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Equation |x| + |y| = 10 encloses a certain region on the coordinate plane. What is the area of this region?

a)20 b)50 c)100 d)200 e)400

Can someone please explain this? Thanks!

Answer is d).

Explanation: |x| + |y| = 10 means there are 4 combination.

x + y = 10 -x - y = 10 x - y = 10 -x + y = 10

Now if you draw these four lines you will get square with diagonal size (20, and half diagonal size = 10) and it will make 4 small triangles with sides are 10, 10 and so its hypotaneous becomes 10 root 2.

Equation |x| + |y| = 10 encloses a certain region on the coordinate plane. What is the area of this region?

a)20 b)50 c)100 d)200 e)400

Can someone please explain this? Thanks!

Answer is d).

Explanation: |x| + |y| = 10 means there are 4 combination.

x + y = 10 -x - y = 10 x - y = 10 -x + y = 10

Now if you draw these four lines you will get square with diagonal size (20, and half diagonal size = 10) and it will make 4 small triangles with sides are 10, 10 and so its hypotaneous becomes 10 root 2.

Square area = 10 root 2 * 10 root 2 = 200.

Since all the sides are the same and you determine that the figure they are referring to is a square, then why wouldn't you just multiply 10*10 to get the area? Please clarify so I may have a better understanding. Thank you!

consider |x| = 2. this can be broken into x= 2 and x = -2. Similary |x| + |y|=10 has to be broken so that all possible values of x,y are taken into consideration. ie. x+y=10 , x-y=10, -x+y=10 , -x-y=10. What you are considering is only the 1st scenario ( x+y=10), but there are 3 more scenarios. Now if you draw all the above 4 graphs in X-Y axis , : x+y=10 ... slope 135 degree in 1st quadrant x-y=10 ... slope 225 degree in 4th quadrant -x+y=10 ... slope 45 degree in 2nd quadrant -x-y=10 ... slope 315 degree in 3rd quadrant .

All of them intersect to make a rhombus of side 10*(2^(1/2)) [Sqrroot( 10^2 + 10^2 )] or if you are looking 45 degree anticlockwise , a square with side 10*(2^(1/2)) . Thus the total area becomes {10*(2^(1/2)) }^2 = 200

I hope I have not made things even more complicated.
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|x/2| + |y/2| = 5 => |x| + |y| = 10...graph of the form y = mx + c [m is the slope] straight line graphs with x and y intercepts at +/- 10 four possibilities: y(0): x: {10, -10} and x(0): y: {10, -10} The shape is a square (two paired parallels at 90 degrees and same size) We only know diagonal = 20 units (from +10 to -10) length of a side: 2a^2 = 400 (a is one side of the square) a^2 = 200 Ans = D
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KUDOS me if you feel my contribution has helped you.

In quadrant 1, this is just the line x+y=10. Which is a line with X-int=10 and Y-int=10. So area enclosed with axes in Q1, is just 0.5*10*10 = 50 (a right angled triangle)

Now notice, if you replace x by -x the equation does not change --> So figure must be symmetric about the Y-axis So Q2, must also have a right angled triangle, of area 50 with the axis

Finally, if you replac y by -y, the equation is still the same --> So the figure must be symmetric about the X-axis So Q3 and Q4 have a mirror image of the above two triangles

So total area = 50*4 = 200

And the figure that this equation form is a diamond shape with (0,0) at the centre of it

Question - can we not generalize that the equation of a square in the form of a diamond around the origin will be -> |x| + |y| = A, where A is the side of the square?

in the explanation it says you can move the denominator of 2 out to the other side of the equation. even though they are in separate abs value... is it okay to apply this to all instances where there are 2 different fractions but with the same denominator each with it's own abs value?
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in the explanation it says you can move the denominator of 2 out to the other side of the equation. even though they are in separate abs value... is it okay to apply this to all instances where there are 2 different fractions but with the same denominator each with it's own abs value?

Generally: \(|xy|=|x|*|y|\), so if you have instead of \(y\) some specific number then you can factor its absolute value from the modulus: \(|-3x|=|3x|=3*|x|\) or \(|\frac{x}{-10}|=|\frac{x}{10}|=\frac{|x|}{10}\).

Hope it's clear.

As for the question itself: If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region? A. 20 B. 50 C. 100 D. 200 E. 400

First of all to simplify the given expression: multiply it be 2: \(|\frac{x}{2}|+|\frac{y}{2}|=5\) --> \(|x|+|y|=10\).

Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0): \(y=0\) --> \(|x|=10\) --> \(x=10\) and \(x=-10\); \(x=0\) --> \(|y|=10\) --> \(y=10\) and \(y=-10\).

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Attachment:

Enclosed region.gif [ 2.04 KiB | Viewed 12501 times ]

You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

@schumi252 i think the point 8,2 will lie within the region graphed by the diamond fig. hence while 8,2 is a true region, the area enclosed by it doesnot include all the possible values. hence the 10's and 0's work the best to graph the total area and hence all the possible solution points.

@calreg11 yes you can multiply inequality by 2. in general the rule is that you can multiply or divide both sides of an inequality by a positive no. without changing the inequality sign. if you multiply or divide by a negative no then the sign of the inequality changes.

If you quickly sketch out the outer boundaries of this equation you quickliy notice that you have a square bounded by (5,5), (-5,5), (-5,-5) and (5,-5). Additionally you also have the points (10,0), (0,10), (-10,0) and (0, -10). These points form 4 equal triangles outside the square formed by the previous coordinates. Therefore, area of square is 10x10 and area of each triangle is (10x5)/(2). Since there are 4 equal triangles we multiply 25x4. Total area is 100+100 = 200 (D)

what I feel whenever we get this type equatn |x| + |y| = A, will always be a SQUARE 2*A will be diagonal of that square and side of square will be A*(sqrt of 2) and area will be 2*A*A

so we have |x/2| +|y/2| = 5 we can write it as |x|+|y| = 10 and area will be 2*10*10 = 200

if u plot the graph once it will be easier for u understand

This one got me at first. Once I realized you could multiply the 2 over to the other side of the equation, it was only a matter of plugging in numbers for x and y to find the area. Answer is D!