The area of each of the 16 square regions in the figure abov : 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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Updated on: 10 Sep 2013, 00:55

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35% (medium)

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73% (02:19) correct

27% (02:11) wrong

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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. \(7T\)

Spoiler: ::

Please provide detailed explanations!

Spoiler: OA

Originally posted by fozzzy on 10 Sep 2013, 00:30.
Last edited by Bunuel on 10 Sep 2013, 00:55, 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, 01:03

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. \(7T\)

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]

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Updated on: 10 Sep 2013, 05:27

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, 01:21.
Last edited by fozzzy on 10 Sep 2013, 05:27, 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, 01:23

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, 23:05

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, 07:43

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, 17:58

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]

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08 Mar 2014, 05:57

satsymbol wrote:

Attachment:

Untitled.png [ 32 KiB | Viewed 5231 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]

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18 Oct 2018, 09:02

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. \(7T\)

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]

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19 Oct 2018, 06:14

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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