Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?
(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
Radius of Circle
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I reached 2 root(5).dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?
(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
Please confirm the option d. is it 2√5 or 5√2 ??
All the answer choices are exactly as written above. Solve and post your reasoning. If there is error in your reasoning others may point it out.Pranay wrote:I reached 2 root(5).dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?
(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
Please confirm the option d. is it 2√5 or 5√2 ??
Hi dtweah,dtweah wrote:All the answer choices are exactly as written above. Solve and post your reasoning. If there is error in your reasoning others may point it out.Pranay wrote:I reached 2 root(5).dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?
(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
Please confirm the option d. is it 2√5 or 5√2 ??
Sorry for bothering you!!!
I missed out a basic out a basic point while drawing a conclusion .. so there an error in my computing .. !!
Anyways ... I am trying to solve it .. and will post if reach the conclusion.
Pranay
Hi ..
The answer is D.
Using the Trignometry concepts, I could get that angle ADP is 45 degrees.
(Applying tan inverse to each of the right angled triangles ADP and BDP)
Using a theorm, the angles subtended by a chord at the edge of a circle is half that of subtended at the centre implies, angle AOB = 90 degrees where O is the centre of the circle.
Now drop a perpendicular from the O to chord AB, at point Z, to form a right angled triangle AZO.
=> AZ=BZ=5 (from question)
angle OZA = 90 degrees, angle ZOA = 45 degrees and angle OAZ = 45 degrees.
=> OA=OZ since, triangle OZA is an isosceles right angled triangle.
let OA = r (radius)
From Pythogaras theorm,
OZ^2 + AZ^2 = OA^2
Thus, by substituting the values we reach r = 5sqrt(2)
which is option D.
Hope the answer is right.
Also, please post the source of the question along with the OA. [:)]
The answer is D.
Using the Trignometry concepts, I could get that angle ADP is 45 degrees.
(Applying tan inverse to each of the right angled triangles ADP and BDP)
Using a theorm, the angles subtended by a chord at the edge of a circle is half that of subtended at the centre implies, angle AOB = 90 degrees where O is the centre of the circle.
Now drop a perpendicular from the O to chord AB, at point Z, to form a right angled triangle AZO.
=> AZ=BZ=5 (from question)
angle OZA = 90 degrees, angle ZOA = 45 degrees and angle OAZ = 45 degrees.
=> OA=OZ since, triangle OZA is an isosceles right angled triangle.
let OA = r (radius)
From Pythogaras theorm,
OZ^2 + AZ^2 = OA^2
Thus, by substituting the values we reach r = 5sqrt(2)
which is option D.
Hope the answer is right.
Also, please post the source of the question along with the OA. [:)]
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