Question

4.5 finding composite perimeter and area
find the perimeter and area of the shapes below. use 3.14 for pi and round to the nearest hundredth. (2 decimal places)
a.
p = ft.
a = ft²
b.
p = in.
a = in²
c.
p = ft.
a = ft²

Explanation:

Step1: Analyze shape a

Separate it into two rectangles.

Step2: Calculate perimeter of shape a

$P_a=(34 + 8+12 + 20+7+(12 + 8 - 7))=94$ ft

Step3: Calculate area of shape a

Area of first rectangle: $A_1 = 34\times8=272$ ft², area of second rectangle: $A_2=20\times12 = 240$ ft², $A_a=272+240 = 512$ ft²

Step4: Analyze shape b

It is composed of two semi - circles and a square.

Step5: Calculate perimeter of shape b

The two semi - circles together form a full circle. $P_b = 2\pi r+2\times2$, with $r = 1$ in, $P_b=2\times3.14\times1+4=10.28$ in

Step6: Calculate area of shape b

Area of the square: $A_{s}=2\times2 = 4$ in², area of the circle: $A_{c}=3.14\times1^{2}=3.14$ in², $A_b=4 + 3.14=7.14$ in²

Step7: Analyze shape c

It is composed of a quarter - circle and a square.

Step8: Calculate perimeter of shape c

$P_c=\frac{1}{4}\times2\pi r+3\times2$, with $r = 2$ ft, $P_c=\frac{1}{4}\times2\times3.14\times2+6=9.14$ ft

Step9: Calculate area of shape c

Area of the square: $A_{s1}=2\times2 = 4$ ft², area of the quarter - circle: $A_{qc}=\frac{1}{4}\times3.14\times2^{2}=3.14$ ft², $A_c=4+3.14 = 7.14$ ft²

Answer:

a. $P = 94$ ft, $A = 512$ ft²
b. $P = 10.28$ in, $A = 7.14$ in²
c. $P = 9.14$ ft, $A = 7.14$ ft²