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CH 11 Study Guide B Needs to be edited!!!!! Study Guide A has not been constructed.
Solve. Show Your work.
1. Find the circumference of a circle with diameter 21 centimeters. Use 22/7 as an approximation for π.
2. Find the area of a quadrant with radius 6 inches. Use 3.14 as an approximation for π.
Area O = πr2
Area quadrant = 1/4(πr2)
= 1/4 x 3.14 x62
= 1/4 x 3.14 x 36
= 3.14 x 9
= 28.26 sq.in.
3. The circumference of a circular table is 18.84 feet. Find the radius of the table. Use 3.14 as an approximation for π.
4. The area of a circular cover is 2,464 square centimeters. What is the diameter of the cover? Use 22/7 as an approximation for π.
5. The diameter of a tricycle wheel is 7 inches. How many revolutions will the wheel need to turn to travel a distance of 11 feet? Use 22/7 as an approximation for π.
6. The figure is made up of two semicircles and a quadrant. Find the area of the shaded part. Use 3.14 as an approximation for π.
7. The figure is made up of a piece of wire. It consists of four semicircles and a straight line. The radius of the largest semicircle is 40 inches. The diameter of the smallest semicircle is 20 inches. The other two semicircles are each a radius of 20 inches. Calculate the length of the wire. Use 22/7 as an approximation for π.
8. Misa draws the design shown below. She wants to paint the shaded part of the design. What is the total area of the shaded parts? Use 22/7 as an approximation for π.
9. The figure is made up of four identical squares of side length 10 inches. A quadrant is drawn inside each square. Find the total area Use 3.14 as an a approximation for π.
ANSWER KEY
1. C = πD
C = 22/7 (21)
C = 22 x 3 = 66 cm.
2. Find the area of a quadrant with radius 6 inches. Use 3.14 as an approximation for π.
Area O = πr2
Area quadrant = 1/4(πr2)
= 1/4 x 3.14 x62
= 1/4 x 3.14 x 36
= 3.14 x 9
= 28.26 sq.in.
3. The circumference of a circular table is 18.84 feet. Find the radius of the table. Use 3.14 as an approximation for π.
C = 2πr
18.84 = 2 (3.14) r
18.84 = 6.28r
18.84/6.28 = r
r = 3
4. The area of a circular cover is 2,464 square centimeters. What is the diameter of the cover? Use 22/7 as an approximation for π.
A= πr2
2,464 = 22/7 x r2
7/22 x 2,464 = r2
7 x 112 = r2
r 2= 784 r = root of 784
r = 28 so D = 28 x 2 = 56 cm.
5. The diameter of a tricycle wheel is 7 inches. How many revolutions will the wheel need to turn to travel a distance of 11 feet? Use 22/7 as an approximation for π.
C = πd
Cwheel = 22/7 x 7
Cwheel = 22 inches.
11 feet x 12 inches = 132 inches.
132 /22inches = 6 full turns of tricycle wheel.
6. The figure is made up of two semicircles and a quadrant. Find the area of the shaded part. Use 3.14 as an approximation for π.
Step 1 Solve for quadrant:
Entire quadrant area: 1/4πr2 = 1/4 x 3.14 x 52
= 1/4 x 3.14 x 25
= 117.75
The shaded semi-circle just counts for the empty semi-circle, so the answer is just the answer for the shaded quadrant: 117.75
7. The figure is made up of a piece of wire. It consists of four semicircles and a straight line. The radius of the largest semicircle is 40 inches. The diameter of the smallest semicircle is 20 inches. The other two semicircles are each a radius of 20 inches. Calculate the length of the wire. Use 22/7 as an approximation for π.
largest semicircle: radius = 40 so 1/2 circumference = 1/2 x 2 xπ x r
= 22/7 x 40 = 125.71 in.
smallest semi circle = 1/2 x 2 xπ x r = 22/7 x 20/2 = 31.4 in. (r= 10)
2 middle semicircles = 1 full circle = 2 x π x r = 2 x 22/7 x 20 = 125.71 in.
total wire length = (125.71 x 2)+31.4 = 282.83
This question is easier on the test because numbers are divisible by 7 for easy factoring and results in whole number answers.
8. Misa draws the design shown below. She wants to paint the shaded part of the design. What is the total area of the shaded parts? Use 22/7 as an approximation for π.
This shape can be broken down in to 4 parts. Solve for 1 part and multiply by 4 for the final answer for all 4 parts.
Take the left quarter:
It is made of two quadrants and a half circle.
The shaded parts of the two quadrants equal a fully shaded square:
212 = 441 sq.cm.
The semicircle is the area of half a circle with a diameter of 21 cm.
A = πr2 so 3.14 x (1/2 x 21)2 = 3.14 x 110.25 = 346.185
Now add the two areas together: 441 + 346.185 = 787.185
This is the area for 1 quarter of the shape. Now multiply by 4 for the entire area of the shaded parts:
787.185 x 4 = 3148.74 cm.sq.
9. The figure is made up of four identical squares of side length 10 inches. A quadrant is drawn inside each square. Find the total area Use 3.14 as an a approximation for π.
Focus only on the middle two squares.
Inside these squares is a large triangle which is empty and is subtracted from the quadrants which would normally be full.
Calculate the full quadrant areas for two quadrants.
1 full quadrant = 1/4 x 3.14 x 102 = 78.5
2 full quadrants = 157
Then calculate the big empty triangle.
A = 1/2 x 20 x 10 = 100.
Subtract the big empty triangle area from the two full quadrants to leave the area of the shaded arcs for two squares.
157 - 100 = 57 in.sq.
Then double this to include the other two squares.
57 x 2 = 114 in.sq.
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