Brianna Zielinski MTH 329 Final Assessment. Geometry:

Transcription

1 Brianna Zielinski MTH 329 Final Assessment Geometry: The figure below is composed of eight circles, seven small circles and one large circle containing them all. Neighboring circles only share one point, and two regions between the smaller circles have been shaded. Each small circle has a radius of 5 cm. Calculate: a. The area of the large circle. b. The area of the shaded part of the figure. a. Since r = 5cm for the small circles, we can say that r = 15cm for the large circle, by adding together the radii of the small circles. a. To find the area of the large circle, we need to find its radius, which we can find by using the radii of the small circles, which is 5cm. Area of large circle = πr 2 A = π*15 2 = 225πcm 2 The length of the radius of the large circle, we can see in this picture, is equal to the radius of one small circle plus the diameter of another small circle. Since the diameter is 2 times the radius, we know that the radius of the large circle is 3 times the radius of the smaller circles. 3 * 5 = 15cm. Using this radius with the formula for the area of a circle, we find that the area of the large

2 circle is 225πcm 2. b. First, we ll find the area of all the spaces between the smaller circles (which includes the shaded area). Area of all small circles = 7(π*5 2 ) = 7(25π) =175πcm 2. Area of space between small circles = 225π - 175π = 50πcm 2. Area of shaded region = 50π / 6 = 25π / 3 cm 2. b. We can first find the area of all the spaces between the small circles by taking the area of the large circle and subtracting the areas of the smaller circles. The area of all the smaller circles is 7 times the area of one small circle with a radius of 5cm. We know the area of the large circle from our answer in part a. To find the area of the shaded region, we can see that the shape of the shaded region is repeated 6 times within the large circle. So, the area of the shaded region is 1/6 the area of the space between the small circles. In other words, dividing the area of the space between the small circles by 6 will give us the area of the shaded region. My own thoughts: This is a good question to students to show their problem solving. I think that the problem could even be expanded for problem based teachings so that they have to use more concepts of circles and area to answer more questions.

3 Data Analysis: Below are the 25 birth weights, in ounces, of all the Labrador retriever puppies born at Kingston Kennels in the last six months: 13, 14, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 18, 18, 19, 20 a. Use an appropriate graph to summarize these birth weights. b. Describe the distribution of birth weights for puppies born at Kingston Kennels in the last six months. Be sure to describe shape, center and variability. c. What is a typical birth weight for puppies born at Kingston Kennels in the last six months? Explain why you chose this value. a. I chose a dot plot to graph the data. a. A dot plot or a bar graph are both good options for representing the data given to see the shape and distributions (weight of puppies in oz.) b. The shape of the data in the graph seems similar to a normal curve, or a bell curve, where the majority lies in the center, gradually getting less and less and you go out to the maximum and minimum. There are different ways to describe the center of the data. The literal center, or median, is 17 oz. The average of the data, or the mean, is oz. The range of the data is 7 oz. and the standard deviation for the data is 1.552, which means that most weights are close to the mean. c. A typical birth weight for a puppy born in the last 6 months would be 18 oz. This is because 18 is the mode of the data, or the one found most often. So out of these puppies, the typical weight was 18 oz. b. Most of these descriptions can be directly seen in the graph or the data, but some numbers, like the mean and standard deviation, would be calculated using the data given. c. Another answer could be 17 oz. since it is the median and also very close to the average. My own thoughts: This problem could have students use different representations to display the data in a graph, if they were learning about different graph types. They also need to communicate what they can see in the graph, which is good for an introduction to analyzing graphs. Part c also gives room for reasoning as students explain their answer and back it up with proof.

4 Numbers & Operation: Without using the square root button on your calculator, estimate to 2 decimal places. (Hint: It is worth noting that 20 2 =400 and 30 2 =900.) as accurately as possible Since we know will be between 20 and 30, we can pick 25 to start = 625, which is too low = 784, still too low = 841, too high = = = is our estimate of the square root of 800. Using what we re given as a hint, we know that our answer will be between 20 and is right in the middle, so we can start there. By squaring 25, we are showing what 25 is square root of. We want to find the number that will get an answer close to 800, to show that it s an estimate of the square root of is too small, so we can pick a higher number like 28. After testing 28 and 29, we see that the answer will be between 28 and 29, so we can start with using tenth decimals. (1 decimal place) 28.3 gets us really close, but just over 800, so we can start using 100 th decimals. (2 decimal places) is still too high, and squared is just under 800. So, our answer is between those two numbers. We can see that gets us just barely closer to 800 than 28.29, so our estimate to 2 decimal places is My own thoughts: This problem is mostly just calculations, but there is room for different representations or problem solving strategies. I ve never seen a problem like this before, which is why I chose it.

5 Algebra: Leila tells Julius to pick a number between one and ten. Add three to your number and multiply the sum by five, she tells him. Next she says, Now take that number and subtract seven from it and tell me the new number. Twenty-three, Julius exclaims. a. Write an expression that records the operations that Julius used. b. What was Julius original number? c. In the next round Leila is supposed to pick a number between 1 and 10 and follow the same instructions. She gives her final result as 108. Julius immediately replies: Hey, you cheated! What might he mean? a. We will use x to represent the number between one and ten that Julius picks. So an expression of the operations used would be 5(x + 3) 7. b. 5(x + 3) 7 = 23 5x = 23 5x + 8 = 23 5x = 15 x = 3. c. 5x + 8 = 108 5x = 100 x = is not between 1 and 10, which was part of the rules. a. First, we add three to the number: x + 3. Then, we multiply this sum by 5: 5(x + 3). Lastly, we subtract 7 from this quantity: 5(x + 3) 7. b. We can find Julius original number by solving for x when we know the answer is 23, which Julius said was the new number he got after his operations on the original number. First, we distribute the 5 through the values in the parentheses. Then, we can combine like terms, or subtract 7 from 15, and then subtract the 8 from both sides. Finally, we divide both sides of our equation by 5 to find that Julius original number was 3. c. We can use the same steps we used in part b to find the original number Leila picked to get an answer of 108. We already did worked through the first couple steps in part b, so we can start with 5x + 8 on one side of our equation and solve for x, getting us and answer of 20, which is not between 1 and 10. You could go on to say that 10, the highest number you can choose, yields an answer of 58 and you can see that the answers get smaller as you count down from 10 as your original number, so 108 would be a clear signal that you had picked a number greater than 10.

6 Choice (Algebra): The students in Mr. Sanchez's class are converting distances measured in miles to kilometers. To estimate the number of kilometers, Abby takes the number of miles, doubles it, then subtracts 20% of the result. Renato first divides the number of miles by 5, then multiplies the result by 8. a. Write an algebraic expression for each method. b. Use your answer to part (a) to decide if the two methods give the same answer. a. m = number of miles Abby s method: 2m -.2(2m) Renato s method: 8( ) a. For Abby s method, she first takes the number of miles and doubles it: 2m. Then, she subtracts, from this value, 20% of the result, so we take 20% of 2m (found by multiplying 2m by.2) and subtract it from 2m: 2m -.2(2m). Renato started by dividing the number of miles by 5:. Then he multiplies that by 8: 8( ). b. We can start with Abby s method and change it to see if it looks like Renato s method. First, we know that 20% is the same as 1/5:.2 = 1/5. So, 2m. And to subtract, we can change 2m into a value using fifths: - = = 8( ). So, yes, the methods give the same answer. b. You can also start with Renato s method and change it to look like Abby s method, or change both of them. To change.2 to 1/5,.2(2m) = 1/5(2m) =. To change 2m into an equal value of fifths, 2m =. My own thoughts: This problem is a good one to see different representations in algebra. There are also connections to real-world problems, which is good to gain an interest in the students.

7 Final reflection: Glows Working with middle school students in the interviews and learning to question for learning Deeper knowledge of mathematical state standards Seeing the standards in classroom settings and incorporating them into a lesson Working with fractions Critiquing other teaching styles (Khan) Reading for understanding Grows Teaching for understanding Writing lessons that really get to the deep meanings behind the math Seeing my own flaws in how I teach Working with a wide variety of middle school math topics