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The area of each of the 16 square regions in the figure abov
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fozzzy
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The area of each of the 16 square regions in the figure abov [#permalink] New post  Updated on: 09 Sep 2013, 23:55
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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. \(7T\)

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Originally posted by fozzzy on 09 Sep 2013, 23:30.
Last edited by Bunuel on 09 Sep 2013, 23:55, edited 1 time in total.
Edited the question.
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Bunuel
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Re: The area of each of the 16 square regions in the figure abov [#permalink] New post  10 Sep 2013, 00:03
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fozzzy wrote:
Image
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. \(7T\)

Show Spoiler
Please provide detailed explanations!



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.

Answer: B.

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] New post  Updated on: 10 Sep 2013, 04:27
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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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Originally posted by fozzzy on 10 Sep 2013, 00:21.
Last edited by fozzzy on 10 Sep 2013, 04:27, edited 1 time in total.
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Bunuel
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Re: The area of each of the 16 square regions in the figure abov [#permalink] New post  10 Sep 2013, 00:23
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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] New post  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] New post  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] New post  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
Area shaded=24^2-(area of 3 unshaded right triangles)
=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] New post  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
Untitled.png [ 32 KiB | Viewed 7622 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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The area of each of the 16 square regions in the figure abov [#permalink] New post  18 Oct 2018, 08:02
fozzzy wrote:
Attachment:
Untitled.png
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. \(7T\)

Show Spoiler
Please provide detailed explanations!


my answer is
The 2nd horizontal line divides the shape into two triangles having the same base and the same ht.

area=2*1/2* sqrt(T)*(1+1+.5)*(sqrt(T)(1+1)
=5T

Is my approach correct??
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The area of each of the 16 square regions in the figure abov [#permalink] New post  19 Oct 2018, 05:14
Rather simple, but maybe more time consuming approach is to simply estimate and count the fraction of the shaded areas in each square:

\(\frac{1}{3}+\frac{2}{3}+\frac{1}{3}+\frac{1}{3}+\frac{2}{3}+\frac{2}{3}+\frac{3}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{15}{3}=5\)
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Re: The area of each of the 16 square regions in the figure abov [#permalink] New post  14 Mar 2021, 22:45
Maybe I am just missing the point of these questions? It's quite easy to see that the shaded pieces fit together to give you 5T.

Am I missing something?
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Re: The area of each of the 16 square regions in the figure abov [#permalink]
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