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Segment AC is the diameter of circle O with the length 10, and AB =
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18 Feb 2022, 08:00
18 Feb 2022, 08:00
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Question Stats:
75% (02:52) correct
25% (01:57) wrong based on 8 sessions
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Segment AC is the diameter of circle O with the length 10, and AB = 5. Segment BD is the altitude drawn from B to AC. What is the length of BD?
GRE Circle and triangle.jpg [ 20.94 KiB | Viewed 1386 times ]
(A) 5
(B) 5√3/2
(C) 2√10
(D) 3√5
(E) 3√10
Attachment:
GRE Circle and triangle.jpg [ 20.94 KiB | Viewed 1386 times ]
(A) 5
(B) 5√3/2
(C) 2√10
(D) 3√5
(E) 3√10
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Re: Segment AC is the diameter of circle O with the length 10, and AB =
[#permalink]
24 Feb 2022, 01:30
24 Feb 2022, 01:30
1
Carcass wrote:
Segment AC is the diameter of circle O with the length 10, and AB = 5. Segment BD is the altitude drawn from B to AC. What is the length of BD?
(A) 5
(B) 5√3/2
(C) 2√10
(D) 3√5
(E) 3√10
Attachment:
The attachment GRE Circle and triangle.jpg is no longer available
(A) 5
(B) 5√3/2
(C) 2√10
(D) 3√5
(E) 3√10
Attachment:
GRE Circle and triangle.jpg [ 16.33 KiB | Viewed 1210 times ]
Applying Pythagoras Theorem in \(ΔABC\);
\(AC^2 = AB^2 + BC^2\)
\(10^2 = 5^2 + BC^2\)
\(100 - 25 = BC^2\)
\(BC = \sqrt{75} = 5\sqrt{3}\)
Now, Area of \(ΔABC\) = \(\frac{1}{2}(AB)(BC)\) = \(\frac{1}{2}(AC)(BD)\)
\((AB)(BC) = (AC)(BO)\)
\((5)(5\sqrt{3}) = (10)(BD)\)
\(BD = \frac{5\sqrt{3}}{2}\)
Hence, option B
gmatclubot
Re: Segment AC is the diameter of circle O with the length 10, and AB = [#permalink]
24 Feb 2022, 01:30
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