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Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!
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Bunuel
In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R
E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure:
\(area_P=\frac{1}{2}*4*1=2\);
\(area_Q=1*2=2\);
\(area_R=\frac{2^2}{2}=2\);
\(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.




Hey Bunuel, How did you find the area of R? Did you divide it into two triangles?
And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D
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Bunuel
In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R
E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure:
\(area_P=\frac{1}{2}*4*1=2\);
\(area_Q=1*2=2\);
\(area_R=\frac{2^2}{2}=2\);
\(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.




Hey Bunuel, How did you find the area of R? Did you divide it into two triangles?
And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D

Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).
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Bunuel
In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R
E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure:
\(area_P=\frac{1}{2}*4*1=2\);
\(area_Q=1*2=2\);
\(area_R=\frac{2^2}{2}=2\);
\(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.




Hey Bunuel, How did you find the area of R? Did you divide it into two triangles?
And just for curiosity is that figure a square?

I got the right answer as I figured out the area of other three and the only option suitable was D

Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

Bunuel for calculating area of R : how did you get 2^2 /2 , where is 2 coming from? Can you explain the counting of boxes?

same question for S and Q
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Bunuel
Bunuel
In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R
E. P, Q, R and S

The area of a rectangle equals to \(area=width*length\);

The area of a triangle equals to \(area=\frac{1}{2}*base*height\) (also the area of a square=diagonal^2/2);

Now it's easy to calculate the areas of given figure:
\(area_P=\frac{1}{2}*4*1=2\);
\(area_Q=1*2=2\);
\(area_R=\frac{2^2}{2}=2\);
\(area_S=\frac{1}{2}*3*1=1.5\).

Answer: D.



Yes, it's a square because its diagonals are equal and perpendicular bisectors of each other. The area of a square=diagonal^2/2.

Does this make sense?

P.S. \(area_{square}=\frac{d^2}{2}\) and \(area_{rhombus}=\frac{d_1*d_2}{2}\).

Bunuel for calculating area of R : how did you get 2^2 /2 , where is 2 coming from? Can you explain the counting of boxes?

same question for S and Q


R is a square --> the area of a square=diagonal^2/2 --> diagonal of R = 2 --> area = 2^2/2.

Q is a rectangle --> the area = length*width = 1*2.

S is a triangle --> the area = 1/2*base*height = 1/2*3*1 (consider vertical side as base).

Hope it's clear.
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Great explanation, but how can I see that diagonal of R is 2? (like can you possibly label this on the diagram or give coordinates) also, I don't see how the width is 2 for the rectangle.

Sorry for the bother.
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monirjewel
Attachment:
Untitled.png
In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R only
E. P, Q, R and S

Attachment:
Clips.jpg

Hi,

By using logic, Calculate the area of P=(1/2)bh=>(1/2)(4)(1)=>2

Now see the options and quickly eliminate A,B,C (doesn't contain P)

Now calculate Are of S=(1/2)(3)(1) is not equal to 2 hence eliminate E

Ans:D
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R is a square --> the area of a square=diagonal^2/2 --> diagonal of R = 2 --> area = 2^2/2.

About this, how did u figure that 2 is a diagonal and not a side length of square?
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People can be confused over R.

Use a right triangle to articulate side length of square R as per my diagram.
Attachments

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monirjewel

In the rectangular coordinate system above, for which of the shaded regions is the area 2?

A. None
B. Q Only
C. Q and R
D. P, Q and R only
E. P, Q, R and S

Attachment:
Untitled.png
Attachment:
Clips.jpg


Observe that the area of each small square formed by the gridlines is 1.

The area of region P is 2 since P is a triangle with base = 4 and height = 1. Thus, the area of P is (1/2)(4)(1) = 2.

The area of region Q is 2 since Q is a rectangle with length = 2 and width = 1. Thus, the area of Q is (2)(1) = 2.

The area of region R is 2 since R can be broken into four right triangles, each of which has an area of exactly 1/2.

The area of region S, however, is not 2. We can see that S is a triangle with base = 3 and height = 1. Thus, the area of S is (1/2)(3)(1) = 3/2.

Answer: D
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Bunuel
Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!


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Bunuel
Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!


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Fixed the lin. Thank you!

Here it is again: GEOMETRY: Shaded Region Problems!
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For R, would it be incorrect to just split it into two triangles? I did for one triangle 1/2*b*h=1/2*2*1 --> *2 triangles = area of 2
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Bunuel
For R, would it be incorrect to just split it into two triangles? I did for one triangle 1/2*b*h=1/2*2*1 --> *2 triangles = area of 2
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Yes, you could this way too, nothing wrong with that.
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Hello from the GMAT Club BumpBot!

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