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10 Aug 2022, 01:30
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Difficulty:
65%
(hard)
Question Stats:
59% (03:15) correct
41%
(03:00)
wrong
based on 39
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In the following figure, AB is the diameter of semicircle with center M. Two semi-circles are drawn with AM and MB as the diameters. A circle is drawn such that it is tangent to all the three semi-circles. If AM = \(r\) units, find the radius of the circle (in terms of \(r\)).
A) \(\frac{r}{2}\)
B) \(\frac{r}{3}\)
C) \(\frac{r}{4}\)
D) \(\frac{r}{{\sqrt{2}}}\)
E) \(\frac{2r}{5}\)
1.png [ 8.92 KiB | Viewed 2370 times ]
Are You Up For the Challenge: 700 Level Questions
A) \(\frac{r}{2}\)
B) \(\frac{r}{3}\)
C) \(\frac{r}{4}\)
D) \(\frac{r}{{\sqrt{2}}}\)
E) \(\frac{2r}{5}\)
Attachment:
1.png [ 8.92 KiB | Viewed 2370 times ]
Are You Up For the Challenge: 700 Level Questions
10 Aug 2022, 07:58
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Bunuel
We can start by drawing in the red triangle.
Rather than working with the variable r, let's assign a value that we can then plug in to the answer choices. r=2 looks like it will work well since it makes the base of the red triangle equal to 1. So, r=2.
Now, the base of the red triangle is 1, the height is 2-x, and the hypotenuse is 1+x. We can plug those into the Pythagorean theorem.
\(1^2+(2-x)^2=(1+x)^2\)
\(1+4-4x+x^2 = 1+2x+x^2\)
\(4=6x\)
\(x=\frac{2}{3}\)
Let's plug r=2 into the answer choices to find which results in \(\frac{2}{3}\).
A) \(\frac{2}{2}\)
Nope.
B) \(\frac{2}{3}\)
Yep!
Answer choice B.
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Picture2.png [ 232.51 KiB | Viewed 2019 times ]
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