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# M06-05

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:27
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75% (hard)

Question Stats:

49% (00:46) correct 51% (01:00) wrong based on 166 sessions

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If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

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16 Sep 2014, 00:27
1
1
Official Solution:

If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50$$.

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19 Oct 2015, 12:46
1
Bunuel wrote:
Official Solution:

If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50$$.

Another way to find the area of the square, if you don't know how to find it from the diagonal, is to take the area of the four triangles that are made from the x and y axis. You end up with $$(\frac{5*5}{2})*4$$.
=
Additionally, you can find the difference between any two points with the formula $$\sqrt{(X2-X1)^2+(Y2-Y1)^2}$$. Plug in (5,0) and (0,5) into that equation, then square the result to find the area.
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20 Jul 2016, 08:06
Is there a reason we can't use abs value of 3 and 2?
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30 Jul 2016, 12:14
Why cant the values of X and Y be 1 and 4 ? or 2 and 3 ? or 0 and 5 ?
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30 Jul 2016, 12:35
devbond wrote:
Why cant the values of X and Y be 1 and 4 ? or 2 and 3 ? or 0 and 5 ?

They can be. You would probably have a much harder time solving the problem using those values though.

Posted from my mobile device
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03 Aug 2016, 08:17
pate13 But you would be getting a different area wouldn't you? For example +/- 3 and +/- 2 solves the equation but would not yield an area of 50. Let me know if I am still missing something.
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03 Aug 2016, 08:37
dmaze01 wrote:
pate13 But you would be getting a different area wouldn't you? For example +/- 3 and +/- 2 solves the equation but would not yield an area of 50. Let me know if I am still missing something.

Hi,

The graph includes all points as mentioned..
the solution of |x|+|y|=5 will be as it has been shown above and x=4 when y=1 and so on are the points which we have joined to get the sketch..

We take values when x= WHAT when y=0 and y=WHAT when x=0 because the line being formed by values of x and y change direction at these points otherwise it continues as straight line..

Hope it helps
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22 Aug 2016, 04:39
I think this is a high-quality question and I agree with explanation.
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19 Jun 2017, 18:21
is there a way to know the sides from the information given? thx
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19 Jun 2017, 21:58
1
cocojiz wrote:
is there a way to know the sides from the information given? thx

Yes.

$$5^2 + 5^2 = side^2$$, which gives $$side = 5\sqrt{2}$$.

Or from the area: $$are=side^2=50$$, which gives $$side = 5\sqrt{2}$$.
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22 Sep 2017, 04:04
I think this is a High quality question and I agree with the explanation.

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!
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22 Sep 2017, 04:38
1
susheelh wrote:
I think this is a High quality question and I agree with the explanation.

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!

|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

Hope it's clear.
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22 Sep 2017, 04:42
A million thanks to you Bunuel for the quick answer. I fully understand this now. Awesome solution to an equally awesome question!

Bunuel wrote:
susheelh wrote:
I think this is a High quality question and I agree with the explanation.

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!

|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

Hope it's clear.

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15 Dec 2017, 09:24
Bunuel wrote:
Official Solution:

If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50$$.

Hey Bunuel,

I dont really get why its a square and not a circle. Couldn't |x|=p^2 and |y|=q^2, and so p^2+q^2 = \sqrt{5} ^2, which would make it a circle instead of a square, and thus change the area of the region.. why is what I'm doing wrong?
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15 Dec 2017, 09:28
13S12 wrote:
Bunuel wrote:
Official Solution:

If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

A. 5
B. 10
C. 25
D. 50
E. 100

X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0) (just make x equal to zero and find y and then make y equal to zero and find x). Now if we join these points we'll get the following region:

The diagonals of the quadrilateral are equal (10 and 10), and also are perpendicular bisectors of each other (as they are on X and Y axis), so the figure must be a square. Area of a square equals to $$\frac{\text{diagonal}^2}{2}=\frac{10^2}{2}=50$$.

Hey Bunuel,

I dont really get why its a square and not a circle. Couldn't |x|=p^2 and |y|=q^2, and so p^2+q^2 = \sqrt{5} ^2, which would make it a circle instead of a square, and thus change the area of the region.. why is what I'm doing wrong?

Why would you substitute x and y by p^2 and q^2? Would you still get a figure in xy-plane?
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21 May 2018, 14:42
Hi Bunuel,

I am also confused as to how you knew to connect the roots in that particular way. Can you please explain?
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29 May 2018, 03:16
Hi,
I got the points as (5,5), (5,-5), (-5,5) and (-5,-5). This makes each side 10 units and is a square. So the area is 100. Where am I going wrong? I got the points by substituting x=0 and y=0 for each equation of the line |x| + |y| = 5.
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29 May 2018, 07:39
Hi,
I got the points as (5,5), (5,-5), (-5,5) and (-5,-5). This makes each side 10 units and is a square. So the area is 100. Where am I going wrong? I got the points by substituting x=0 and y=0 for each equation of the line |x| + |y| = 5.

How did you get those values if you substituted x = 0 and y = 0?

|x| + |y| = 5:

If x = 0, then y is 5 or -5 --> two y-intercepts (0, 5), and (0, -5).
If y = 0, then x is 5 or -5 --> two x-intercepts (5, 0), and (-5, 0).

So, X and Y intercepts are (0, 5), (0, -5), (5, 0), and (-5, 0)
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23 Jun 2018, 20:47
There are mutiple of questions that in this format i.e
|X/A| + |Y/B| = C

In the question above
A=1, B=1 and C=5

Use the forumlae- 2C^2AB =50
M06-05 &nbs [#permalink] 23 Jun 2018, 20:47

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