The Question has missing figure. The figure is in the attachment.
Answer:
The Area of the figure is [tex]66.2\ units^2[/tex].
Step-by-step explanation:
The given figure is composed of a rectangle and a triangle.
So ABCDE is a pentagon in which ABDE is a rectangle and BCD is atriangle.
Length of AB = 4.8
Length of AE = 9.6
Now draw a line segment from C to AE that is Point M on AE.
So, Length of AM = 9
[tex]\therefore Height\ of\ triangle=9-4.8=4.2[/tex]
Now,
Area of rectangle = [tex]length\times width[/tex]
[tex]Area\ of\ rectangle = 9.6\times4.8 = 46.08\ units^2[/tex]
Area of Triangle = [tex]\frac{1}{2}\times base\times height[/tex]
[tex]Area\ of\ triangle =\frac{1}{2}\times9.6\times 4.2=20.16\ units^2[/tex]
Thus area of figure=Area of rectangle+Area of Triangle
Area of the figure = [tex]46.08+20.16=66.24\ units^2[/tex]
Round to the nearest is 66.2 sq. units.
Hence The Area of the figure is [tex]66.2\ units^2[/tex].
