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In the graph above, RS^2 + ST^2 =

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In the graph above, RS^2 + ST^2 = [#permalink]

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New post 17 Dec 2017, 23:16
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Image
In the graph above, RS^2 + ST^2 =

(A) 4
(B) 17
(C) 26
(D) 32
(E) 41

[Reveal] Spoiler:
Attachment:
2017-12-18_1010_001.png
2017-12-18_1010_001.png [ 11.95 KiB | Viewed 703 times ]
[Reveal] Spoiler: OA

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Re: In the graph above, RS^2 + ST^2 = [#permalink]

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New post 18 Dec 2017, 05:39
Bunuel wrote:
Image
In the graph above, RS^2 + ST^2 =

(A) 4
(B) 17
(C) 26
(D) 32
(E) 41

[Reveal] Spoiler:
Attachment:
2017-12-18_1010_001.png



RS^2 + ST^2 = RT^2

R = (1,4)
T = (5,3)

i.e. RT \(= (\sqrt{(3-4)^2 + (5-1)^2}\)

i.e. RT = √17

i.e. RS^2 + ST^2 = 17

Answer: Option B
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Re: In the graph above, RS^2 + ST^2 = [#permalink]

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New post 18 Dec 2017, 14:48
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Bunuel wrote:
Image
In the graph above, RS^2 + ST^2 =

(A) 4
(B) 17
(C) 26
(D) 32
(E) 41

[Reveal] Spoiler:
Attachment:
The attachment 2017-12-18_1010_001.png is no longer available

Attachment:
2017-12-18_1010_001ed.png
2017-12-18_1010_001ed.png [ 25.28 KiB | Viewed 429 times ]

Another way (not using distance formula):

Imagine or draw another right triangle that shares a hypotenuse, RT, with ∆ RST - see diagram

Original ∆ RST:
RS\(^2\) + ST\(^2\) = RT\(^2\)

New ∆ QRT:
QR\(^2\) + QT\(^2\) = RT\(^2\)
• QR = 4 (subtract x-coordinates of Q and R)
• QT = 1 (subtract y-coordinates of Q and T)
(4\(^2\) + 1\(^2\)) = (16 + 1) = 17

Answer B
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Re: In the graph above, RS^2 + ST^2 =   [#permalink] 18 Dec 2017, 14:48
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