The area of each of the 16 square regions in the figure abov : GMAT Problem Solving (PS)
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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]

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09 Sep 2013, 23:30
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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]

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10 Sep 2013, 00:03
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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]

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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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image modified.png [ 21.29 KiB | Viewed 2329 times ]

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Last edited by fozzzy on 10 Sep 2013, 04:27, edited 1 time in total.
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Re: The area of each of the 16 square regions in the figure abov [#permalink]

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10 Sep 2013, 00:23
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]

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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
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Re: The area of each of the 16 square regions in the figure abov [#permalink]

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07 Mar 2014, 06:43
Bunuel.. can you please elaborate on "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)."
I not able to follow this part.
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Re: The area of each of the 16 square regions in the figure abov [#permalink]

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07 Mar 2014, 16:58
Option B.
Let T be any convenient no. say 36.
Side of each smaller square=6
Side of bigger square=6*4=24
=576-36-144-216
=180
Which is 5*36
Ans:5T

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Re: The area of each of the 16 square regions in the figure abov [#permalink]

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08 Mar 2014, 04:57
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satsymbol wrote:
Bunuel.. can you please elaborate on "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)."
I not able to follow this part.

Attachment:

Untitled.png [ 32 KiB | Viewed 1851 times ]

The areas of three unshaded triangles are =
Half of the area of blue rectangle with area of 2T , thus T;
Half of the area of black rectangle with area of 8T , thus 4T.
Half of the area of red rectangle with area of 12T , thus 6T.

T + 4T + 6T = 11T.

Hope it's clear.
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Re: The area of each of the 16 square regions in the figure abov [#permalink]

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06 Jan 2016, 05:46
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Re: The area of each of the 16 square regions in the figure abov   [#permalink] 06 Jan 2016, 05:46
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