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

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Math Expert
Joined: 02 Sep 2009
Posts: 55277
In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of  [#permalink]

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08 Aug 2017, 10:23
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:36) correct 21% (02:52) wrong based on 17 sessions

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In the right-angled ∆ABC above, y/3 = 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 452 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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08 Aug 2017, 11:04
Bunuel wrote:

In the right-angled ∆ABC above, y/3 = 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

Using Pythagorean Triplets Rule, $$\triangle$$ $$ABC$$ have sides in ratio $$= 3:4:5$$

Given $$BC = 3$$ and $$AB = 4$$

Therefore $$AC = 5$$

$$AC = x + y => 5$$

$$\frac{y}{3} = \frac{3}{(x + y)}$$

$$\frac{y}{3} = \frac{3}{5}$$

$$y = \frac{3*3}{5} = \frac{9}{5}$$

Re: In the right-angled ∆ ABC above, y/3 = 3/(x + y). What is the value of   [#permalink] 08 Aug 2017, 11:04
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