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04 May 2020, 07:02
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[GMAT math practice question]
The figure shows a quadrangle \(ABCD\) with \(∠ADC = 90°, ∠DCB = 90°\) and \(AD = 6, AB = 8\) and \(BC = 10.\) What is the length \(BD\)?
5.4ps.png [ 14.35 KiB | Viewed 2189 times ]
A. \(2\sqrt{37}\)
B. \(10 \)
C. \(3\sqrt{39 }\)
D. \(14\)
E. \(5\sqrt{41}\)
The figure shows a quadrangle \(ABCD\) with \(∠ADC = 90°, ∠DCB = 90°\) and \(AD = 6, AB = 8\) and \(BC = 10.\) What is the length \(BD\)?
Attachment:
5.4ps.png [ 14.35 KiB | Viewed 2189 times ]
A. \(2\sqrt{37}\)
B. \(10 \)
C. \(3\sqrt{39 }\)
D. \(14\)
E. \(5\sqrt{41}\)
04 May 2020, 07:14
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Draw a perpendicular from A to BC and make a square. Call it P. Now AB is the hypotenuse and BP is the base. AP is the height.
Now AP = \(\sqrt{AB^2-BP^2}\) = \(\sqrt{64-16} = 4\sqrt{3}\)
Now, AP = DC = 4\(\sqrt{3}\)
Now, BCD is a right angled triangle. Hypotenuse BD = \(\sqrt{100+48} = 2\sqrt{37}\)
A is the answer.
Now AP = \(\sqrt{AB^2-BP^2}\) = \(\sqrt{64-16} = 4\sqrt{3}\)
Now, AP = DC = 4\(\sqrt{3}\)
Now, BCD is a right angled triangle. Hypotenuse BD = \(\sqrt{100+48} = 2\sqrt{37}\)
A is the answer.
Attachments
5.4ps.png [ 15.66 KiB | Viewed 2127 times ]
06 May 2020, 01:21
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=>
5.4PS(A).png [ 18.47 KiB | Viewed 2043 times ]
From the above figure, since we have \(EC = 6\), we have \(BE = 4\). Then we use Pythagorean Theorem to solve for \(AE\), giving us\( AE = \sqrt{8^2- 4^2} = \sqrt{64-16} = \sqrt{48} = 4√3\) and \(AE = DC = 4√3.\)
Thus, we have \(x = \sqrt{10^2+(4√3)^2} = \sqrt{100+48} = 2√37.\)
Therefore, A is the answer.
Answer: A
Attachment:
5.4PS(A).png [ 18.47 KiB | Viewed 2043 times ]
From the above figure, since we have \(EC = 6\), we have \(BE = 4\). Then we use Pythagorean Theorem to solve for \(AE\), giving us\( AE = \sqrt{8^2- 4^2} = \sqrt{64-16} = \sqrt{48} = 4√3\) and \(AE = DC = 4√3.\)
Thus, we have \(x = \sqrt{10^2+(4√3)^2} = \sqrt{100+48} = 2√37.\)
Therefore, A is the answer.
Answer: A
