Determine the area of each of the following figures. Assume that all angles that look like right angles are right angles.​

Answer:Part a) The area of the figure is [tex]31\ ft^2[/tex]Part b) The area of the figure is [tex]46\ ft^2[/tex]Step-by-step explanation:Part a) we know thatThe area of the figure is equal to the area of rectangle plus the area of trapezoidstep 1The area of rectangle is[tex]A_1=(b)(h)[/tex]we have[tex]b=8\ ft\\h=2\ ft[/tex]substitute the given values[tex]A_1=(8)(2)=16\ ft^2[/tex]step 2The area of trapezoid is[tex]A=\frac{1}{2}(b_1+b_2)(h)[/tex]we have[tex]b_1=4\ ft\\b_2=8-(1+1)=6\ ft\\h=3\ ft[/tex]substitute[tex]A=\frac{1}{2}(4+6)(3)[/tex][tex]A_2=15\ ft^2[/tex]step 3 Find the area of the figure[tex]A=A_1+A_2[/tex]substitute[tex]A=16+15=31\ ft^2[/tex]Part b) we know thatThe area of the figure is equal to the area of rectangle plus the area of two trianglesstep 1The area of rectangle is[tex]A_1=(b)(h)[/tex]substitute the given values[tex]A_1=(5)(6)=30\ ft^2[/tex]step 2The area of triangle at the top is[tex]A_2=\frac{1}{2}(b)(h)[/tex]we have[tex]b=4\ ft\\h=6\ ft[/tex]substitute[tex]A_2=\frac{1}{2}(4)(6)[/tex][tex]A_2=12\ ft^2[/tex]step 3The area of triangle at right is[tex]A_3=\frac{1}{2}(b)(h)[/tex]we have[tex]b=4\ ft\\h=2\ ft[/tex]substitute[tex]A_3=\frac{1}{2}(4)(2)[/tex][tex]A_3=4\ ft^2[/tex]step 4 Find the area of the figure[tex]A=A_1+A_2+A_3[/tex]substitute[tex]A=30+12+4=46\ ft^2[/tex]