The area of the region bounded by the curves y = |x - 2|, x = 1, x =

y=x-2: y=0 x=2, and x=0 y=-2, points (2,0;0,-2)
y=-x+2: y=0 x=2, and x=0 y=2, points (2,0;0,2)
x=1 intersects y=x-2 at y=(1)-2=-1, point (1,-1)
x=1 intersects y=-x+2 at y=-(1)+2=1, point (1,1)
x=5 intersects y=x-2 at y=(5)-2=3, point (5,3)
x=5 intersects y=-x+2 at y=-(5)+2=-3, point (5,-3)

plot these points on the plane and you will get two triangles:
triangle a: 1•1/2=0.5
triangle b: 3•3/2=4.5
total area: 5

Answer (E)

can anyone explain this question? in details?

Hello Baraa900.


Diagram 1, shows the lines x = 1 and x = 5, which are 2 vertical lines.


Diagram 2 shows the line y = |x - 2|. Since this is a modulus, it denotes 2 different lines.

When x = 2, y = |0| = 0. This is the common point of intersection

When x = 1, y = |-1| = 1. At this point this line intersects the horizontal x = 1

When x = 5, y = |3| = 3. At this point this line intersects the horizontal x = 5


Diagram 2 shows the 2 triangle formed by y = |x - 2|, x = 1, x = 5 and the x axis.


The base of Triangle 1 = 2 - 1 = 1 and the height = 1. Area = 1/2 * 1 * 1 = 0.5

The base of Triangle 2 = 5 - 2 = 3 and the height = 3. Area = 1/2 * 3 * 3 = 4.5


The required area = 0.5 + 4.5 = 5


Option E

Arun Kumar

Asked: The area of the region bounded by the curves y = |x - 2|, x = 1, x = 5 and the x-axis is _________ sq. units



The area of the region bounded by the curves y = |x - 2|, x = 1, x = 5 and the x-axis = The area of triangle ABC + The area of triangle CDE
The area of triangle ABC = 1/2 * AB * BC = 1/2 * 1 * 1 = 1/2
The area of triangle CDE = 1/2 * CD * DE = 1/2 * 3 *3 = 9/2

The area of the region bounded by the curves y = |x - 2|, x = 1, x = 5 and the x-axis = 1/2 + 9/2 = 10/2 = 5

IMO E

Area of two triangle formed= 0.5*1*1+0.5*3*3=5

To understand the graphs of absolute values, check this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... h-to-mods/

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