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Bunuel
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@ Bunuel- is there any possibility value of z is missing.
Because by no means we can calculate value 3 values from 2 equations.
The above equation will form a square but to find value of sides of square we need any value or passing point- Z, X or Y.

If my reason is wrong, waiting for your explanation on this question.
Or it is a 3d figure.

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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Varunsawhney8
@ Bunuel- is there any possibility value of z is missing.
Because by no means we can calculate value 3 values from 2 equations.
The above equation will form a square but to find value of sides of square we need any value or passing point- Z, X or Y.

If my reason is wrong, waiting for your explanation on this question.

Posted from my mobile device

Fixed. It should have been "What is the area of the region bounded by the lines y = |x| - 2 and y = 2 - |x| ?"
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area of triangle enclosed by y = |x| - 2

Case 1: y = x - 2
when y = 0 , x = 2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 -------------(1)

Case 2: y = -x - 2
when y = 0 , x = -2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 --------------(2)

and y = 2 - |x|

Case 1: y = 2 - x
when y = 0 , x = 2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 -------------(3)

Case 2: y = 2 + x
when y = 0 , x = -2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 --------------(4)

Adding all four area = 8
Answer is C
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After Graphing (the hard part),

Can also be solved by Taking the Area of a Square = (Diagonal)^2 / 2

Square because Diagonals are Perpendicular Bisectors in a Square --- in this Case the Diagonals running along the X-Axis and Y-Axis are Perpendicular Bisectors.

(4)^2 / 2 = 8

Answer C
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yashikaaggarwal
area of triangle enclosed by y = |x| - 2

Case 1: y = x - 2
when y = 0 , x = 2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 -------------(1)

Case 2: y = -x - 2
when y = 0 , x = -2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 --------------(2)

and y = 2 - |x|

Case 1: y = 2 - x
when y = 0 , x = 2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 -------------(3)

Case 2: y = 2 + x
when y = 0 , x = -2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 --------------(4)

Adding all four area = 8
Answer is C


Hi yashikaaggarwal,

How did you figure out it is a triangle?

Thank you :)

Posted from my mobile device
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yashikaaggarwal
area of triangle enclosed by y = |x| - 2

Case 1: y = x - 2
when y = 0 , x = 2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 -------------(1)

Case 2: y = -x - 2
when y = 0 , x = -2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 --------------(2)

and y = 2 - |x|

Case 1: y = 2 - x
when y = 0 , x = 2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 -------------(3)

Case 2: y = 2 + x
when y = 0 , x = -2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 --------------(4)

Adding all four area = 8
Answer is C


Hi yashikaaggarwal,

How did you figure out it is a triangle?

Thank you :)

Posted from my mobile device
Remember, I told you you have to simply put ± sign in place of modulus.
I solved using that strategy only. Plus you can say every line which have x and y variable and is in form of y = MX+c will definitely have x and y intercept by putting significant other as zero.
Such as if y = x+4
X will be -4 when y = 0
Y = 4, when x = 0
Joining them will form a triangle and you can find area easily after that.
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yashikaaggarwal
Nups1324
yashikaaggarwal
area of triangle enclosed by y = |x| - 2

Case 1: y = x - 2
when y = 0 , x = 2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 -------------(1)

Case 2: y = -x - 2
when y = 0 , x = -2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 --------------(2)

and y = 2 - |x|

Case 1: y = 2 - x
when y = 0 , x = 2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 -------------(3)

Case 2: y = 2 + x
when y = 0 , x = -2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 --------------(4)

Adding all four area = 8
Answer is C


Hi yashikaaggarwal,

How did you figure out it is a triangle?

Thank you :)

Posted from my mobile device
Remember, I told you you have to simply put ± sign in place of modulus.
I solved using that strategy only. Plus you can say every line which have x and y variable and is in form of y = MX+c will definitely have x and y intercept by putting significant other as zero.
Such as if y = x+4
X will be -4 when y = 0
Y = 4, when x = 0
Joining them will form a triangle and you can find area easily after that.


Oh my goodness. I just visualised your explanation and now your original answer makes clear sense to me and it's amazing.

Yes I did as you had told me but I was failing to understand the role of Y Intercept Form.

Now it's clear.

Thank you :)
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yashikaaggarwal
area of triangle enclosed by y = |x| - 2

Case 1: y = x - 2
when y = 0 , x = 2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 -------------(1)

Case 2: y = -x - 2
when y = 0 , x = -2
when x = 0 , y = -2
area of triangle = 1/2*2*2 = 2 --------------(2)

and y = 2 - |x|

Case 1: y = 2 - x
when y = 0 , x = 2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 -------------(3)

Case 2: y = 2 + x
when y = 0 , x = -2
when x = 0 , y = 2
area of triangle = 1/2*2*2 = 2 --------------(4)

Adding all four area = 8
Answer is C
i have a question, why there is positive and negative case fore each equation?
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