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Bunuel
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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 04:06
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

68% (02:02) correct 32% (01:59) wrong based on 505 sessions

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What is the area of the region bounded by the lines y = |x| - 2 and y = 2 - |x| ?

A. 2
B. 6
C. 8
D. 12
E. 16
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C
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Satishqwerty
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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 04:52
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As we can see in the graph that the region bounded by lines y = |x| - 2 and y = 2 - |x| is a square ABCD.

Area of ABCD = \(2 \times Area( \triangle ABD)\)

Area of \( \triangle ABD = \frac{1}{2} \times BD \times AO\)

\(= \frac{1}{2} \times 4 \times 2 = 4\)

Area of ABCD \( = 2 \times 4 = 8\)

OA, C
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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 04:20
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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.

Posted from my mobile device
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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 04:21
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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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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 04:35
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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
Show more

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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What is the area of the region bounded by the lines y = |x| - 2 and y 23 Jul 2020, 05:07
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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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What is the area of the region bounded by the lines y = |x| - 2 and y 25 Aug 2020, 20:42
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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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What is the area of the region bounded by the lines y = |x| - 2 and y 30 Aug 2020, 23:01
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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
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Hi yashikaaggarwal,

How did you figure out it is a triangle?

Thank you :)

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What is the area of the region bounded by the lines y = |x| - 2 and y 30 Aug 2020, 23:13
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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
Show more
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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What is the area of the region bounded by the lines y = |x| - 2 and y 30 Aug 2020, 23:34
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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.
Show more


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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What is the area of the region bounded by the lines y = |x| - 2 and y 25 Feb 2026, 12:03
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