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, y4are 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, y4of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below. Add the diagonal products x1y2, x2y3x3y4 and x4yare shown in the dark arrows.

(x1y2 + x2y+ x3y4 + x4y1) -----(1)

Add the diagonal products x2y1, x3y2, x4y3 and x1yare shown in the dotted arrows.

(x2y+ x3y+ 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 + x2y+ x3y4 + x4y1)

(x2y+ x3y+ 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.

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