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Give the right triangle ABC, find th slope of the segment
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11 Mar 2024, 06:19
11 Mar 2024, 06:19
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Give the right triangle ABC, find the slope of the segment \(\overline{AB}\) and \(\overline{BC}\)
GRE xy.jpg [ 50.1 KiB | Viewed 1074 times ]
A. \(\frac{2}{3}\) and \(- \frac{3}{2}\)
B. \(\dfrac{2}{\sqrt{6}}\) and \(\dfrac{-3}{\sqrt{6}}\)
C. \(\dfrac{2}{\sqrt{5}}\) and \(\dfrac{-\sqrt{5}}{2}\)
D. \(\dfrac{-3}{\sqrt{7}}\) and \(\dfrac{\sqrt{7}}{3}\)
E. cannot be determined
Attachment:
GRE xy.jpg [ 50.1 KiB | Viewed 1074 times ]
A. \(\frac{2}{3}\) and \(- \frac{3}{2}\)
B. \(\dfrac{2}{\sqrt{6}}\) and \(\dfrac{-3}{\sqrt{6}}\)
C. \(\dfrac{2}{\sqrt{5}}\) and \(\dfrac{-\sqrt{5}}{2}\)
D. \(\dfrac{-3}{\sqrt{7}}\) and \(\dfrac{\sqrt{7}}{3}\)
E. cannot be determined
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: Give the right triangle ABC, find th slope of the segment
[#permalink]
09 Apr 2024, 08:57
09 Apr 2024, 08:57
Expert Reply
OE
B. This problem tests your understanding of the proportions on a right triangle. Because of the orientation of the triangle, you can determine the rise for each segment. The missing part of the slope is the run, which corresponds to the height of the triangle (the unlabeled segment that intersects the hypotenuse at a right angle). Fortunately, this value can be found using the ratios of the proportions of a right triangle. Let the height of the triangle be h.
Use the ratios of the hypotenuses to find h:
short leg of the small triangle/hypotenuse of the small triangle=short leg of the large triangle/hypotenuse of the large triangle
\(\frac{2}{AB}=\frac{AB}{5}\)
\(10=AB^2\)
\(2^2+h^2=AB^2\)
\(h=\sqrt{6}\)
Now
slope \(\overline{AB}=\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}\)
slope \(\overline{BC}=\dfrac{-3}{\sqrt{6}}=\dfrac{- \sqrt{6}}{2}\)
B. This problem tests your understanding of the proportions on a right triangle. Because of the orientation of the triangle, you can determine the rise for each segment. The missing part of the slope is the run, which corresponds to the height of the triangle (the unlabeled segment that intersects the hypotenuse at a right angle). Fortunately, this value can be found using the ratios of the proportions of a right triangle. Let the height of the triangle be h.
Use the ratios of the hypotenuses to find h:
short leg of the small triangle/hypotenuse of the small triangle=short leg of the large triangle/hypotenuse of the large triangle
\(\frac{2}{AB}=\frac{AB}{5}\)
\(10=AB^2\)
\(2^2+h^2=AB^2\)
\(h=\sqrt{6}\)
Now
slope \(\overline{AB}=\dfrac{2}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}\)
slope \(\overline{BC}=\dfrac{-3}{\sqrt{6}}=\dfrac{- \sqrt{6}}{2}\)
gmatclubot
Re: Give the right triangle ABC, find th slope of the segment [#permalink]
09 Apr 2024, 08:57
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