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# In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of

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Joined: 02 Sep 2009
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In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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24 Oct 2019, 00:46
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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

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2017-08-08_2112_002.png [ 7.11 KiB | Viewed 422 times ]

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Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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24 Oct 2019, 02:36
4²+3²=(x+y)²=25
x+y=5
y/3=3/5
5y=9,y=9/5

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In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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24 Oct 2019, 03:50
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!
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Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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26 Oct 2019, 08:59
Bunuel wrote:

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

We see that triangle ABC is a 3-4-5 right triangle. Therefore, x + y = 5. Since we are given that y/3 = 3/(x + y), we have:

y/3 = 3/5

5y = 9

y = 9/5

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Re: In the right-angled ∆ABC above, y/3 = 3/(x + y). What is the value of   [#permalink] 26 Oct 2019, 08:59
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