Q28 of 52 Page 1

Using integration, find the area of triangle ABC, whose vertices are A(2, 5), B(4, 7), and C(6, 2).

OR


Find the area of the region lying above the x-axis and included between the circle x2 + y2 = 8x and inside of the parabola y2 = 4x.


Plotting the points, we get,



Let us first find the equation of sides of the triangle,


Equation of line passing from points P(x1, y1) and Q(x2, y2) is given by:



Therefore,


Equation of AB:


y – 5 = (x – 2)


y = x + 3


Equation of BC:




Equation of AC:




Now, let us look at the limits that we have to take,



Therefore,





Area = 21 square units.


OR


Given Curves: C1: x2 + y2 = 8x


C2: y2 = 4x


Let us find the intersection point of the curves.


Putting the value of y2 from second curve in first curve,


x2 + 4x = 8x


x2 – 4x = 0


x(x – 4) = 0


x = 0 or x = 4


And,


y2 = 4.0, y = 0


y2 = 4.4, y = ± 4


On plotting we will get,



Area required is shared in the given figure,


Area required will be found by:



Therefore,


Area required = Area OPC + Area PCQ


For parabola, y = ±√4x


y = ±2√x


Area OPC





Now, for circle,


Area of PCQ





Now, we know that,



Therefore,


Area PCQ





= 4π



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