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05 Dec 2014, 07:33
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E
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Tough and Tricky questions: Geometry.
In the right triangle ABD below, AC is perpendicular to BD. If AB = 5 and AD = 12 , then AC is equal to?
Attachment:
2014-12-05_1831.png [ 4.12 KiB | Viewed 6579 times ]
A) 30/13
B) \(\sqrt{12}\)
C) 4
D) \(2 \sqrt{5}\)
E) 60/13
Kudos for a correct solution.
Source: Chili Hot GMAT
05 Dec 2014, 11:03
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Bunuel
Tough and Tricky questions: Geometry.
In the right triangle ABD below, AC is perpendicular to BD. If AB = 5 and AD = 12 , then AC is equal to?
Attachment:
2014-12-05_1831.png
A) 30/13
B) \(\sqrt{12}\)
C) 4
D) \(2 \sqrt{5}\)
E) 60/13
Kudos for a correct solution.
Source: Chili Hot GMAT
\(BD^2 = AB^2+AD^2\)
now, put \(AB=5, AD=12\)
we have\(BD^2= 25+144 =169\)
thus \(BD=13\)
now area of triangle ABD =\(\frac{1}{2} AB.AD\)
= \(\frac{1}{2}5.12\)
=30-----------------1)
Also, AC is perpendicular to BD. thus. area of triangle ABD can also be given as
\(\frac{1}{2}\) AC.BD
= \(\frac{1}{2}\) AC.13
but we have already calculated the area in step 1). thus we have
\(\frac{1}{2}\) AC.13 =30
or AC=60/13 hence E.
Updated on: 06 Dec 2014, 02:39
Originally posted by arunspanda on 05 Dec 2014, 11:30.
Last edited by arunspanda on 06 Dec 2014, 02:39, edited 1 time in total.
Last edited by arunspanda on 06 Dec 2014, 02:39, edited 1 time in total.
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In the right triangle ABD, AC is perpendicular to BD.
AB = 5 and AD = 12.
Therefore, BD = 13.
The triangle ABD and CBA are similar.
\(\frac{BC}{AB}= \frac{AB}{BD}\) or, \(BC = \frac{(AB)^2}{BD} =\frac{5^2}{13}\).....(1)
Similarly, the traingles ABD and CAD are similar.
\(\frac{CD}{AD}= \frac{AD}{BD}\) or \(CD=\frac{(AD)^2}{BD}=\frac{12^2}{13}\)...........(2)
The triangles CAD and CBA are similar.
\(\frac{AC}{CD}= \frac{BC}{AC}\)or \((AC)^2 = BC.CD\)
Substituting values of BC and CD from (1) and (2), \((AC)^2 = \frac{12^2*5^2}{13^2}\)
Or, \(AC = \frac{12*5}{13}= \frac{60}{13}\)
Answer: E
AB = 5 and AD = 12.
Therefore, BD = 13.
The triangle ABD and CBA are similar.
\(\frac{BC}{AB}= \frac{AB}{BD}\) or, \(BC = \frac{(AB)^2}{BD} =\frac{5^2}{13}\).....(1)
Similarly, the traingles ABD and CAD are similar.
\(\frac{CD}{AD}= \frac{AD}{BD}\) or \(CD=\frac{(AD)^2}{BD}=\frac{12^2}{13}\)...........(2)
The triangles CAD and CBA are similar.
\(\frac{AC}{CD}= \frac{BC}{AC}\)or \((AC)^2 = BC.CD\)
Substituting values of BC and CD from (1) and (2), \((AC)^2 = \frac{12^2*5^2}{13^2}\)
Or, \(AC = \frac{12*5}{13}= \frac{60}{13}\)
Answer: E
