# AREA OF TRIANGLE WORKSHEET 1

## About "Area of triangle worksheet 1"

Area of triangle worksheet 1 :

Area of triangle worksheet 1 is much useful to the students who would like to practice problems on area of triangles using the given three vertices.

## Area of triangle worksheet 1 - Problems

1)  Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

2)  If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

3)  Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

4)  If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b  =  1. where  a ≠ b.

## Area of triangle worksheet 1 - Answers

Problem 1 :

Find the area of the triangle whose vertices are (1, 2), (-3, 4) and (-5, -6)

Solution :

Plot the given points in a rough diagram as given below and take them in order (counter clock wise) Let the vertices be A(1, 2), B(-3, 4) and C(-5, -6)

Then, we have

(x₁, y₁)  =  (1, 2)

(x₂, y₂)  =  (-3, 4)

(x₃, y₃)  =  (-5, -6)

Area of triangle ABC is

=  (1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }

=  (1/2) x  { [1.4 + (-3).(-6) + (-5).2] - [(-3).2 + (-5).4 + 1.(-6)] }

=  (1/2) x  { [4 + 18 - 10] - [-6 - 20 -6] }

=  (1/2) x  {  - [-32] }

=  (1/2) x  { 12 + 32 }

=  (1/2) x  {  44 }

=  22 square units.

Hence, are of triangle ABC is 22 square units.

Let us look at the next problem on "Area of triangle worksheet 1"

Problem 2 :

If the area of the triangle ABC is 68 square units and the vertices are A(6, 7), B(-4, 1) and C(a, -9) taken in order, then find the value of "a".

Solution :

Let (x₁, y₁)  =  (6, 7),  (x₂, y₂)  =  (-4, 1) and (x₃, y₃)  =  (a, -9)

Given :  Area of triangle ABC   =  68 square units

(1/2) x  { [x₁y₂ + xy + xy₁] - [xy + xy + xy₃] }  =  68

Multiply by 2 on both sides,

{ [x₁y₂ + x₂y₃ + x₃y₁] - [x₂y₁ + x₃y₂ + x₁y₃] }  =  136

{ [6 + 36 + 7a] - [-28 + a - 54] }  =  136

[42 + 7a] - [a - 82]  =  136

42 + 7a -a +82  =  136

6a + 124  =  136

6a  =  12

a  =  2

Hence, the value of "a" is 2.

Let us look at the next problem on "Area of triangle worksheet 1"

Problem 3 :

Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.

Solution :

Using the concept of area of triangle, if the three points A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃)  are collinear, then

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃

Comparing the given points to A(x₁, y₁), B(x₂, y₂) and C (x₃, y₃), we get

(x₁, y₁)  =  (5, -2)

(x₂, y₂)  =  (4, -1)

(x₃, y₃)  =  (1, 2)

x₁y₂ + x₂y₃ + x₃y₁  =  5x(-1) + 4x2 + 1x(-2)

x₁y₂ + x₂y₃ + x₃y₁  =  -5 + 8 -2

x₁y₂ + x₂y₃ + x₃y₁  =  1 --------(1)

xy + x₃y₂ + xy₃  =  4x(-2) + 1x(-1) + 5x(2)

xy + x₃y₂ + xy₃  =  -8 - 1 + 10

xy + x₃y₂ + xy₃  =  1 --------(2)

From (1) and (2), we get

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃

Hence, the three points A, B and C are collinear.

Let us look at the next problem on "Area of triangle worksheet 1"

Problem 4 :

If P(x, y) is any point on the line segment joining the points (a, 0), and (0, b), then prove that x/a + y/b  =  1. where  a ≠ b.

Solution :

Clearly, the points (x, y), (a, 0), and (0, b) are collinear.

Then, area of the triangle  =  0

Since, area of the triangle is zero, we have

x₁y₂ + x₂y₃ + x₃y   =   xy + x₃y₂ + xy₃ ------(1)

(x₁, y₁)  =  (x, y)

(x₂, y₂)  =  (a, 0)

(x₃, y₃)  =  (0, b)

Plugging the above points in (1),

x.0 + a.b + 0.y   =   a.y + 0.0 + x.b

0 + a.b + 0   =   a.y + 0 + x.b

a.b   =   a.y + x.b

Divide by ab on bothe sides,

1  =  y/b + x/a

or

x/a + y/b  =  1

After having gone through the stuff given above, we hope that the students would have understood "Area of triangle worksheet 1".

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