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06 Jun 2017, 11:13
Kudos
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A
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:
62% (02:57) correct
38%
(03:09)
wrong
based on 159
sessions
History
Date
Time
Result
Not Attempted Yet
In the circle pictured above, AB=2√5, CD=1, and AC is a diameter. What is the circumference of the circle?
A. 5π
B. 6π
C. 7π
D. 8π
E. 9π
Attachment:
ABCD_diameter.PNG [ 11.3 KiB | Viewed 6397 times ]
General Discussion
06 Jun 2017, 12:12
Kudos
Bookmarks
Let's take the Right Angled Triangle ADB
\(AB^2\) = \(BD^2\) + \(AD^2\)
\((2√5)^2\) = \(BD^2\) +\((r+r-1)^2\)
20 = \(BD^2\) + 4\(r^2\)+1-4r --------> Equation 1
Let's take the Right Angled Triangle ABC
\(AC^2\) = \(AB^2\)+ \(BC^2\)
4\(r^2\) = \((2√5)^2\) + \(BC^2\)
4\(r^2\) = 20 + \(BC^2\) --------> Equation 2
Let's take the Right Angled Triangle BDC
\(BC^2\) = \(BD^2\) + \(DC^2\)
\(BC^2\)= \(BD^2\)+ 1--------> Equation 3
Replace \(BC^2\) from equation 3 to Equation 2
4\(r^2\) = 20 + \(BD^2\)+ 1
4\(r^2\) = 21 +\(BD^2\)
\(BD^2\) = 4\(r^2\) - 21 --------> Equation 4
Replace \(BD^2\) from equation 4 to Equation 1
20 = 4\(r^2\) - 21 + 4\(r^2\) + 1 - 4r
8\(r^2\) - 4r - 40 = 0
2\(r^2\) - r - 10 = 0
Solving for r we get
r = -2 or 5/2
Circumference of the circle = 2 * π *\(\frac{5}{2}\) = 5π
Answer is A
\(AB^2\) = \(BD^2\) + \(AD^2\)
\((2√5)^2\) = \(BD^2\) +\((r+r-1)^2\)
20 = \(BD^2\) + 4\(r^2\)+1-4r --------> Equation 1
Let's take the Right Angled Triangle ABC
\(AC^2\) = \(AB^2\)+ \(BC^2\)
4\(r^2\) = \((2√5)^2\) + \(BC^2\)
4\(r^2\) = 20 + \(BC^2\) --------> Equation 2
Let's take the Right Angled Triangle BDC
\(BC^2\) = \(BD^2\) + \(DC^2\)
\(BC^2\)= \(BD^2\)+ 1--------> Equation 3
Replace \(BC^2\) from equation 3 to Equation 2
4\(r^2\) = 20 + \(BD^2\)+ 1
4\(r^2\) = 21 +\(BD^2\)
\(BD^2\) = 4\(r^2\) - 21 --------> Equation 4
Replace \(BD^2\) from equation 4 to Equation 1
20 = 4\(r^2\) - 21 + 4\(r^2\) + 1 - 4r
8\(r^2\) - 4r - 40 = 0
2\(r^2\) - r - 10 = 0
Solving for r we get
r = -2 or 5/2
Circumference of the circle = 2 * π *\(\frac{5}{2}\) = 5π
Answer is A
Attachments
Solution.PNG [ 10.28 KiB | Viewed 5841 times ]
