# Using integration, find the area of ΔABC, the equation of whose sides AB,BC and AC are given byY=4x+5,x+y=5 and 4y=x+5 respectively.

Given,

• ABC is a triangle

• Equation of side AB of y = 4x + 5

• Equation of side BC of x + y = 5

• Equation of side CA of 4y = x+5

By solving AB & BC we get the point B,

AB : y = 4x+5 , BC: y = 5-x

4x + 5 = 5-x

5x = 0

x = 0

by substituting x = 0 in AB we get y = 5

The point B = (0,5)

By solving BC & CA we get the point C,

AC : 4y = x+5 , BC: y = 5-x

4y - 5 = 5-y

5y = 10

y = 2

by substituting y = 2 in BC we get x = 3

The point C = (3,2)

By solving AB & AC we get the point A,

AB : y = 4x+5 , AC : 4y = x+5

16x + 20 = x+5

15x = -15

x = -1

by substituting x = -1 in AB we get y = 1

The point A = (-1,1)

These points are used for obtaining the upper and lower bounds of the integral.

From the given information, the area under the triangle (colored) can be given by the below figure.

From above figure we can clearly say that, the area between ABC and DEF is the area to be found.

For finding this area, the line equations of the sides of the given triangle are considered. By calculating the area under these lines we can find the area of the complete region.

Consider the line AB, y = 4x+5

The area under line AB:

From the above figure, the area under the line AB will be given by,

[ using the formula, and ]

= {[2(02) + 5(0)] – [2(-1)2 + 5 (-1)]}

= (0) – (2-5) = 0+3 = 3

Area under AB = 3 sq. units. ---- (1)

Consider the line BC, y = 5-x

Consider the area under BC:

From the above figure, the area under the line BC will be given by,

[ using the formula, and ]

Area under BC sq. units. ---- (2)

Consider the line AC,

Consider the area under AC:

From the above figure, the area under the line AC will be given by,

[ using the formula, and ]

Area under AC = 6 sq.units ---- (3)

If we combined, the areas under AB, BC and AC in the below graph, we can clearly say that the area under AC (3) is overlapping the previous twoareas under AB & BC.

Now, the combined area under the rABC is given by

Area under rABC

=Area under AB + Area under BC - Area under AC

From (1), (2) and (3), we get

Therefore, area under rABC sq.units.

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