Question

Find the area of each figure. Round your answer to 2 decimal places if required.
(Use π = 3.14)​

Answer 1

The values areas are:

Figure 1 = 69.81 in²

Figure 2 = 192.50 ft²

Figure 3 = 153.50 yd²

Figure 4 = 296.00 in²

Figure 5 = 126.00 ft²

Figure 6 = 26.30 ft²

Area of Compound Shapes

This exercise requires your knowledge about the area of compound shapes. For solving this, you should:

• Identify the basic shapes;
• Calculate your individual areas;
• Sum each area found.

The steps and solutions for each given figure are presented below.

• STEP 1 - Calculate the area for the figure 1

The figure 1 is composed by a rectangle and a semicircle. Therefore, you should sum the area of these geometric figures.

Area of rectangle  - $A_{rectangle}=l.w$, where:

l= length (12 in)and w=width (5 in).

$A_{rectangle}=l.w=12*5=60 in^{2}$

Area of semicircle- $A_{semicircle}=\frac{Area circle}{2}=\frac{\pi*r^{2} }{2}$, where:

r= radius ( $\frac{w}{2} =\frac{5}{2} =2.5$) and π = 3.14

$A_{semicircle}=\frac{Area circle}{2}=\frac{\pi*2.5^{2} }{2}=9.81 in^{2}$

Therefore, $A_{fig1}= 60 + 9.81=69.81 in^{2}$.

• STEP 2 - Calculate the area for the figure 2

The figure 2 is composed by a parallelogram and a trapezoid. Therefore, you should sum the area of these geometric figures.

Area of parallelogram - $A_{parallelogram}=b.h$, where:

b= length of the base (11 ft)and h=height (7 ft).

$A_{parallelogram}=b.h=11*7=77ft^{2}$

Area of trapezoid - $A_{trapezoid}=\frac{(a+b)*h}{2}$, where:

a= long base (20-7=13 ft ), b = short base (8 ft) and height (11 ft)

$A_{trapezoid}=\frac{(a+b)*h}{2}=\frac{(13+8)*11}{2}=\frac{21*11}{2}=\frac{231}{2}=115.50$

Therefore, $A_{fig2}= 77+ 115.5=192.50 ft^{2}$.

• STEP 3 - Calculate the area for the figure 3

The figure 3 is composed by a triangle and a trapezoid. Therefore, you should sum the area of these geometric figures.

Area of triangle - $A_{triangle}=\frac{b*h}{2}$, where:

b= length of the base (19 yd) and h=height (7 yd).

$A_{triangle}=\frac{b*h}{2} =\frac{19*7}{2} =\frac{133}{2}= 66.5 yd^{2}$

Area of trapezoid - $A_{trapezoid}=\frac{(a+b)*h}{2}$, where:

a= long base (19 yd ), b = short base (10 yd) and height (13-7=6 yd)

$A_{trapezoid}=\frac{(a+b)*h}{2}=\frac{(19+10)*6}{2}=\frac{29*6}{2}=29*3=87 yd^2$

Therefore, $A_{fig3}= 66.5+ 87=153.50 yd^{2}$.

• STEP 4 - Calculate the area for the figure 4

The figure 4 is composed by two rectangles. Therefore, you should sum the area of these geometric figures.

Area of rectangle 1  - $A_{rectangle}=l.w$, where:

l= length (16+5=21 in)and w=width (8 in).

$A_{rectangle}=l.w=21 *8=168 in^{2}$

Area of rectangle 2  - $A_{rectangle}=l.w$, where:

l= length (16 in)and w=width (5 in).

$A_{rectangle}=l.w=16*8=128 in^{2}$

Therefore, $A_{fig4}= 168+ 128=296.00 in^{2}$.

• STEP 5 - Calculate the area for the figure 5

The figure 5 is composed by a square and a parallelogram. Therefore, you should sum the area of these geometric figures.

Area of square - $A_{square}=l^2$, where:

l= length (9 ft).

$A_{square}=l^{2}=9^2=81 ft^{2}$

Area of parallelogram - $A_{parallelogram}=b.h$, where:

b= length of the base (9 ft)and h=height (14-9=5 ft).

$A_{parallelogram}=b.h=9*5=45ft^{2}$

Therefore, $A_{fig5}= 81+ 45=126.00 ft^{2}$

• STEP 6 - Calculate the area for the figure 6

The figure 5 is composed by a triangle and a semicircle. Therefore, you should sum the area of these geometric figures.

Area of triangle - $A_{triangle}=\frac{b*h}{2}$, where:

b= length of the base (6 yd) and h=height (4 yd).

$A_{triangle}=\frac{b*h}{2} =\frac{6*4}{2} =\frac{24}{2}= 12 yd^{2}$

Area of semicircle- $A_{semicircle}=\frac{Area circle}{2}=\frac{\pi*r^{2} }{2}$, where:

r= radius ( $\frac{6}{2} =3$) and π = 3.14

$A_{semicircle}=\frac{Area circle}{2}=\frac{\pi*3^{2} }{2}=14.3 yd^{2}$

Therefore, $A_{fig6}= 12+ 14.3=26.3 yd^{2}$

Learn more about the area of compound shapes here:

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