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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. \(\frac{8}{1+\sqrt{2}}\)


Attachment:
1045966_f946153db51a405bb0a4234ae0109dc2.png


Ballpark, please!

The radius of the quarter circle is 4. Is the diameter of the inscribed circle larger than that or smaller? Smaller. How much smaller? Is it 10%? 50%? 90% I don't know, let's say it's 3.5, maybe a touch less.
Awesome, that's the diameter of the inscribed circle, which is what we are asked to find.
Let's look at the answer choices. All of the denominators are the same and are roughly 1+1.4, so denominator is 2.4. We need a numerator that's 2.4 * 3.5 = 8.4. Only one answer choice is close.

Answer choice E.


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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
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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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