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# The area of each of the 16 square regions in the figure abov

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The area of each of the 16 square regions in the figure abov [#permalink]  09 Sep 2013, 23:30
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Difficulty:

45% (medium)

Question Stats:

62% (03:06) correct 37% (01:35) wrong based on 29 sessions
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The area of each of the 16 square regions in the figure above is T. What is the area of the shaded region?

A. \frac{13T}{3}

B. 5T

C. \frac{16T}{3}

D. \frac{11T}{2}

E. 7T

[Reveal] Spoiler:
[Reveal] Spoiler: OA

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Last edited by Bunuel on 09 Sep 2013, 23:55, edited 1 time in total.
Edited the question.
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Re: The area of each of the 16 square regions in the figure abov [#permalink]  10 Sep 2013, 00:03
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Expert's post
fozzzy wrote:

The area of each of the 16 square regions in the figure above is T. What is the area of the shaded region?

A. \frac{13T}{3}

B. 5T

C. \frac{16T}{3}

D. \frac{11T}{2}

E. 7T

[Reveal] Spoiler:

The whole area = 16T.

The areas of three unshaded triangles are = T + 6T + 4T = 11T (half of a rectangle with area of 2 + half of a rectangle with area of 12 + half of a rectangle with area of 8).

The area of the shaded region = 16T - 11T = 5T.

Similar question to practice: what-is-the-area-of-a-triangle-with-the-following-vertices-125983.html
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Re: The area of each of the 16 square regions in the figure abov [#permalink]  10 Sep 2013, 00:21
So basically what you have done is count the squares and calculated the area's I have modified the image and added color ( Red,Black and blue) so

Black has 4*2*1/2
Red has 4*3*1/2
Blue has 2*1*1/2

is this correct and then subtract from the total?
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Last edited by fozzzy on 10 Sep 2013, 04:27, edited 1 time in total.
Math Expert
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Posts: 15204
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Kudos [?]: 15808 [0], given: 1572

Re: The area of each of the 16 square regions in the figure abov [#permalink]  10 Sep 2013, 00:23
Expert's post
fozzzy wrote:
So basically what you have done is count the squares and took the area's I have modified the image and added color ( Red,Black and blue) so

Black has 4*2*1/2
Red has 4*3*1/2
Blue has 2*1*1/2

is this correct and then subtract from the total?

Yes, that's correct.
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Re: The area of each of the 16 square regions in the figure abov [#permalink]  19 Nov 2013, 22:05
16T-[(\frac{1}{2}*2\sqrt{T}*\sqrt{T})+(\frac{1}{2}*4\sqrt{T}*3\sqrt{T})+(\frac{1}{2}*2\sqrt{T}*4\sqrt{T})] = 5T
Re: The area of each of the 16 square regions in the figure abov   [#permalink] 19 Nov 2013, 22:05
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