# Math in Focus Grade 6 Chapter 11 Lesson 11.3 Answer Key Real-World Problems: Circles

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.3 Real-World Problems: Circles detailed solutions for the textbook questions.

## Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.3 Answer Key Real-World Problems: Circles

### Math in Focus Grade 6 Chapter 11 Lesson 11.3 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
The circumference of the moon is the approximate distance around a circle with radius 1,736 kilometers. Find the circumference of the moon.
Circumference of moon = 2πr

= km
The circumference of the moon to the nearest 10 kilometers is kilometers.
The circumference of the moon to the nearest 10 kilometers is 10,900 kilometers,

Explanation:
Given the radius of circle as 1,736 kilometers, So circumference of the moon is 2πr = 2 X 3.14 X 1,736 kilometers = 10,902.08 kilometer nearest 10 is 10,900 kilometers.

The circumference of the moon to the nearest 1,000 kilometers is kilometers.
The circumference of the moon to the nearest 1,000 kilometers is 11,000 kilometers,

Explanation:
We got the circumference of the moon as 10,902.08 to the nearest 1,000 kilometers rounding we get 11,000 kilometers.

Question 2.
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. How much ribbon does Sarah need? Round your answer to the nearest inch.

Length of semicircular arc AB = $$\frac{1}{2}$$ • 2πr

= 1 •
= in.
Semicircular arcs AO and OB have the same length.
Total length of semicircular arcs AO and OB
= 2 • $$\frac{1}{2}$$ • πd

= 1 •
= in.
Distance around the card
= length of semicircular arc AB + total length of semicircular arcs AO and OB
= +
= in.
in.
Sarah needs approximately inches of ribbon
Sarah needs approximately 49 inches of ribbon,

Explanation:
A greeting card is made up of three semicircles. O is the center of the large semicircle. Sarah wants to decorate the distance around the card with a ribbon. Ribbon does Sarah need length of semicircular arc AB = $$\frac{1}{2}$$ • 2πr = πr, radius is 5.2 in. so length of semicircular arc AB = 3.14 X 5.2 in. = 16.328 inches, Total length of semicircular arcs AO and OB = 2 X $$\frac{1}{2}$$ X πd as diameter is 5.2 in. + 5.2 in. = 10.4 inches, So total length of semicircular arcs AO and OB = 3.14 X 10.4 inches = 32.656 inches. Now distance around the card is length of semicircular arc AB + total length of semicircular arcs AO and OB = 16.328 inches + 32.656 inches = 48.984 inches approximately 49 inches.

Question 3.
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. Find the length of the wire.

Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ
= + + +
= cm
The length of the wire is approximately centimeters.
The length of the wire is approximately 81 centimeters,

Explanation:
As part of her artwork, Sally bends a length of wire into the shape shown. The shape is made up of a semicircle and a quadrant. To find the length of the wire first we calculate length of semicircular arc $$\frac{1}{2}$$ X 2πr = πr as r = 12 cm so it is 3.14 X 12 cm = 37.68 cm, Length of arc RO = $$\frac{1}{4}$$ X 2πr = $$\frac{1}{2}$$ X πr = $$\frac{1}{2}$$ X 3.14 X 12 cm = 18.84 cm, Now Distance around the shape
= length of semicircular arc PQ + length of arc RO + RP + OQ = 37.68 cm + 18.84 cm + 12 cm + 12 cm = 80.52 cm approximately 81 centimeters.

Complete. Use $$\frac{22}{7}$$ as an approximation for π.

Question 4.
Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. Find the area of the remaining pizza.

Area of remaining pizza = area of quadrant + area of semicircle
= +
= cm2
The area of the remaining pizza is approximately square centimeters.
The area of the remaining pizza is approximately 2,885 square centimeters,

Explanation:
Given Judy baked a pizza and had part of it for lunch. After the meal, the shape of the remaining pizza is made up of a semicircle and a quadrant. So to find the area of the remaining pizza first we calculate area of quadrant $$\frac{1}{4}$$ X πras radius is 35 cm it is $$\frac{1}{4}$$ X 3.14 X 35 cm X 35 cm = 961.625 square centimeters, Second we calculate area of semicircle = $$\frac{1}{2}$$ X πr2 = $$\frac{1}{2}$$ X 3.14 X 35 cm X 35 cm = 1,923.25 square centimeters, So area of remaining pizza = area of quadrant + area of semicircle =
961.625 square centimeters + 1,923.25 square centimeters = 2,884.875 square centimeters approximately 2,885 square centimeters.

Question 5.
A rug is made up of a quadrant and two semicircles. Find the area of the rug.

Total area of two semicircles = 2 • $$\frac{1}{2}$$ • πr2

= 1 •
= in.2
Area of figure = area of quadrant + total area of two semicircles
= +
= in.2
The area of the rug is approximately square inches.
The area of the rug is approximately 2,770 square inches,

Explanation:
A rug is made up of a quadrant and two semicircles. To calculate the area of the rug first we calculate area of quadrant $$\frac{1}{4}$$ X πras radius is 42 in. $$\frac{1}{4}$$ X 3.14 X 42 in. X 42 in. = 1,384.74 square inches, radius of semicircle is half the diameter 42 in/2 = 21 in. now total area of two semicircles = 2 X $$\frac{1}{2}$$ X πr2 = 3.14 X 21 in. X 21 in. = 1,384.74 square inches therefore area of figure = area of quadrant + total area of two semicircles = 1,384.74 square inches + 1,384.74 square inches = 2,769.48 square inches
approximately 2,770 square inches.

Question 6.
The diameter of a bicycle wheel is 60 centimeters. How far does the wheel travel when it makes 35 revolutions? Give your answer in meters.
Circumference of wheel = πd

= cm
Distance traveled = circumference of wheel • number of revolutions
=
= cm
= m Divide by 100 to convert to meters.
The wheel travels approximately meters.
The wheel travels approximately 66 meters,

Explanation:
Given the diameter of a bicycle wheel is 60 centimeters. Circumference of wheel = πd = 3.14 X 60 centimeters = 188.4 centimeters, Distance traveled = circumference of wheel X number of revolutions = 188.4 centimeters X 35 revolutions = 6,594 now we convert to meters by dividing 100 so we get 6,594/100 = 65.94 meters approximately
66 meters.

Question 7.
A park is shaped like the diagram below. It is a rectangle with semicircles at the two ends. There is a running track around the park.

a) The total length of the track is 220 yards. Find the length of $$\overline{P S}$$.
The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal.
Total length of semicircular arcs PQ and SR
= 2 • $$\frac{1}{2}$$ • πd
≈ 1 •
= yd
The length of $$\overline{P S}$$ and $$\overline{Q R}$$ are equal.
Total length of track = 220
PS + QR + total length of semicircular arcs PQ and SR = 220
PS + QR + = 220
PS + QR = 220
2 • PS =
PS =
= ys
The length of $$\overline{P S}$$ is approximately yards.
The length of $$\overline{P S}$$ is approximately 45 yards.

Explanation:
The total length of the track is 220 yards. The length of $$\overline{P S}$$, The track is made up of semicircular arcs PQ and SR, and sides PS and QR. Semicircular arcs PQ and SR are equal. Total length of semicircular arcs PQ and SR ith diameter 35 yd is 2 X $$\frac{1}{2}$$ X πd = 3.14 X 35 yd = 109.9 yards,
The length of $$\overline{P S}$$ and $$\overline{Q R}$$ are equal.
Total length of track = 220 yards,
PS + QR + total length of semicircular arcs PQ and SR = 220 yards,
PS + QR + 109.9 yards = 220 yards,
PS + QR = 220 yards – 109.9 yards = 90.1 yards, (PS = QR),
2 PS = 90. 1 yards,
PS = 90.1 yards/2 = 45.05 yards approximately 45 yards.

b) A jogger runs once around the track in 125 seconds. What is his average speed in yards per second?
The jogger runs yards in 125 seconds.
125 seconds →
1 second →
= yd
The average speed of the jogger is yards per second.
The average speed of the jogger is 1.76 yards per second,

Explanation:
Given a jogger runs once around the track in 125 seconds, The jogger runs 220 yards in 125 seconds.
125 seconds = 220 yards, 1 second = 220/125 yards = 1.76 yards, So the average speed of the jogger is 1.76 yards per second.

c) A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. How many minutes will he take to water the entire park? Round your answer to the nearest minute.
= ÷ 2
= yd
The areas of the two semicircles are equal.
Total area of two semicircles = 2 • $$\frac{1}{2}$$ • πr2
= 2 • $$\frac{1}{2}$$ •
= 1 •
= yd2
Area of rectangle PQRS = lw
=
= yd2
Area of park = area of rectangle PQRS + total area of two semicircles
= +
= yd2
Time taken = area of park ÷ rate of watering park
= ÷
= s
= min
min Round to the nearest minute.

The gardener will take approximately minutes to water the entire park.
The gardener will take approximately 11 minutes to water the entire park,

Explanation:
Given A gardener is hired to water the grass in the park. Using a machine, he waters 4 square yards per second. So many minutes will he take to water the entire park, Radius = diameter ÷ 2 = 35 yd ÷ 2 = 17.5 yd, The areas of the two semicircles are equal. Total area of two semicircles = 2 X $$\frac{1}{2}$$ X πr2 = 3.14 X 17.5 yd X 17.5 yd = 961.625 yd2,
Area of rectangle PQRS = length X width,
= 45  yd X 35 yd,
= 1,575 yd2,
Area of park = area of rectangle PQRS + total area of two semicircles
= 1,575 yd2 + 961.625 yd2 = 2,536.625 yd2,
Time taken = area of park ÷ rate of watering park
= 2,536.625 yd2 X seconds ÷ 4 yd2 ,
=  634.15625 s
= 634.15625 / 60 minutes = 10.5692 min
≈ 11 min round to the nearest minute.

Math in Focus Course 1B Practice 11.3 Answer Key

Question 1.
The radius of a circular pond is 8 meters. Find its area and circumference. Use 3.14 as an approximation for π.
Area = 200.96 square meters,
Circumference = 50.24 meters,

Explanation:
Given the radius of a circular pond is 8 meters so area is πr2 = 3.14 X 8 mts X 8 mts = 200.96 square meters,
Circumference = 2πr = 2 X 3.14 X 8 mts = 50.24 meters.

Question 2.
The diameter of a metal disc is 26 centimeters. Find its area and circumference. Use 3.14 as an approximation for π.
Area = 530.66 square centimeters,
Circumference = 81.64 centimeters,

Explanation:
Given the diameter of a metal disc is 26 centimeters so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms, so area is πr2 = 3.14 X 13 cms X 13 cms = 530.66 square centimeters, Circumference = 2πr = 2 X 3.14 X 13 cms = 81.64 centimeters.

Question 3.
The shape of a carpet is a semicircle. Use $$\frac{22}{7}$$ as an approximation for π.
a) Find its area.
Area of semicircle carpet is 76.93 square feet,

Explanation:
Area of the carpet which is semi-circle with diameter 14 ft first radius = 14 ft/2 = 7 ft and area of carpet is $$\frac{1}{2}$$ X πr2 = $$\frac{1}{2}$$ X 3.14 X 7 ft X 7 ft = 76.93 square feet.

b) Janice wants to put a fringed border on all sides of the carpet. How many feet of fringe are needed?

21.98 feet of fringe is needed,

Explanation:
Given Janice wants to put a fringed border on all sides of the carpet. So feet of fringe are needed $$\frac{1}{2}$$ X 2πr = πr and we have diameter as 14 ft radius = diameter/2 = 14 ft/2 = 7 ft now 3.14 X 7 ft = 21.98 feet.

Question 4.
The circumference of the rim of a wheel is 301.44 centimeters. Find the diameter of the rim. Use 3.14 as an approximation for π.
The diameter of the rim is 96 centimeters,

Explanation:
Given the circumference of the rim of a wheel is 301.44 centimeters. So it is 301.44 cms = 2πr ,
r = 301.44 cms/2 X 3.14 = 310.44 cms/6.28 = 48 cms, Now diameter = 2r = 2 X 48 cms = 96 centimeters.

Question 5.
A Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants. What is the area of the portion that is made of cloth? Use $$\frac{22}{7}$$ as an approximation for π.

The area of the portion that is made of cloth is 1,846.32 square centimeters,

Explanation:
Given a Japanese fan is made out of wood and cloth. The shape of the fan is made up of two overlapping quadrants.
First we calculate area of cloth and wood with radius 28 cm as we know area πr2 = 3.14 X 28 cm X 28 cm = 2,461.76 square cms, Area of wood with radius 14 cms is 3.14 X 14 cms X 14 cms = 615.44 square centimeters, Therefore area of the cloth is area of cloth and wood – area of the wood = 2,461.76 square centimeters – 615.44 square centimeters = 1,846.32 square centimeters.

Question 6.
A pancake restaurant serves small silver-dollar pancakes and regular-size pancakes. Use 3.14 as an approximation for π.

a) What is the area of a small silver dollar-pancake? Round your answer to the nearest tenth of an inch.
The area of a small silver dollar-pancake to the nearest tenth of an inch is 10 square inches,

Explanation:
The area of small silver dollar-pan cake with diameter 3.5 in. as radius = 3.5 in./2 = 1.75 in. now area = πr2 = 3.14 X 1.75 in. X 1.75 in. = 9.61625 square inches to the nearest tenth is 10 square inches.

b) What is the area of a regular-size pancake? Round your answer to the nearest tenth of a square inch.
The area of a regular-size pancake to the  nearest tenth of a square inch is 28 square inches,

Explanation:
The area of a regular-sizepan cake with diameter 6 in. as radius = 6 in./2 = 3 in. now area = πr2 = 3.14 X 3 in. X 3 in. = 28.26 square inches to the nearest tenth is 28 square inches.

c) If the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes, which is a better deal?
3 regular-size pancakes is a better deal,

Explanation:
6 X area of small silver dollar-pan cake = 6 X 9.61625 square inches = 57.6975 sqaure inches and 3 X area of a regular-sizepan cake = 3 X 28.26 square inches = 84.78 sqaure inches if the total price of 6 small silver-dollar pancakes is the same as the total price of 3 regular-size pancakes then the better deal is 3 regular-size pancakes as it has bigger area.

Question 7.
A park is shaped like a rectangle with a semicircle on one end, and another semicircle cut out of one side.
a) Find the distance around the park.
The distance around the park is 480 m,

Explanation:
The semicircle on one end is equal to another semicircle which is cut out of one side, so the distance around the park is the rectangle so it is 2 X (170 m + 70 m) = 2 X 240 m = 480 m.

b) Find the area of the park. Use $$\frac{22}{7}$$ as an approximation for π.

Explanation:
As the semicircle on one end is equal to another semicircle which is cut out of one side, so it the complete rectangle so the area of rectangle is length X breadth = 170 m X 70 m = 11,900 square meters.

Question 8.
The diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain. Use $$\frac{22}{7}$$ as an approximation for π.
a) Find the area of the sidewalk.
The area of the sidewalk is 346.185 square feet,

Explanation:
Given the diameter of a circular fountain in a city park is 28 feet. A sidewalk that is 3.5 feet will be built around the fountain so first area of the circular fountain with diameter 28 ft, radius = 28 ft/2 = 14 ft so area of the cirular fountain is πr2 = 3.14 X 14 ft X 14 ft = 615.44 square feet now the area of circular fountain with the side walk is
radius = 28 ft + 7 ft /2= 35 ft/2 = 17.5 ft so area is πr2 = 3.14 X 17.5 ft X 17.5 ft =
If the sidewalk is 3.5 feet means it is radius so the area = πr2 = 3.14 X 3.5 feet X 3.5 feet = 961.625 square feet, So only the side walk area = area of circular fountain with side walk – area of circular fountain = 961.625 square feet – 615.44 square feet = 346.185 square feet.

b) 0.8 bag of concrete will be needed for every square foot of the new sidewalk. What is the minimum number of bags needed?

The minimum number of bags needed are 277 bags,

Explanation:
The area of the sidewalk is 346.185 square feet and 0.8 bag of concrete will be needed for every square foot of the new sidewalk. So the minimum number of bags needed are 346.185 X 0.8 = 276.948 bags approximately 277 bags.

Question 9.
The diagram shows an athletic field with a track around it, The track is 4 feet wide. The field is a rectangle with semicircles at the two ends. Find the area of the track. Use 3.14 as an approximation for π.

The area of the track is 1,110.4 square feet,

Explanation:
The field is a rectangle with semicircles at the two ends. First we calculate the area with out the track so we have
rectangle area as length = 84 ft – 8 ft = 76 ft and width is 36 ft so area is 76 ft X 36 ft = 2,736 square feet,
Now area of two semicircles with diameter 36 ft so radius is 36 ft/2 = 18 ft, so area is 2 X $$\frac{1}{2}$$ X πr2 = 3.14 X 18 ft X 18 ft = 1,017.36 square feet, Field area is 2,736 sq ft + 1,017.36 sq ft = 3,753.36 square feet,
Now field area with track area of rectangle is length 76 ft and width is 36 ft + 8 ft = 44 ft is 76 ft X 44 ft = 3,344 square feet, area of two semicircles with diameter 44 ft so radius is 44 ft/2 = 22 ft, so area is 2 X $$\frac{1}{2}$$ X πr2 = 3.14 X 22 ft X 22 ft = 1,519.76 square feet, Field area with track is 3,344 square feet + 1,519.76 square feet = 4,863.76 square feet. Now the area of the track is field area with track – field area = 4,863.76 square feet – 3,753.36 square feet = 1,110.4 square feet.

Question 10.
The petal of a paper flower is created by cutting along the outlines of two overlapping quadrants within a square. Use 3.14 as an approximation for π.
a) Find the distance around the shaded part.
The distance around the shaded part is 15.7 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given diameter = 10 cm, So distance around the shaded part  =  2 X πd ÷ 4 = 2 X 3.14 X 10 cm ÷ 4 = 62.8 cm ÷ 4 = 15.7 cm.

b) Find the area of the shaded part.

39.25 square centimeters,

Explanation:
The area of the 2 quardants is 2 X $$\frac{1}{4}$$ X πr2 as diameter is 10 cm radius is 10 cm/2 = 5 cm,
Now area is 2 X $$\frac{1}{4}$$ X 3.14 X 5 cm X 5 cm = 39.25 square centimeters.

Question 11.
Wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Suppose each wheel makes 15 revolutions. Find the distance between the wheels after they have made these 15 revolutions.
Use $$\frac{22}{7}$$ as an approximation for π.

The distance between the wheels after they have made these 15 revolutions is 494.55 inches,

Explanation:
Given wheels A and B are placed side by side on a straight road. The diameter of wheel A is 56 inches. The diameter of wheel B is 35 inches. Radius of wheel A is diameter of wheel A/2 = 56 inches/2 = 28 inches, Radius of wheel B is diameter of wheel B/2 = 35 inches/2 = 17.5 inches, Circumference of wheel A = πr = 3.14 X 28 inches = 87.92 inches,
Circumference of wheel B = πr = 3.14 X 17.5 inches = 54.95 inches, If Wheel A makes 15 revolutions the distance is
87.92 inches X 15 = 1,318.8 inches and If wheel B makes 15 revolutions the distance is 54.95 inches X 15 = 824.25 inches so the distance between the wheels after they have made these 15 revolutions is distance of 15 revolutions of wheel A – distance of 15 revolutions of wheel B = 1,318.8 inches – 824.25 inches = 494.55 inches.

Question 12.
Nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, what is the area of the paper that is left after all the circles are cut out? Use 3.14 as an approximation for π.
The area of the paper that is left after all the circles are cut out is 278.64 square centimeters,

Explanation:
Given nine identical circles are cut from a square sheet of paper whose sides are 36 centimeters long. If the circles are as large as possible, So the area of the paper that is left after all the circles are cut out is area of the square – area of 9 circles, The area of the square is 36 cms X 36 cms = 1,296 square cms, diameter = 38 cms/3 approximately 12 cms, radius is 12 cms / 2 cms = 6 cms, Area of 9 circles is 9 X πr2 = 9 X 3.14 X 6 cms X 6 cms = 1,017.36 square centimeters, therefore the area of the paper that is left after all the circles are cut out = 1,296 sq cms – 1,017.36 sq cms = 278.64 square centimeters.

Question 13.
A designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If $$\overline{A B}$$ is 28 millimeters, find the area of the shaded part. Use $$\frac{22}{7}$$ as an approximation for π.

The area of the shaded part is 307.72 square millimeters,

Explanation:
Given a designer drew an icon as shown below. O is the center of the circle, and AB is a diameter. Two semicircles are drawn in the circle. If $$\overline{A B}$$ is 28 millimeters, if we see the design the shaded region is half of the circle so the area of the shaded part as diameter is 28 mm its radius is 28 mm/2 = 14 mm, Now area of shaded region = $$\frac{1}{2}$$ X πr2 = $$\frac{1}{2}$$ X 3.14 X 14 mm X 14 mm = 307.72 square millimeters.

Use graph paper. Solve.

Question 14.
Mary wants to draw the plan of a circular park on graph paper. The coordinates of the center of the park are A (3, 4). The circle has a radius of 3 units.
a) Use a compass and draw the plan of the circular park on graph paper.

Explanation:
Used a compass and drawn the plan of the circular park on graph paper as shown above.

b) Assume that the y-axis points north and south. A barbecue pit is located at the northernmost part of the park. Plot and label the location of the barbecue pit as point B. Give the coordinates of point B.

c) Connect points A, B, and the origin to form a triangle. Find the area of the triangle.

Question 15.
A wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. Use $$\frac{22}{7}$$ as an approximation for π.
a) Find the length of the wire.
The length of the wire is 112 cm,

Explanation:
Given to find the length of the wire it is perimeter of the square and length of square is 28 cm therfore length of the wire is 4 X 28 cm = 112 cms.

b) Find the area of the whole shape.

The area of the whole space is 2,014.88 square centimeters,

Explanation:
Given a wire is bent to make the shape below. The shape is made up of four identical circles. Each circle intersects two other circles. The four circles meet at a common point T, which is the center of square PQRS. So the area of the whole shape is (4 X $$\frac{1}{2}$$ the area of circle) + area of the square, radius = diameter/2 = 28 cm/2 = 14 cm, So [4 X ($$\frac{1}{2}$$ X 3.14 X 14 cm X 14 cm)] + (28 cm X 28 cm) = (2 X 615.44 square centimeter) +  784 square centimeters = 1,230.88n square centimeters + 784 square centimeters = 2,014.88 square centimeters.

Brain @ Work

Question 1.
The figure shows two identical overlapping quadrants. Find the distance around the shaded part. Use 3.14 as an approximation for π. Round your answer to the nearest tenth of a centimeter.

The distance around the shaded part is 47.1 cm,

Explanation:
Distance around the shaded part is of the 2 quadrants = 2 X circumference ÷ 4 and C = πd or 2πr as given radius = 15 cm, So distance around the shaded part  =  2 X 2 X 3.14 X 15  cms ÷ 4 = 3.14 X 15 cm = 47.1 cm.

Question 2.
A cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. Find the total area of the shaded parts of the design. Use $$\frac{22}{7}$$ as an approximation for π.

The total area of the shaded parts of the design is 42.14 square inches,

Explanation:
Given a cushion cover design is created from a circle of radius 7 inches, and 4 quadrants. The total area of the shaded parts of the design is area of square – 4 X area of the quadrant = 14 in X 14 in – 4 X $$\frac{1}{4}$$ X 3.14 X 7 in X 7 in = 196 square inches – 153.86 square inches = 42.14 square inches.

Question 3.
Two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. They are pushed towards each other at the same time, each making one revolution per second. How long does it take for them to knock into each other? Use $$\frac{22}{7}$$ as an approximation for π.

Long does it will take for them to knock into each other is 4.0926 meters,

Explanation:
Given two identical wheels are placed along a straight path so that their centers are 9.31 meters apart. The radius of each wheel is 3.5 centimeters. Circumfrence of the one wheel is 2 X 3.14 X 3.5 cms = 21.98 cms converting into meters 21.98 X 0.01 m = 0.2198 meters so 2 wheels means 0.2198 m X 2 =  0.4396 m their centers are apart of 9.31 meters so the distance for the 2 wheels to knock into each other is 9.31 meters X 0.4396 = 4.0926 meters.

Question 4.
A stage prop is made up of a semicircle and a quadrant. Its area is 924 square inches, Find the value of x. Use $$\frac{22}{7}$$ as an approximation for π.

924 sq in = $$\frac{1}{4}$$ X 3.14 X 2x in X 2x in + $$\frac{1}{2}$$ X 3.14 X 2x in X 2x in,