Question

check your answers by referring to the checkpoint 4 materials located at the back of your book.
ideally, at this point you are comfortable working with these types of problems and can solve them correctly. if you feel that you need more confidence when solving these types of problems, then review the checkpoint 4 materials and try the practice problems provided. from this point on, you will be expected to do problems like these correctly and with confidence.
this problem is a checkpoint for area and perimeter of circles and complex figures. it will be referred to as checkpoint 4.
find the area and perimeter or circumference of each figure.
a. circle with radius 3 cm.
b. circle with diameter 10 feet.
c. only the shaded region (each sector has equal area).
d.

Explanation:

Step1: Find the area of circle a

The formula for the area of a circle is \(A = \pi r^{2}\). Given \(r = 3\) cm.

$$A=\pi\times3^{2}=9\pi\approx9\times3.14 = 28.26\space cm^{2}$$

The formula for the circumference of a circle is \(C = 2\pi r\).

$$C=2\pi\times3=6\pi\approx6\times3.14 = 18.84\space cm$$

Step2: Find the area of circle b

Given \(d = 10\) ft, then \(r=\frac{d}{2}=5\) ft.
The area formula \(A=\pi r^{2}\), so \(A=\pi\times5^{2}=25\pi\approx25\times3.14 = 78.5\space ft^{2}\)
The circumference formula \(C=\pi d\) (since \(C = 2\pi r=\pi\times(2r)=\pi d\)), so \(C=\pi\times10 = 10\pi\approx10\times3.14=31.4\space ft\)

Step3: Find the area of the shaded sector in circle c

The circle is divided into 3 equal - sector. The formula for the area of a circle is \(A=\pi r^{2}\), with \(r = 12\) ft.
The area of the whole circle \(A=\pi\times12^{2}=144\pi\)
The area of one - sector \(A_{sector}=\frac{1}{3}\times144\pi = 48\pi\approx48\times3.14=150.72\space ft^{2}\)

Step4: Find the perimeter and area of figure d

The figure is composed of a rectangle and a semicircle.
For the area:
The area of the rectangle: \(A_{rectangle}=l\times w\). The diameter of the semicircle \(d = 8\) ft, so \(r = 4\) ft. The length of the rectangle \(l=22 - 4=18\) ft, \(w = 8\) ft. \(A_{rectangle}=18\times8 = 144\space ft^{2}\)
The area of the semicircle \(A_{semicircle}=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi\times4^{2}=8\pi\approx8\times3.14 = 25.12\space ft^{2}\)
The total area \(A=144 + 25.12=169.12\space ft^{2}\)
For the perimeter:
The perimeter of the rectangle part: \(P_{rectangle\space part}=22+(22 - 4)+8=48\) ft
The length of the semicircle arc \(L=\pi r=\pi\times4\approx12.56\) ft
The total perimeter \(P=48+12.56 = 60.56\space ft\)

Answer:

• a. Area: \(28.26\space cm^{2}\), Circumference: \(18.84\space cm\)
• b. Area: \(78.5\space ft^{2}\), Circumference: \(31.4\space ft\)
• c. Area of the shaded sector: \(150.72\space ft^{2}\)
• d. Area: \(169.12\space ft^{2}\), Perimeter: \(60.56\space ft\)