Let us consider the quadrilateral ABCD shown below.
In the above quadrilateral, A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) and D(x_{4}, y_{4}) are the vertices.
To find area of the quadrilateral ABCD, now we have take the vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) and D(x_{4}, y_{4}) of the quadrilateral ABCD in order (counter clockwise direction) and write them column-wise as shown below.
Add the diagonal products x_{1}y_{2}, x_{2}y_{3}, x_{3}y_{4} and x_{4}y_{1 }are shown in the dark arrows.
(x_{1}y_{2} + x_{2}y_{3 }+ x_{3}y_{4} + x_{4}y_{1}) -----(1)
Add the diagonal products x_{2}y_{1}, x_{3}y_{2}, x_{4}y_{3} and x_{1}y_{4 }are shown in the dotted arrows.
(x_{2}y_{1 }+ x_{3}y_{2 }+ x_{4}y_{3} + x_{1}y_{4}) -----(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) ⋅ {(x_{1}y_{2} + x_{2}y_{3 }+ x_{3}y_{4} + x_{4}y_{1})
- (x_{2}y_{1 }+ x_{3}y_{2 }+ x_{4}y_{3} + x_{1}y_{4})}
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
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