Unit 4 - Practice Problems
Pavan Govu's Practice Problem #1
Question: Bob leaves his house to walk to school. To get around the woods in front of his neighborhood, Bob walks 6 blocks north and 8 blocks east. In his school, he finds a magical suit of armor in the janitor's closet that allows him to walk through anyone or anything, almost like a ghost. Bob lacks integrity and decides to keep the magical suit for himself. When he's coming home from school that day, he puts on his suit, and walks through the woods in a straight line from school to his house. If Bob did not find the suit, or if he had returned it to the janitor like he should have, and had to walk home from school the same way he walked to school, how many more blocks would he have had to walk?
How to Solve: The first step to solving this problem is to assign a variable for any missing or unknown value. In this scenario, we do not know the number of blocks that Bob has to walk if he goes from school to his house in a straight line. Let's represent this value with x. The next step, is to create an equation to solve for this variable using any or all of the given information. From the diagram given in the problem, we find that Bob's routes to school form a triangle, which the diagram further specifies as a right triangle with that little square. This means that the Pythagorean Theorem can be used to solve for x, which is the hypotenuse of the right triangle. The equation turns out to be: 6^2+8^2=x^2. Once an equation has been formulated to solve for the variable, the next step is to simply solve the equation. By following the PEMDAS procedure, we find that the first thing we have to solve are the exponents: 6^2=36 and 8^2=64. After exponents, we should add the squares: 64+36=100. Now, the equation is down to: 100=x^2. Once you're in this position, all you have to do is isolate the variable, and in this case, that means finding the square root of 100. Now, you should know off the top of your head that the square root of 100 is 10, and if you don't, then... you've got problems. Anyway, we have concluded that x=10, but we're still not done. The question asks us the difference in the number of blocks Bob has to walk with the suit, and without, and answering this question is our final step. So, we just found that with the suit, Bob only has to walk 10 blocks, and now, we've to find how many blocks he has to walk without the suit. No equation is required for this either because it is simple, one-digit addition. We know that without the magical suit, Bob normally walks 6 blocks north and 8 blocks east to get to school, which adds up to 14 blocks total. And to answer the problem's question, we just have to subtract the smaller value from the larger value or basically do: 14-10. This gives us 4, and 4 is our answer. If Bob did not find the suit, or if he had returned it to the janitor like he should have, and had to walk home from school the same way he walked to school, he would've had to walk 4 more blocks. Lazy kid!
Pavan Govu's Practice Problem #2
Question: Long ago, Bob had found a magical suit of armor that allowed him to pass through anyone or anything without an effort. For example, when he wore it, he could pass through walls without having to use a door, walk through large crowds without having to say, "excuse me," and could grab the little stuffed animals out of those arcade claw games without having to spend a single penny. Anyway, when he got this suit, he dropped out of school and dedicated his life to stopping crime and bringing justice to his town with his new powers. Over time, he earned many certificates and medals, which he hung on a square wall in his house. Eventually, when the wall was nearly full, he decided that he wanted to and a border along the edges of the wall to make it look truly magnificent. If the square wall has an area of 256 feet, what is the length of the border that Bob will need?
How to Solve: The first step to solving this problem is to assign a variable for any missing or unknown value. In this scenario, we do not know the length of the border that Bob will need, or basically, the perimeter of Bob's wall. Let's represent this value with x. Also, to make calculations simpler, we can say that the length of one side of the wall is y. The next step, is to create an equation to solve for this variable, or variables in this case, using any of the given information or your own knowledge of mathematics. In the problem, it is stated twice that the wall Bob wants to put a border on is in the shape of a square. This means that all 4 edges of the wall are congruent. It also reveals that the length of one side is the square root of the area of the entire square. Since there are 2 variables, it is required that we create 2 equations. Our first equation can be based on the fact that the perimeter of the entire square is just the length of one side quadrupled: 4y=x. The other equation will use the postulate that the length of one side of a square is just the square root of the square's area: sqrt(256)=y. Now that we have our equations, we can solve for the variables. Now, fortunately for us, there is a simple method to solve 2 equations at once; since the value of y is equal to the square root of 256, then we can plug in that value for the y variable in the equation, "4y=x," giving us: 4[sqrt(256)]=x. The rest is simple. Now we are only left with one variable, which if we solve for, we will get the answer. So if we follow PEMDAS, we find that the first thing we have to simplify is the parentheses. In our equation, that means finding the square root of 256. If you don't know the square root of the top of your head, you can use this trick. We know that 10 squared is 100, and 20 squared is 400. Since 256 is between 100 and 400, that means that the square root of 256 is between 10 and 20. But there's another strategy to narrow down the numbers. Notice that the last digit of 256 is a 6. This means that the square of the last digit of the square root of 256 must end in a 4 or 6, because those are the only numbers between 10 and 20 that, when squared, produce numbers that end in a 6. In the end, we are left with 2 possible numbers, 14 and 16. Now, we can square each one and see which one gives us 256. 14 squared equals 196 which is not 256, so by the process of elimination, 16 should be the square root of 256. And if we check by squaring 16, we do indeed get 256. Anyway, so now that we've found the square root of 256, our equation becomes: 4(16)=x. And all that's left to do, according to PEMDAS, is multiply 4 and 16 to get our answer: 16*4=64. That is our answer. Bob will need 64 feet of bordering for his wall. Can we fix it? Yes we can!
Prashanth Gowda Practice Problem #1
Question- Jeff bought a 70 inch TV from Best Buy. He wanted to know the length and width of the TV screen. All he knows is that the length of the screen is 3 times the width of the screen. What's the area of the entire TV screen?
Answer- 1470 inches squared
Work-
Answer- 1470 inches squared
Work-
a^2 + b^2 = c^2 Pythagorean Theorem
(x)^2 + (3x)^2 = 70^2
x^2 + 9x^2 = 4900
10x^2 = 4900
x^2 = 490
x = √490
x = √49 x √10
x = 7√10 = width
3x = 21√10 = length
Area of Square = lw
Area of Square = 7√10 x 21√10
Area of Square = 1470 in. squared
Explanation- I first saw that the it was a 70 inch TV. This means the diagonal of the screen is 70 inches. I made a diagram and drew the diagonal. I noticed that this formed 2 Right Triangles. Next I saw that the length:width was 3:1. This made me label the length 3x and the width x. Next I solved for x using the Pythagorean Theorem. I got x to be 7√10 after simplifying the radical √490. After getting x, I plugged in x to get the length to be 21√10 and the width to be 7√10. Finally I multiplied the length and the width of the TV screen to get the area. I got the area of the TV screen to be 1470 inch squared
(x)^2 + (3x)^2 = 70^2
x^2 + 9x^2 = 4900
10x^2 = 4900
x^2 = 490
x = √490
x = √49 x √10
x = 7√10 = width
3x = 21√10 = length
Area of Square = lw
Area of Square = 7√10 x 21√10
Area of Square = 1470 in. squared
Explanation- I first saw that the it was a 70 inch TV. This means the diagonal of the screen is 70 inches. I made a diagram and drew the diagonal. I noticed that this formed 2 Right Triangles. Next I saw that the length:width was 3:1. This made me label the length 3x and the width x. Next I solved for x using the Pythagorean Theorem. I got x to be 7√10 after simplifying the radical √490. After getting x, I plugged in x to get the length to be 21√10 and the width to be 7√10. Finally I multiplied the length and the width of the TV screen to get the area. I got the area of the TV screen to be 1470 inch squared
Prashanth Gowda Practice Problem #2
Question- Juan is wanting to go from the pink point labeled Start to the yellow point labeled end. He doesn't know all far it is though. All of a sudden he realized that he could use the lengths of the street to form a right triangle. He drew a blue line to represent this. He knew that the base is 15 miles and the height is 8 miles. Using this knowledge, how far apart is his start point to his destination?
Answer- 17 miles
Work- a^2 + b^2 = c^2 Pythagorean Theorem
15^2 + 8^2 = c^2
225 + 64 = c^2
289 = c^2
17 = c
Explanation- I first noticed in the problem that the blue lines helped form a right triangle. This made me realize I was going to use the Pythagorean Theorem to find my answer. Next I saw that one blue lines was 15 miles long and the other one was 8 miles long. This helped me set my equation. Finally after solving this equation, I got that distance from his start point to his destination is 17 miles long.
Work- a^2 + b^2 = c^2 Pythagorean Theorem
15^2 + 8^2 = c^2
225 + 64 = c^2
289 = c^2
17 = c
Explanation- I first noticed in the problem that the blue lines helped form a right triangle. This made me realize I was going to use the Pythagorean Theorem to find my answer. Next I saw that one blue lines was 15 miles long and the other one was 8 miles long. This helped me set my equation. Finally after solving this equation, I got that distance from his start point to his destination is 17 miles long.
Meredith's Practice Problems
Question #1:
Given the square below, find the length of the diagonal, the perimeter, and find the area.
Given the square below, find the length of the diagonal, the perimeter, and find the area.
Explanation/Steps:
In order to find anything, you first have to gather information you know about the figure. We know it is a square because it said so in the question. We can also find out that the figure splits into two equal triangles. The angles of these triangles are 45-45-90 because it splits a 90° angle into two equal angles. What we know about 45-45-90 triangles is that when taking a leg, you can multiply it by the √2 ( square root of 2) to find it's length. When you do that in this problem, you get a diagonal/hypotenuse length of 6√2.
In order to find the perimeter, you just take 6 and multiply it by 4 (number of sides, since sides are congruent) so it would look like this: 6 x 4 = 24. Therefore, the perimeter is 24. To find the area, you multiply the two side lengths, so it would look like this: 6 x 6 = 36. Therefore, the area is 36.
Answers:
Diagonal/Hypotenuse: 6√2.
Perimeter: 24
Area: 36
In order to find anything, you first have to gather information you know about the figure. We know it is a square because it said so in the question. We can also find out that the figure splits into two equal triangles. The angles of these triangles are 45-45-90 because it splits a 90° angle into two equal angles. What we know about 45-45-90 triangles is that when taking a leg, you can multiply it by the √2 ( square root of 2) to find it's length. When you do that in this problem, you get a diagonal/hypotenuse length of 6√2.
In order to find the perimeter, you just take 6 and multiply it by 4 (number of sides, since sides are congruent) so it would look like this: 6 x 4 = 24. Therefore, the perimeter is 24. To find the area, you multiply the two side lengths, so it would look like this: 6 x 6 = 36. Therefore, the area is 36.
Answers:
Diagonal/Hypotenuse: 6√2.
Perimeter: 24
Area: 36
Question #2:
Do the lengths given in the triangle below create a triangle? If so, classify what type of triangle it is by its sides and angles.
Do the lengths given in the triangle below create a triangle? If so, classify what type of triangle it is by its sides and angles.
Explanation/Steps:
To find out if the lengths of the triangle above can actually form a triangle, you have to do a + b > c. In this case it would be 25 + 8 >36. When you solve that, it is 33 > 36, which is false. When the equation does not come out true, then the lengths cannot form a triangle.
Answer:
The lengths in the image given cannot create a triangle.
To find out if the lengths of the triangle above can actually form a triangle, you have to do a + b > c. In this case it would be 25 + 8 >36. When you solve that, it is 33 > 36, which is false. When the equation does not come out true, then the lengths cannot form a triangle.
Answer:
The lengths in the image given cannot create a triangle.
Patrick's (practice) Problems
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