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04 Feb 2018, 13:03
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D
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% (02:47) correct
38%
(03:02)
wrong
based on 150
sessions
History
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Time
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Not Attempted Yet
In trapezium ABCD, AB || CD and AC = BD. If area of the trapezium is 60 sq units, height of the trapezium is 10 units, what is the length of diagonal BD?
(A) 10 units
(B) \(\sqrt{116}\) units
(C) \(5\sqrt{5}\) units
(D) \(\sqrt{136}\) units
(E) 12 units
(A) 10 units
(B) \(\sqrt{116}\) units
(C) \(5\sqrt{5}\) units
(D) \(\sqrt{136}\) units
(E) 12 units
06 Jun 2018, 20:50
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GMATSkilled
Kindly refer the affixed diagram,
Given AB||CD and BD=AC(diagonals), hence it's a isosceles trapezium. hence AM=BN.
ADM, BNC & DMB are right angled triangles.
Given area of trapezium=60 sq unit
Or, ½*10(AB+CD)=60
Or, AB+CD=12
Or, X+AM+X+BN=12
Or, 2X+2BN=12
Or, X+BN=6
Or, MB=6 unit
Applying Pythagoras theorem in the right angled triangle DMB, we have
\(BD^2\)=\(DM^2\)+\(MB^2\)=\(10^2\)+\(6^2\)=136
So, BD=\(\sqrt{136}\) units
Ans Option (D)
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Trapezoid.JPG [ 18.15 KiB | Viewed 16559 times ]
General Discussion
rahul16singh28
Joined: 31 Jul 2017
Last visit: 09 Jun 2020
Posts: 428
Given Kudos: 752
Location: Malaysia
Schools: INSEAD Jan '19
GPA: 3.95
WE:Consulting (Energy)
04 Feb 2018, 14:24
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DHAR
Area of Trapezium = \(\frac{1}{2}h(a+b)\)
\(a, b = Length of parallel sides.\)
\(h = Height between the two parallel sides.\)
Attachments
New Doc 32_4.jpg [ 344.55 KiB | Viewed 17276 times ]
