Unit 6 Assessment Questions:
In 225 BC, Archimedes created this proof of the area of a circle in his book, Measurement of
a Circle.
Source: http://mathworld.wolfram.com/Circle.html
In the proof, he suggested that you cut a circle into small wedges and stack those wedges
next to each other. Then, cut the first wedge in half and place it at the other end to make a
shape similar to a rectangle.
How can this idea help prove the area of a circle?
Solution:
If the circle is instead cut into wedges, as the number of wedges increases to infinity,
a rectangle results, so A = (πr)r = πr2.
Given that the volume formula for a cone is V = 1/3πr2h,what is the approximate volume of
the cone below?
Solution:
183 cm3
Examine these two drinking glasses full of water.
What would need to be true about these glasses in order for the volume of water in the tall
glass to equal the volume of water in the short glass?
Solution:
The height of the water must be equal in both glasses. The radius of each glass must also be
equal.
Janine is planning on creating a water-based centerpiece for each of the 30 tables at her
wedding reception. She has already purchased a cylindrical vase for each table. The radius of
the vases is 6 cm and the height is 28 cm. She intends to fill them half way with water and
then add a variety of colored marbles until the waterline is approximately three-quarters of
the way up the cylinder. She can buy bags of 100 marbles in 2 different sizes, with radii of 9
mm or 12 mm. A bag of 9 mm marbles costs $3 and a bag of 12 mm marbles costs $4.
If Janine only bought 9 mm marbles how much would she spend on marbles for the whole
reception? What if Janine only bought 12 mm marbles? (Note: 1 cm3=1 ml)
Solution: (part a only) http://www.illustrativemathematics.org/illustrations/514
Jared is scheduled for some tests at his doctor’s office tomorrow. His doctor has instructed
him to drink 3 liters of water today to clear out his system before the tests. Jared forgot to
bring his water bottle to work and was left in the unfortunate position of having to use the
annoying paper cone cups that are provided by the water dispenser at his workplace. He
measures one of these cones and finds it to have a diameter of 7cm and a slant height
(measured from the bottom vertex of the cup to any point on the opening) of 9.1cm.
Note: 1 cm3=1 ml
How many of these cones of water must Jared drink if he typically fills the cone to within
1cm of the top and he wants to complete his drinking during the work day?
Suppose that Jared drinks 25 cones of water during the day. When he gets home he measures
one of his cylindrical drinking glasses and finds it to have a diameter of 7cm and a height of
15cm. If he typically fills his glasses to 2cm from the top, about how many glasses of water
must he drink before going to bed?
Solution: http://www.illustrativemathematics.org/illustrations/527
An Olympic-sized swimming pool is a rectangle that is 164 feet long, 82 feet wide and 10
feet deep.
Bowling balls have a diameter of 8.5 inches.
So, how many bowling balls could fit in an Olympic-sized swimming pool?
Solution:
Step 1:
We know that the swimming pool is a rectangular prism and that the bowling ball is a sphere.
We are asked how many bowling balls could fit inside the Olympic-sized swimming pool.
Step 2:
If we find the volume of the swimming pool and the volume of one ball, we could divide the
volume of the pool by the volume of one ball to get the number of balls in the pool.
Step 3:
Volume of pool:
V = 164 feet x 82 feet x 10 feet
V = 134480 ft3
Volume of ball:
V = 4/3πr3
We need to convert the measurements of the bowling ball into feet to match the dimensions
of the swimming pool
Diameter = 8.5 in = 0.70833 ft.
r = 0.35416 ft
V = 4/3π(0.35416)3
V = 0.186 ft3
Now, we divide the volume of the pool by the volume of a ball
134480 / 0.186 = 723010
So, an Olympic-sized swimming pool can hold approximately 723,010 bowling balls!
Step 4:
We were asked how many bowling balls could fit inside an swimming pool, which is what
we found. If we multiply the number of bowling balls we got by the volume of one bowling
ball, we get the original volume of the pool. So, we have solved the problem correctly.