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by sanju09 » Sat Mar 06, 2010 2:23 am
A circle with center O is inscribed in a quadrilateral ABCD such that it touches the length AB at P. If AP = PB = OP, with OC = 5 and OD = 12, what is the length of CD?
(A) 13
(B) less than 13
(C) more than 13
(D) less than 7
(E) more than 17
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by rohan_vus » Sat Mar 06, 2010 4:09 am
Let Circle touch AD at Q , CD at R and BC at S

From question stem angle PAO = angle POA = angle PBO = angle POB = 45

Angle AOQ = angle Angle BOS = 45 ( as length of tangents from an external point to a circle are equal )

Angle QOS = 180 ( a strg line)

Angle QOD = Angle DOR ( say x)
Angle ROC = Angle SOC ( say y)
Now since angle QOS = 180 , so 2x + 2y = 180 , gives x + y = 90

X + y is equal to angle of DOC = 90 , hence length of CD = 12^2 + 5^2 = 13

A
Last edited by rohan_vus on Sat Mar 06, 2010 4:30 am, edited 1 time in total.
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by sanju09 » Sat Mar 06, 2010 4:23 am
rohan_vus wrote:Let Circle touch AD at Q , CD at R and BC at S

From question stem angle PAB = angle POA = angle PBO = angle POB = 45

Angle AOQ = angle Angle BOS = 45 ( as length of tangents from an external point to a circle are equal )

Angle QOS = 180 ( a strg line)

Angle QOS = Angle DOR ( say x)
Angle ROC = Angle SOC ( say y)
Now since angle QOS = 180 , so 2x + 2y = 180 , gives x + y = 90

X + y is equal to angle of DOC = 90 , hence length of CD = 12^2 + 5^2 = 13

A
Please see if the bold contents need some attention.
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by rohan_vus » Sat Mar 06, 2010 4:31 am
Edited , but ans still remains the same
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by sanju09 » Sat Mar 06, 2010 4:40 am
rohan_vus wrote:Edited , but ans still remains the same
no issue with your answer, mate
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