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In the figure above, V represents an observation point at one end of

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Joined: 02 Sep 2009
Posts: 42545

Kudos [?]: 135229 [0], given: 12674

In the figure above, V represents an observation point at one end of [#permalink]

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New post 01 Sep 2017, 07:41
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In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

[Reveal] Spoiler:
Attachment:
2017-09-01_1836_002.png
2017-09-01_1836_002.png [ 4.89 KiB | Viewed 397 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 135229 [0], given: 12674

Expert Post
Math Expert
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D
Joined: 02 Aug 2009
Posts: 5336

Kudos [?]: 6088 [0], given: 121

Re: In the figure above, V represents an observation point at one end of [#permalink]

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New post 01 Sep 2017, 09:42
Bunuel wrote:
Image

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

[Reveal] Spoiler:
Attachment:
2017-09-01_1836_002.png



Let the point of right angle be A..
So VRA is a right angle triangle where VR = hypotenuse=10
Perpendicular=VA=5, so \(AR^2=10^2-5^2=100-25=75....AR=5√3\)...

But AS=10, so RS=AS-AR=10-5√3

A
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Kudos [?]: 6088 [0], given: 121

VP
VP
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P
Joined: 22 May 2016
Posts: 1105

Kudos [?]: 394 [0], given: 640

In the figure above, V represents an observation point at one end of [#permalink]

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New post 01 Sep 2017, 10:13
Bunuel wrote:
Image

In the figure above, V represents an observation point at one end of a pool. From V, an object that is actually located on the bottom of the pool at point R appears to be at point S. If VR = 10 feet, what is the distance RS, in feet, between the actual position and the perceived position of the object.

(A) 10 – 5√3
(B) 10 – 4√3
(C) 5
(D) 10 – 5√2
(E) 6

[Reveal] Spoiler:
Attachment:
2017-09-01_1836_002.png

To find perceived distance, use right triangle properties. VR is the hypotenuse of a right triangle.

Let the right angle in the diagram be X.

XR = perceived distance, from R to X, of object

For ∆ XRV, find length of leg XR using Pythagorean theorem

Length of hypotenuse VR, given, = 10
Length of leg XV, given, = 5

\(5^2 + (XR)^2 = 10^2\)
\((XR)^2 = 75\)

\(XR\) = \(\sqrt{75}\), so

\(XR\) = 5\(\sqrt{3}\)

Length of XS = 10 (actual distance of object from S to A)

Length of XR = 5\(\sqrt{3}\) (perceived distance of object from R to A)

Difference between actual and perceived is XS - XR =

10 - 5\(\sqrt{3}\)

Answer A

Kudos [?]: 394 [0], given: 640

In the figure above, V represents an observation point at one end of   [#permalink] 01 Sep 2017, 10:13
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