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24 Oct 2019, 00:46
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C
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Difficulty:
15%
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Question Stats:
88% (01:26) correct
13%
(02:35)
wrong
based on 48
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In the right-angled ∆ABC above, \(\frac{y}{3} = \frac{3}{x + y}\). What is the value of y?
(A) 5/2
(B) 3
(C) 9/5
(D) 5/9
(E) 7/2
Attachment:
2017-08-08_2112_002.png [ 7.11 KiB | Viewed 1899 times ]
24 Oct 2019, 02:36
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4²+3²=(x+y)²=25
x+y=5
y/3=3/5
5y=9,y=9/5
Posted from my mobile device
x+y=5
y/3=3/5
5y=9,y=9/5
Posted from my mobile device
24 Oct 2019, 03:50
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Using Pythagoras theorem:
\(4^2 + 3^2 = (x+y)^2\)
x+y = +5 or -5 -> should be positive so +5
Given equation:
\(\frac{y}{3} = \frac{3}{(x + y)}\) => y(x+y) = 9 => y(5) = 9
\(y = \frac{9}{5}\)
IMO Option C it is!
\(4^2 + 3^2 = (x+y)^2\)
x+y = +5 or -5 -> should be positive so +5
Given equation:
\(\frac{y}{3} = \frac{3}{(x + y)}\) => y(x+y) = 9 => y(5) = 9
\(y = \frac{9}{5}\)
IMO Option C it is!
