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705-805 (Hard)|   Geometry|      
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MathRevolution
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In the figure, ABCD is a square, EFG is a right triangle and H is the 24 Jul 2019, 01:12
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E
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Difficulty:

85% (hard)

Question Stats:

52% (03:10) correct 48% (03:17) wrong based on 63 sessions

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[GMAT math practice question]

In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\). What is the value of \(x\)?

Attachment:
7.24ps.png
7.24ps.png [ 8.57 KiB | Viewed 2916 times ]

\(A. 200\)

\(B. 216\)

\(C. 230\)

\(D. 236\)

\(E. 250\)
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Official Answer
E
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In the figure, ABCD is a square, EFG is a right triangle and H is the 24 Jul 2019, 01:24
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MathRevolution
[GMAT math practice question]

In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\). What is the value of \(x\)?

Attachment:
7.24ps.png

\(A. 200\)

\(B. 216\)

\(C. 230\)

\(D. 236\)

\(E. 250\)
Show more

Given: In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\).
Asked: What is the value of \(x\)?

Using the similarity of triangle approach, we get
\(\frac{34+x}{1775} = \frac{20}{(x/2)} = \frac{40}{x}\)
\(34x + x^2 = 1775*40\)
x=250 satisfies the equation.

IMO E
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In the figure, ABCD is a square, EFG is a right triangle and H is the 25 Jul 2019, 02:20
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Kinshook
MathRevolution
[GMAT math practice question]

In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\). What is the value of \(x\)?

Attachment:
7.24ps.png

\(A. 200\)

\(B. 216\)

\(C. 230\)

\(D. 236\)

\(E. 250\)

Given: In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\).
Asked: What is the value of \(x\)?

Using the similarity of triangle approach, we get
\(\frac{34+x}{1775} = \frac{20}{(x/2)} = \frac{40}{x}\)
\(34x + x^2 = 1775*40\)
x=250 satisfies the equation.

IMO E
Show more
Hello
Could you please elaborate how did you get 40?
and how did you assume the measurements given as lengths?
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In the figure, ABCD is a square, EFG is a right triangle and H is the 26 Jul 2019, 01:32
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=>

Since triangles \(EFG\) and \(EAH\) are similar,

\((34 + x):20 = 1775: (\frac{x}{2})\). So,

\(x^2 + 34x – 1775*20*2 = 0.\)

Factoring yields \((x+284)(x-250) = 0\). Since \(x\) must be positive, \(x = 250.\)

Therefore, E is the answer.
Answer: E
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In the figure, ABCD is a square, EFG is a right triangle and H is the 26 Jul 2019, 03:49
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mohikajain
Kinshook
MathRevolution
[GMAT math practice question]

In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\). What is the value of \(x\)?

Attachment:
7.24ps.png

\(A. 200\)

\(B. 216\)

\(C. 230\)

\(D. 236\)

\(E. 250\)

Given: In the figure, \(ABCD\) is a square, \(EFG\) is a right triangle and \(H\) is the midpoint of \(AD\). Moreover, point \(A\) lies on the line \(EF\).
Asked: What is the value of \(x\)?

Using the similarity of triangle approach, we get
\(\frac{34+x}{1775} = \frac{20}{(x/2)} = \frac{40}{x}\)
\(34x + x^2 = 1775*40\)
x=250 satisfies the equation.

IMO E
Hello
Could you please elaborate how did you get 40?
and how did you assume the measurements given as lengths?
Show more


\(\frac{20}{(x/2)} = \frac{40}{x}\)
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In the figure, ABCD is a square, EFG is a right triangle and H is the 26 Mar 2020, 18:41
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Is there an easy way to know how to factor x2+34x–1775∗20∗2=0? My first guess wouldn't have been to use 284 and 250- in fact, these numbers are so big, how would we be able to compute this in the timeframe required to answer this question?

Also, how can you say that those triangles are similar? If you can say the small triangle at the top (20/x) and the big triangle (34+x)/1775 are similar, then can't you say the small triangle and the diagonal triangle of the square (x/x) are similar? In that case, you would get x = 40, which is clearly wrong.
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In the figure, ABCD is a square, EFG is a right triangle and H is the 01 Mar 2023, 23:42
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