Find the area of quadrilateral is formed by the point (8,6) (5,11) (-5,12)(-4,3)
Answers
The area of the quadrilateral is 79 sq. units.
Step-by-step explanation:
We have to find the area of the quadrilateral that is formed by the point (8,6), (5,11), (-5,12), and (-4,3).
Let the vertices of the quadrilateral be A() = A(8, 6)
B() = B(5, 11)
C() = C(-5, 12)
D() = D(-4, 3)
As we know that the area of the quadrilateral having vertices is given by the following formula;
Area of the quadrilateral ABCD =
Here; = 8,
.
Now, putting these values in the formula we get;
Area of the quadrilateral ABCD =
=
=
= = 79 sq. units
Hence, the area of the quadrilateral is 79 sq. units.
The area of the quadrilateral is 79 sq. units.
Step-by-step explanation:
We have to find the area of the quadrilateral that is formed by the point (8,6), (5,11), (-5,12), and (-4,3).
Let the vertices of the quadrilateral be A(x_1,y_1x
1
,y
1
) = A(8, 6)
B(x_2,y_2x
2
,y
2
) = B(5, 11)
C(x_3,y_3x
3
,y
3
) = C(-5, 12)
D(x_4,y_4x
4
,y
4
) = D(-4, 3)
As we know that the area of the quadrilateral having vertices is given by the following formula;
Area of the quadrilateral ABCD = \frac{1}{2}[(x_1-x_3) (y_2-y_4)- (x_2-x_4)(y_1-y_3)]
2
1
[(x
1
−x
3
)(y
2
−y
4
)−(x
2
−x
4
)(y
1
−y
3
)]
Here; x_1x
1
= 8, x_2= 5, x_3=-5, x_4= -4 , y_1 = 6, y_2=11,y_3=12 \text{ and } y_4=3x
2
=5,x
3
=−5,x
4
=−4,y
1
=6,y
2
=11,y
3
=12 and y
4
=3 .
Now, putting these values in the formula we get;
Area of the quadrilateral ABCD = \frac{1}{2}[(8-(-5)) (11-3)- (5-(-4))(6-12)]
2
1
[(8−(−5))(11−3)−(5−(−4))(6−12)]
= \frac{1}{2}[(13 \times 8)- (9 \times (-6))]
2
1
[(13×8)−(9×(−6))]
= \frac{1}{2}[104+54]
2
1
[104+54]
= \frac{1}{2} \times 158
2
1
×158 = 79 sq. units
Hence, the area of the quadrilateral is 79 sq. units.