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04 May 2020, 04:18
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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:
65%
(hard)
Question Stats:
62% (02:27) correct
38%
(02:46)
wrong
based on 92
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In the figure above, angles QPA and BAP are right, PQ = 3, AB = 5 and AP = 8. If N is an intersection of BP and AQ, then what is the length of the perpendicular NM?
A. 5/8
B. 1
C. 15/8
D. 2
E. 5/2
Are You Up For the Challenge: 700 Level Questions
Attachment:
Untitled.png [ 10.9 KiB | Viewed 3866 times ]
04 May 2020, 04:49
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Bunuel
Strategy:
- Find two pairs of Similar Triangles
- Use the ratio of the corresponding sides
- Find two pairs of Similar Triangles
- Use the ratio of the corresponding sides
Answer: Option C
Please find the solution as attached.
Attachments
Screenshot 2020-05-04 at 5.17.51 PM.png [ 958.74 KiB | Viewed 3657 times ]
General Discussion
05 May 2020, 05:23
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Given: PQ = 3, AB = 5 and AP = 8
then what is the length of the perpendicular NM?
Assume NM = h, PM = x, and MA = y, so that x+y = 8
Projection of N on line AB is N', so AN' = h, and N'B = 5-h
Now, angle BNN' = angle NPM
\(\frac{(5-h)}{y }= \frac{h}{x}\) ----- (1)
solving (1), we get \(h = \frac{5x}{8}\)
Again, projection of N on line PQ is N'', so PN'' = h, and N''Q = 3-h
Now, angle QAP = angle QNN''
\(\frac{3}{8} = \frac{(3-h)}{x}\) ---- (2)
substituting 'h' (1) in (2), and solving (2) for x, we get x = 3,
finally h = NM = 15/8
then what is the length of the perpendicular NM?
Assume NM = h, PM = x, and MA = y, so that x+y = 8
Projection of N on line AB is N', so AN' = h, and N'B = 5-h
Now, angle BNN' = angle NPM
\(\frac{(5-h)}{y }= \frac{h}{x}\) ----- (1)
solving (1), we get \(h = \frac{5x}{8}\)
Again, projection of N on line PQ is N'', so PN'' = h, and N''Q = 3-h
Now, angle QAP = angle QNN''
\(\frac{3}{8} = \frac{(3-h)}{x}\) ---- (2)
substituting 'h' (1) in (2), and solving (2) for x, we get x = 3,
finally h = NM = 15/8
