FIND AREA OF QUADRILATERAL WITH VERTICES

Let us consider the quadrilateral ABCD shown below. 

In the above quadrilateral, A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄) are the vertices.

To find area of the quadrilateral ABCD, now we have take the vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) and D(x₄, y₄) of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below. 

Add the diagonal products x₁y₂, x₂y₃, x₃y₄ and x₄y₁ are shown in the dark arrows.

(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) -----(1)

Add the diagonal products x₂y₁, x₃y₂, x₄y₃ and x₁y₄ are shown in the dotted arrows.

(x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄) -----(2)

Subtract (2) from (1) and multiply the difference by 1/2 to get area of the quadrilateral ABCD.

So, area of the quadrilateral ABCD is

=  (1/2) ⋅ {(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁)

- (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)}

Example 1 :

Find the area of the quadrilateral whose vertices are at

(–9, –2), (–8, –4), (2, 2) and (1, –3)

Solution :

By taking the points in counter clock wise, we have to find the area of CABD.

= (1/2)[(-4 + 36 + 24 + 2) - (-18 + 16 - 4 - 6)]

= (1/2)[(62-4) - (-28 + 16)]

= (1/2)[58 - (-12)]

 = (1/2)(58 + 12)

= (1/2)(70)

= 35 square units.

Example 2 :

Find the area of the quadrilateral whose vertices are at

(–9, 0), (–8, 6), (–1, –2) and (–6, –3)

Solution :

By plotting the given points in the graph paper, we get

By taking the points in counter clock wise direction, we find the area of BADC.

= (1/2)[(0 + 27 + 12 - 6) - (-54 + 0 + 3 + 16)]

= (1/2)[(39 - 6) - (-54 + 19)]

= (1/2)[(33) - (-35)]

= (1/2)(33 + 35)

= (1/2) (68)

  = 34 square units.

Example 3 :

Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (–4, –2), (–3, k), (3, –2) and (2, 3)

Solution :

Area of quadrilateral = 28 square units

[(-4k  + 6 + 9 - 4) - (6 + 3k - 4 - 12)] = 28(2)

(-4k + 11) - (3k - 10) = 56

-4k + 11 - 3k + 10 = 56

-7k = 56 - 21

-7k = 35

k = 35/(-7)

k = -5

Example 4 :

If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.

Solution :

a + b = 1 ----(1)

Because the given points are collinear, 

area of triangle = 0

[(-3b - 5a + 36) - (9a + 4b + 15)] = 0

-3b - 5a + 36 - 9a - 4b - 15 = 0

-5a - 9a - 3b - 4b + 21 = 0

-14a - 7b + 21 = 0

2a + b = 3 ----(2)

(2) - (1) : 

2a + b - (a + b) = 3 - 1

2a - a + b - b = 2

a = 2

Substitute a = 2 in (1).

2 + b = 1

b = -1

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.