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11 Oct 2013, 22:50
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (03:08) correct
40%
(03:21)
wrong
based on 366
sessions
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In triangle PQR, the angle Q = 90 degree, PQ = 6 cm, QR = 8 cm. X is a variable point on PQ. The line through X parallel to QR, intersects PR at Y and the line through Y, parallel to PQ, intersects QR at Z. Find the least possible length of XZ
A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm
A. 3.6 cm
B. 2.4 cm
C. 4.8 cm
D. 2.16 cm
E. 3.2 cm
12 Oct 2013, 06:23
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Qoofi
Good question. +1.
Look at the diagram below:
Notice that QZYX is a rectangle, thus diagonals XZ and QY are equal. So, we need to minimize QY. QY (the distance from Q to the hypotenuse) will be minimized when QY is perpendicular to the hypotenuse.
Now, in case when QY is perpendicular to PR, two right triangles PQR and PQY are similar: QY:QP=QR:PR --> QY:6=8:10 --> QY=4.8.
Answer: C.
Hope it's clear.
Attachment:
Untitled2.png [ 3.28 KiB | Viewed 68266 times ]
23 Oct 2013, 01:06
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... another approach:
I'm gonna refer to the graphics bunuel provided:
From the information given in the question, we can compute the area of the triangle = 6*8*(1/2)=24
As bunuel said ZX and QY are equal, and QY is also PR's height of the triangle.
Since we know the triangles area and the length of PR we can compute:
24=10*x*(1/2)
x=4,8
I'm gonna refer to the graphics bunuel provided:
From the information given in the question, we can compute the area of the triangle = 6*8*(1/2)=24
As bunuel said ZX and QY are equal, and QY is also PR's height of the triangle.
Since we know the triangles area and the length of PR we can compute:
24=10*x*(1/2)
x=4,8
