VIGHNESHKAMATH
TestPrepUnlimited
Bunuel
A quarter circle with radius 4 cm has a circle inscribed in it as shown above. What is the diameter of the inscribed circle ?
A. \(\frac{3}{1+\sqrt{2}}\)
B. \(\frac{4}{1+\sqrt{2}}\)
C. \(\frac{5}{1+\sqrt{2}}\)
D. \(\frac{6}{1+\sqrt{2}}\)
E. \(\frac{8}{1+\sqrt{2}}\)
The radius of the big circle is consisted of two segments OA and AB. Let AB = x, then \(OA = \sqrt{2}*x\) and \(AB = x\).
Then OB = \(\sqrt{2}x + x = 4\)
\(x = \frac{4}{\sqrt{2} + 1}\) however the question is asking for diameter which is 2x.
Ans: E
How did you get OA?
Thank You
Vighnesh
(1st)
The center of the inscribed circle (call it point O) is also the geometric center of the quarter circle
Draw a line segment from vertex B to the point of Tangency on the circumference (call it point D)
BD passes through center O of the inscribed circle
OD = r = radius of inscribed circle
(2nd)
Concept: a radius drawn to a tangent line at the point of Tangency creates a 90 degree angle with the tangent line
Draw 2 line segments from center O to sides BA and BC
This quadrilateral is a square with side = r
And diagonal of this square = BO = r * sqrt(2)
(3rd)
BO + OD = radius of the quarter circle = 4
r * sqrt(2) + r = 4
r * [(sqrt(2) + 1)] = 4
r = 4 / (sqrt(2) + 1)
Double it to get the diameter
8 / (sqrt(2) + 1)
*E*
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