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14 Feb 2022, 13:59
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51% (02:18) correct
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A street has parallel curbs 40 feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is 15 feet and each stripe is 50 feet long. Find the distance, in feet, between the stripes.
A. 9
B. 10
C. 12
D. 15
E. 25
Chinese GMAT study manual
A. 9
B. 10
C. 12
D. 15
E. 25
Chinese GMAT study manual
KabirKanha
GMAT 1: 780 Q51 V46
Posts: 32
08 Jul 2022, 12:07
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Let us illustrate the scenario detailed in the question.
PS0.png [ 6.15 KiB | Viewed 2706 times ]
We see that using Pythagoras' Theorem, we can easily deduce the length \(30\) along the base.
Since, the two strips are parallel, the angles formed with Curb 2 are also the same.
PS1.png [ 8.89 KiB | Viewed 2701 times ]
From the larger triangle, we see that \(\sin \theta = \frac{4}{5}\)
So in triangle ABC, \(\sin \theta = \frac{4}{5} = \frac{AC}{15}\)
\(\Rightarrow AC = 12\) feet, which is the distance between the stripes.
Attachment:
PS0.png [ 6.15 KiB | Viewed 2706 times ]
We see that using Pythagoras' Theorem, we can easily deduce the length \(30\) along the base.
Since, the two strips are parallel, the angles formed with Curb 2 are also the same.
Attachment:
PS1.png [ 8.89 KiB | Viewed 2701 times ]
So in triangle ABC, \(\sin \theta = \frac{4}{5} = \frac{AC}{15}\)
\(\Rightarrow AC = 12\) feet, which is the distance between the stripes.
10 Jul 2022, 13:31
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I am uploading a sketch of a different approach. I calculated the surface area of the parallelogram which is formed by the crosswalk by forming a bigger rectangle that constitutes from the area of the aforementioned parallelogram and the sum of two identical right angle triangles. The given sides of the triangles are a leg of length 40 and a hypothenuse of length 50. The other leg is calculated by applying Pythagoras theorem and equals 30; therefore area of parallelogram is solved from the following
Pxzca=45*40 - 2*30*40/2 = 600
Height of parallelogram = Surface area/Base = 600/50 = 12
Pxzca=45*40 - 2*30*40/2 = 600
Height of parallelogram = Surface area/Base = 600/50 = 12
Attachments
Qaltern.jpg [ 115.34 KiB | Viewed 2521 times ]
