Question

Find the area of quadrilateral is formed by the point (8,6) (5,11) (-5,12)(-4,3)

Answer 1

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($x_1,y_1$) = A(8, 6)

B($x_2,y_2$) = B(5, 11)

C($x_3,y_3$) = C(-5, 12)

D($x_4,y_4$) = 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 = $\frac{1}{2}[(x_1-x_3) (y_2-y_4)- (x_2-x_4)(y_1-y_3)]$

Here; $x_1$ = 8, $x_2= 5, x_3=-5, x_4= -4 , y_1 = 6, y_2=11,y_3=12 \text{ and } y_4=3$.

Now, putting these values in the formula we get;

Area of the quadrilateral ABCD = $\frac{1}{2}[(8-(-5)) (11-3)- (5-(-4))(6-12)]$    

                                                     = $\frac{1}{2}[(13 \times 8)- (9 \times (-6))]$

                                                     = $\frac{1}{2}[104+54]$

                                                     = $\frac{1}{2} \times 158$  = 79 sq. units

Hence, the area of the quadrilateral is 79 sq. units.

Answer 2

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.