Answer:
14 units
Solution,
A( 1,4)
B(3,6)
C(8,4)
D(3,2)
Now,
Finding the distance of A and C
A(1,4)----->(X1, y1)
C (8,4)---->(x2, y2)
AC :
√(x2,X1)^2+ (y2-y1)^2
![\sqrt{ {(8 - 1)}^(2) + {(4 - 4)}^(2) }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/l215lp1261gs9lzq1jqn5h4x4t3uycjacc.png)
![\sqrt{ {(7)}^(2) + {(0)}^(2) }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/rsxu60pppujtad6g6c9bbwmicwdppxcql1.png)
![√(49)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/uv918v2ezww58tbp3g9ezr3e3smc8gabw9.png)
![\sqrt{ {(7)}^(2) }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7xg5c8llpect9fm9a53i589jys7bwkozkj.png)
7 units
AC = 7 units
By connecting BD, we get 2 triangles
i.e ∆ABD and ∆ BCD
Total area of polygon
= Area of ∆ ABD + Area of ∆ BCD
Area of ∆ABD
= 1/2 * base *height
= 1/2* BD * AO
= 1/2*4*2
= 4 units
Again,
Area of ∆BCD
= 1/2* base* height
= 1/2 * BD*CO
= 1/2*4*5
= 10 units
Total area:
= Area of ∆ ABD + Area of ∆ BCD
= 4 units +10 units
= 14 units
Hope this helps...
Good luck on your assignment..