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hiranmay
What is the area enclosed of the region enclosed by y = |x – 1| + 2, x = 1, x–axis and y–axis?

A. 2.5 --> correct: y = |x – 1| + 2, so (1) x=1, y=2 & if you increase or decrease x, y will always increase linearly with slop of 1, and intersect y-axis (2) @x=0, y=3, so enclosure enclosed by x = 1, x–axis and y–axis looks like trapezium with height=1, and two unequal side length = 2&3, so the area= (1/2)*(2+3)*1=5/2=2.5
B. 5
C. 7
D. 9
E. 10


Hi,

How do we deal with the absolute sign in the equation above. We should be having 2 equations to remove the absolute sign on? One >0 and the other one <0?
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While i am trying to put values in the x and y , i am not getting a trapezium. Can someone explain the plotting bit.
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Attachment:
Untitled.png

Area \(= (2*1)+\frac{1}{2} *1*1 = 2.5\)

Bunuel
What is the area enclosed of the region enclosed by y = |x – 1| + 2, x = 1, x–axis and y–axis?

A. 2.5
B. 5
C. 7
D. 9
E. 10


Are You Up For the Challenge: 700 Level Questions

Thanks a lot! I understood my mistake, I did not understand the problem correctly.
x=1 was another line (represented as red in the graph).
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Please explain the trapezium.
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Please explain the trapezium.

You can ignore this question. This type of geometry question is no longer tested on the GMAT.
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