Area of ∆ABC
= Area of ∆ABD  + Area of trapezium
    BDEC – Area of ∆AEC
Now equation of the sides AB, BC and
CA are given by
y = 2(x – 1), y = 4 – x, y = 1
2  (x – 1), respectively Hence,
area of ∆ ABC =
2
3
3
1
2
1
1
2 (
1)
(4
)
2
x
x
dx
x dx
−dx
−
+
−
−
∫
∫
∫
=
2
3
3
2
2
2
1
2
1
1
2
4
2
2
2
2
x
x
x
x
x
x






−
+
−
−
−












=
2
2
2
2
1
3
2
2
2
1
4 3
4 2
2
2
2
2












−
−
−
+
× −
−
×
−
























–
2
1
3
1
3
1
2
2
2






−
−
−












= 3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2  = 4
and (x – 2)2 + y2 = 4 Solution Equations of the given circles are
x2 + y2 = 4 (1)
and
(x – 2)2 + y2 = 4