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
[tex] \sqrt{ {(8 - 1)}^{2} + {(4 - 4)}^{2} } [/tex]
[tex] \sqrt{ {(7)}^{2} + {(0)}^{2} } [/tex]
[tex] \sqrt{49} [/tex]
[tex] \sqrt{ {(7)}^{2} } [/tex]
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..
