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Using Integration, Find the Area of the Triangle Whose Vertices Are (2, 3), (3, 5) and (4, 4). - Mathematics

Sum

Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).

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Solution 1

The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)

Equation of line segment AB is

`("y" - 5) = (5-3)/(3-2) ("x" -3)`

=`"y" -5 = 2 ("x" -3)`

=`"y" = 2"x" -1`

Equation of line segment BC is

`("y" - 5) = (5-3)/(3-4) ("x" -3)`

=`"y" -5 = -1 ("x" -3)`

=`"y" = -"x" + 8`

Equation of line segment AC is

`("y" - 4) = (4-3)/(4-2) ("x" -4)`

=`"y" -4 = (1)/(2)  ("x" -4)`

= `"y" = ("x")/(2) + 2`

 

∴ Area of ΔABC = `int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"`

= `int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"`

= `[(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4`

= `(27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)`

= `(3)/(2) "sq. units"`.

Solution 2

The Vertices of ΔABC are A (2, 3), B (3, 5), and C (4, 4)

Equation of line segment AB is

`("y" - 5) = (5-3)/(3-2) ("x" -3)`

=`"y" -5 = 2 ("x" -3)`

=`"y" = 2"x" -1`

Equation of line segment BC is

`("y" - 5) = (5-3)/(3-4) ("x" -3)`

=`"y" -5 = -1 ("x" -3)`

=`"y" = -"x" + 8`

Equation of line segment AC is

`("y" - 4) = (4-3)/(4-2) ("x" -4)`

=`"y" -4 = (1)/(2)  ("x" -4)`

= `"y" = ("x")/(2) + 2`

Area of ΔABC = `int_2^3 [(2"x" -1) - ("x"/2+2)] d"x" + int_3^4 [(-"x" + 8) - ("x"/2 + 2)] . d"x"`

= `int_2^3 ((3"x")/2 -3) . d"x" + int_3^4 ((-3"x")/2 + 6) . d"x"`

= `[(3"x"^2)/(4) - 3"x"]_2^3 + [ (-3"x"^2)/4 + 6"x"]_3^4`

= `(27/4 - 9) - (3 -6) + (-12 + 24) - (-27/4 + 18)`

= `(3)/(2) "sq. units"`.

  Is there an error in this question or solution?
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