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17 Jun 2020, 05:01
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
48% (02:05) correct
52%
(02:33)
wrong
based on 21
sessions
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What is the radius of the circle shown above in terms of a, b, and c?
A. \(\frac{\sqrt{(b^2 + c^2)(b^2 + (a + c)^2)}}{b}\)
B. \(\frac{\sqrt{(b^2 + c^2)(b^2 + (a + c)^2)}}{2b}\)
C. \(\frac{\sqrt{(b^2 + c^2)(b^2 + (a + c)^2)}}{3b}\)
D. \(\frac{\sqrt{(b^2 + c^2)(b^2 + (a + c)^2)}}{4b}\)
E. \(\frac{\sqrt{(b^2 + c^2)(b^2 + (a + c)^2)}}{5b}\)
Are You Up For the Challenge: 700 Level Questions
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17 Jun 2020, 09:00
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AO=BO {radius of the circle}
AO^2 = BO^2
\(x^2+\frac{a^2}{4} = (b-x)^2 + (c+\frac{a}{2})^2\)
\(x^2+\frac{a^2}{4} = b^2+x^2-2bx +c^2 +\frac{a^2}{4} +ca \\
\)
\(2bx = b^2+c^2+ca\)
\(x= \frac{b^2+c^2+ca}{2b}\)
\(r^2 = x^2+\frac{a^2}{4}\)
\(r^2 = \frac{(b^2+c^2+ca)^2}{4b^2} + \frac{a^2}{4}\)
\(r^2 = \frac{1}{4b^2}[(b^2+c^2+ca)^2 + a^2b^2]\)
\(r^2 = \frac{1}{4b^2} [(b^4+b^2c^2 +b^2c^2+c^4 +c^2a^2 +2b^2ca +2c^2ca+a^2b^2)\)
\(r^2 = \frac{1}{4b^2} [(b^2(b^2+c^2) +c^2(b^2+c^2) + +2ca(b^2 +c^2)+a^2(b^2 +c^2)]\)
\(r^2 = \frac{1}{4b^2}[ (b^2+c^2) (b^2+c^2+2bc+a^2)]\)
\(r^2 = \frac{1}{4b^2} [(b^2+c^2) ((b+c)^2+a^2)]\)
\(r = \sqrt{\frac{[(b^2+c^2) ((b+c)^2+a^2)]}{2b}}\)
17 Jun 2020, 09:51
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If there's a formula that will give us the radius r for any values of a, b and c, then that formula will need to work when c= 0. If c = 0, then the lines of lengths a and b become chords of the circle. If we make a triangle using those two lines, the third side will be a diameter (so has length 2r), because a and b meet at a 90 degree angle. Now we can use Pythagoras:
(2r)^2 = a^2 + b^2
r^2 = (a^2 + b^2) / 4
r = [√(a^2 + b^2) ] / 2
and plugging c=0 into each answer choice, we see that B is right.
(2r)^2 = a^2 + b^2
r^2 = (a^2 + b^2) / 4
r = [√(a^2 + b^2) ] / 2
and plugging c=0 into each answer choice, we see that B is right.
