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31 Aug 2021, 05:06
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
37% (02:02) correct
63%
(02:27)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
In a triangle ABC, AB=8, BC=7 and CA=6. Extend BC to P such that △PAB is similar to △PCA, then the length of PC is
A. 2
B. 5
C. 6
D. 9
E. 12
A. 2
B. 5
C. 6
D. 9
E. 12
31 Aug 2021, 18:18
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v12345
From similar triangles, we have \(\frac{PA}{PC} = \frac{PB}{PA} = \frac{AB}{CA}\).
Set PC = 7. Then from \(\frac{PA}{PC} = \frac{PB}{PA}\), we can find \(PA^2 = PC*PB = x*(x + 7)\) then \(PA = \sqrt{x*(x + 7)}\).
Finally from \(\frac{PA}{AB} = \frac{PC}{CA}\) we can get \(\frac{\sqrt{x*(x+ 7)}}{8} = \frac{x}{6}\).
\(x*(x + 7) = \frac{16}{9}*x^2\)
\(7x = \frac{7}{9}x^2\)
\(x = 7*\frac{9}{7} = 9\) is one solution.
Ans: D
10 Sep 2021, 10:23
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Let PC = x
△PAB is similar to △PCA
\(\frac{AB}{AC} = \frac{PA}{PC} = \frac{PB}{PA}\)
Capture.PNG [ 20.44 KiB | Viewed 4198 times ]
\(PA = \frac{AB * PC }{ AC}\)
\(PA^2 = PB * PC\)
\(\frac{AB^2 * PC^2 }{ AC^2} = PB * PC\)
\(PC = \frac{PB * AC^2 }{ AB^2}\)
\(x = \frac{(x+7) * 6^2 }{ 8^2}\)
\(x = \frac{(x+7) * 9 }{ 16}\)
\(16x =9x + 63\)
\(x = PC = 9\)
Hence, OA is (D).
△PAB is similar to △PCA
\(\frac{AB}{AC} = \frac{PA}{PC} = \frac{PB}{PA}\)
Attachment:
Capture.PNG [ 20.44 KiB | Viewed 4198 times ]
\(PA = \frac{AB * PC }{ AC}\)
\(PA^2 = PB * PC\)
\(\frac{AB^2 * PC^2 }{ AC^2} = PB * PC\)
\(PC = \frac{PB * AC^2 }{ AB^2}\)
\(x = \frac{(x+7) * 6^2 }{ 8^2}\)
\(x = \frac{(x+7) * 9 }{ 16}\)
\(16x =9x + 63\)
\(x = PC = 9\)
Hence, OA is (D).
