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In the figure, ABC and ADC are right triangles. Which of the [#permalink]
17 Mar 2019, 03:02
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#GREpracticequestion In the figure, ABC and ADC are right triangle..jpg [ 16.2 KiB  Viewed 680 times ]
In the figure, ABC and ADC are right triangles. Which of the following could be the lengths of AD and DC, respectively? (I) \(\sqrt{3}\) and \(\sqrt{4}\) (II) 4 and 6 (III) 1 and \(\sqrt{24}\) (IV) 1 and \(\sqrt{26}\) (A) I and II only (B) II and III only (C) III and IV only (D) IV and I only (E) I, II and III only
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Re: In the figure, ABC and ADC are right triangles. Which of the [#permalink]
20 Mar 2019, 10:54
We can use the Pythagorean theorem to find the length of AC:
\(3^{2} + 4^{2} = (AC)^{2}\)
\(9 + 16 = (AC)^{2}\)
\(25 = (AC)^{2}\)
\(5 = AC\)
We should also recognize this as a 345 triangle, one of the Pythagorean triplets.
In any case, we know that ADC is a right triangle, but we don't know which angle is the right angle. So, any of these three possibilities could be correct, depending on which side is the hypotenuse:
\(AD^{2} + DC^{2} = AC^{2}\)
\(AD^{2} + AC^{2} = DC^{2}\)
\(AC^{2} + DC^{2} = AD^{2}\)
Plug in AC = 5, and we get these possible equations:
\(AD^{2} + DC^{2} = 25\)
\(AD^{2} + 25 = DC^{2}\)
\(25 + DC^{2} = AD^{2}\)
I want to isolate the known value, 25:
\(25 = AD^{2} + DC^{2}\)
\(25 = DC^{2}  AD^{2}\)
\(25 = AD^{2}  DC^{2}\)
Now we can just plug in the answer choices. Clearly the first statement won't work, as 3 and 4 are much too small to add up to 25. Already that eliminates choices A, D, and E.
The second statement is pretty easy to eliminate as well, since 4² and 6² = 16 and 36, respectively. But no combination of adding or subtract 16 and 36 will give us an answer of 25. So Statement II must be false, and the answer must be C.



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Re: In the figure, ABC and ADC are right triangles. Which of the [#permalink]
10 May 2019, 08:01
Can someone please explain C (especially IV option) is correct ? Thank you



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Re: In the figure, ABC and ADC are right triangles. Which of the [#permalink]
10 May 2019, 10:33
In the case AC is the hypotenuse of the triangle, we have by The Pythagorean Theorem, \(AC^2 = AD^2 + DC^2\) \(5^2 = AD^2 + DC^2\) This equation is satisfied by III since \(5^2 = 1^2 + (\sqrt{24})^2\) . Therefore, III is possible. Hope is more clear now to you Sir. regards
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Re: In the figure, ABC and ADC are right triangles. Which of the [#permalink]
11 May 2019, 06:32
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JelalHossain wrote: Can someone please explain C (especially IV option) is correct ? Thank you The trick here is to recognize that the right angle in triangle ACD could be in 3 different places. ∠CAD could be 90° ∠CDA could be 90° ∠DCA could be 90° Cheers, Brent
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Re: In the figure, ABC and ADC are right triangles. Which of the
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11 May 2019, 06:32





