: 55803
In the figure above, O is the center of the circle. If AB has a length  [#permalink]

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10 Jan 2019, 03:21
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Difficulty:

25% (medium)

Question Stats:

94% (01:23) correct 6% (01:43) wrong based on 24 sessions

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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 [ 16.83 KiB | Viewed 384 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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10 Jan 2019, 04:49
Bunuel wrote:

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

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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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10 Jan 2019, 05:42
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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10 Jan 2019, 07:08

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

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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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10 Jan 2019, 07:42
[quote="Bunuel"]
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   [#permalink] 10 Jan 2019, 07:42
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