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Contents

Nearing the End

If you made it this far in the course, congratulations, you should really be proud of all your hard work. You've encountered so much algebra, and my hope is that you've learned a lot along the way. We have just two more units left, and I hope you'll stay with us to complete the course. In this unit we'll learn about the Pythagorean theorem and it's applications. Then we'll move on to see the different ways that we can slice a cone. Now these cones will produce some intereting patterns. And I can't wait for you to see them. For now let's look closely at a special right triangle.

A Special Right Triangle

You may have heard about special right triangles before. They're special, because they have a 90 degree angle for one of their angles. We're going to take a closer look at this one. It's a 3, 4, 5 right triangle. We call it that, since the ratio of the legs is 3 to 4 and the hypotenuse is 5. The legs are just the sides of the triangle that form the right angle. And the hypotenuse is the side that is directly across from the 90 degree angle. Not only is the hypotenuse across from the 90 degree angle, it's also the longest side of the entire triangle. We're going to look at an interesting relationship between the legs of the triangle and the hypotenuse. I'm going to shrink the triangle, and then I'm going to draw boxes off of each of the legs and the hypotenuse. These boxes are actually squares, so this square has side lengths of three, this square has side lengths of four, and this square has side lengths of five. What I want you to do is to tell me the area of each of these squares. You can type those numbers here.

A Special Right Triangle

This square has an area of 9 units squared. This one has an area of 16 units squared. And this one has an area of 25 units squared. Great work for finding these numbers. To find the area of any square, we simply square its side. So we know 3 squared equals 9. For this square, we'll have 4 times 4, which equals is 25.

The Pythagorean Theorem

So, we know if we square 3 we'll get 9, and if we square 4, we'll get 16, and if we square 5, we'll get 25. But what's even more amazing if we add these two together we get 25. It turns out that the area of the squares of the legs of the triangle sum to the the area of the square of the hypotenuse. Pretty amazing. It turns out that this relationship is true for any right triangle. So, lets take our triangle and give it unknown lengths. This side will be a, this side will be b and this hypotenuse will be c. What equation could you write to relate the variables a, b, and c? When you think you have that equation, type it in here.

The Pythagorean Theorem

Well, if we square a, and if we square b, and add those two together, we'll get c squared. So a squared, the area of this square, plus b squared, the area of this squared, equals c squared, the area of our largest square. Now that's great thinking if you found this equation.

Special Triangle Visual Proof

So when examining the 3-4-5 triangle, we saw the relationship of a squared plus b squared equals c squared. We'll square each of the legs of the triangles and then add the sum together to get the square of the hypotenuse. Here's another visual of what's going on. We'll have b squared which is 16, a squared which is

Visual Proof of the Pythagorean Theorem

So we said this works for any right triangle. We can square the leg, square another leg, and add those together to get the hypotenuse squared. But, why does this work? Well, let's see this by starting off with a square. Here's one large square, and then this small square has a side length of a, and this larger square has a side length of b. So, we know that the area of this square is a squared. And the area of this square is b squared. And now, what I'm going to do is break this white space up into four congruent rectangles. If this length is a, then we know that this length is also a. And we know if this length is b, then this length is also b. If our entire shape is a square, then we know that if this side is a plus b, then this side must also be a plus b. That's how we know that this side length is b. We can use the same reasoning to say that this side length here, is a. Since we know the entire side must be a plus b. So, notice that all four of these triangles are identical. They all have a leg of a, a leg of b, and a hypotenuse of c. The four triangles are just oriented in different ways. Like for this one, here's a, here's b, and here's they hypotenuse across the 90 degree angle c. So now what we're going to do is take this triangle and fix it at this point. We're going to rotate the triangle around this point. We're also going to take this triangle here and fix it at this point. So we can rotate the triangle around this point. So let's move these two triangles to a different position. We're going to swing this triangle up, around this point, and we're going to swing this triangle down, around this point. When we rearrange the triangles, we get this new arrangement. Notice that I have the same four triangles, with legs a and b, and a hypotenuse of c, and the area inside here is c squared. So, what do we know must be true about this area and this new area? Take some time to think and then fill in the missing statements to this sentence. Choose one of these statements to make this sentence true. And then, check any of these boxes that provide the correct reasoning.

Visual Proof of the Pythagorean Theorem

We know the new area c squared, is equal to the old area, a squared plus b squared. This of course Pythagoras' theorem, but the reason why it worked in that proof was because the areas of the 4 triangles did not change. In other words the light blue area that we had for a squared plus b squared, was the same after we rotated the triangle. After we rotate the 4 orange triangles the light blue square is still the same area. It's just need recombined to form c squared the new area.

Missing Triangles Sides

So, we know we can use the Pythagorean theorem for any right triangle. Any triangle that has a 90 degree angle, we can square the legs and then add them together to find the square of the hypotenuse. So, lets use the Pythagorean theorem to problem solve. What do you think would be the link of this hypotenuse? Write your answer here.

Missing Triangles Sides

This hypotenuse here would be 40 units in length. Great job if you've found this number. Remember, these are the legs of the triangle, since they form the the hypoteneuse. We don't know the length of the hypoteneuse, so we'll call it x. We'll start with our Pathagorean Theorem and we'll plug in the numbers for the legs here and then hypotenuse here. So, we'll have 32 squared plus 24 squared equals x squared. We'll square each of these numbers and then we'll have these two added together equals x squared. This gives us 1600 equals x squared. Now we want to take the square root of both sides of the equation to solve for x. When we take the square root, we'll need to include the plus or minus symbol. But, in this case, since the distance is positive, we can disregard the negative root. We only want the positive root. This gives us x equals 40, or 40 units in length. Now, there's another way to solve this problem. And it involves a 3, 4, 5 right triangle. This is one of our special triangles. For our special right triangle, the side links are the ratio of 3 to ratio. We know that the legs are 24 and 32. So, I'm going to put those here. If we multiply this leg by eight, we would end up with 24. And if we multiply this leg by eight, we would wind up with 32. So, we can see that eight would be the multiplier for this ratio. So, to get the side length of a hypotenuse of this triangle we would just need to multiple 5 by 8. This is another way without using the Pythagorean Theorem that we can find this hypotenuse.

Pythagorean Triples

There are some special right triangles that can help us problem solve even more quickly than using the Pythagorean Theorem. If you were able to solve the last problem without using the Pythagorean Theorem, then great work. If you didn't, I would go back and check that solution, see how we can do it. For the last problem, we could actually use this special ratio 3 to 4 to 5 to find a missing side length. If the side lengths are on the ratio of 3 to 4 to 5, then any multiple of that would also be a right triangle. So, if I multiplied each of these by 2, I could also have a 6, 8, 10 right triangle. We know this is a right triangle since 6 squared is 36, 8 squared is 64, and 10 squared is 100. A squared plus b squared equals c squared. So multiplying through by any number for this ratio would give us a right triangle. We could multiply through by 3 to get a different right triangle. In fact, that triangle would be a 9-12-15 right triangle. So by knowing a special right triangle, we can generate an infinite amount of larger right triangles. That's why this ratio is so important. In fact, it's called a Pythagorean Triple. There are a couple of Pythagorean Triples that appear often enough that it's helpful to know these. If you're able to quickly recognize these, then you can easily find missing side links. For example if we had this right triangle that had a leg of 35. And a hypotenuse of 125 units. What would be this missing leg? Now you could use a pythagorean theorem to solve for this missing leg. But I really want you to think about this in terms of ratios. Which of these ratios can help you find this missing leg. Choose the one that you think it corresponds to and then fill in the missing number based on the multiplier. Good luck.

Pythagorean Triples

This triangle has it's sides in a ratio of 7 to 24 to 25. Great thinking if you chose that ratio. We know that sense if we multiply this leg by 5 we would get leg, we simply simplify multiply 24 by 5. This gives us a side length of 120 units.

Pythagorean Theorem Check 1

Now not every right triangle will have a special ratio. For example this right triangle will not be in the ratio of the Pythagorean triples that we've seen before. But using Pythagorean Theorem, I know you can find this missing side length, what is it.

Pythagorean Theorem Check 1

This hypotenuse is the square root of 241. This square root would be about 15 and a half. Great thinking if you got this exact answer. For this problem, we know the legs are 4 and 15, and the hypotenuse is x. It's unknown, since we don't know it. So we'll have 4 squared plus 15 squared equals x squared. We'll square 4 to get 16 and we'll square 15 to get 225. Adding these two together, we'll get 241 equals X squared. We'll take the square root of both sides, keeping only the positive root, since this is a distance; which gives us x equals the square root of 241. Again, great work, if you got this correct.

Pythagorean Theorem Check 2

Here's another problem using the Pythagorean Theorem. See if you can find this missing side length. Also be sure that you give the exact answer. I don't want you to give an approximation that has a decimal point.

Pythagorean Theorem Check 2

This side lenght is 3 times the square root of 39. Great problem solving for getting that one right. Now 7, 24, 25 is a special right triangle that we could use. But since this hypotenuse isn't 25, we won't use that. Instead, we'll just need to use the Pythagorean Theorem. A squared plus b squared equals c squared. In this case, 7 is the leg, 20 is the hypotenuse, since it's across from the right angle, and x is the unknown leg we're trying to find. So, we'll have 7 squared plus x squared equals 20 squared. Squaring 7, we get 49 and squaring equals 351. We take the square root of both sides keeping only the positive root. So, x is equal to the square root of 351. We can simplify this square root by breaking it into 3 times 117. Then, we break it down one more time, to get 3 times 3 times 39. We have a repeated factor of 3. So, a 3 comes outside of our square root, and 39 stays inside. This gives us our final answer of 3 times the square root of 39.

Missing Sides of a Triangle

Here's a more challenging problem with the Pythagorean Theorem. This time, you have a right triangle whose hypotenuse is 5 inches longer than its longer leg. If the shorter leg is 5 inches shorter than the longer leg, what are the side lengths of the triangle? Before you solve for each of the sides of the triangle, I want you to try and use the information in the problem to come up with expressions for each of these. So you want to write an expression for the hypotenuse the shorter leg and the longer leg. Be sure you use the variable x when you write the expressions. You'll need to think carefully about how each of these is related to one another. Good luck here.

Missing Sides of a Triangle

When writing the expressions, it's best to use x for the longer leg. We know that since the hypotenuse is 5 inches longer than the longer leg and the shorter leg is 5 inches shorter than that longer leg. The longer leg is the connecting thread that links the relationships of the sides together. So, if we let X equal the longer leg, then the hypotenuse will be 5 units longer or x plus 5 and the shorter leg will be x minus 5. 5 units less than the longer leg. Great thinking if you found these 3 expressions this one was challenging.

Missing Sides of a Triangle Solved

Now that we have expressions for the sides of the triangle. Let's use the Pythagorean theorem to solve for x so we can find each of those sides. When you have each of the sides, enter them here.

Missing Sides of a Triangle Solved

The hypotenuse is 25, the longer leg is 20, and the shorter leg is 15 units. Great work for finding those three numbers. For a squared we'll have x minus 5 squared, for b squared we'll have x squared and for c squared we'll have the quantity x plus 5 squared. Keep in mind a and b are the legs, and c is our hypotenuse. Now we have one equation with one variable, which means we can solve. Squaring the quantity x minus 5, we get x squared minus 10x plus 25, and squaring the quantity x plus 5 squared, we get x squared plus 10x plus 25. This x squared just comes straight down here. Next, we'll add like terms on the left hand side of our equation to get 2x squared minus 10x plus 25. The right hand side of our equation stays the same. And since we have a quadratic equation, we want to move all of these terms to the left hand side so that we can get a quadratic expression set equal to 0. We move each of these terms to the other side of the equation by subtracting the appropriate amount. We'll subtract x squared, subtract 10x, and subtract 25 from both sides of the equation. We do this so that with the x squared term remains positive, and then we can factor. This leaves us with x squared minus 20x equals 0. We can factor an x from these two terms since it is the greatest common factor. So, of x times the quantity x minus 20. We set each factor equal to 0. And then we solve for x. X can equal 0 or x can equal positive 20. Since we know we're talking about the side length the longer leg must be 20. The 0 doesn't make sense in this case. So our longer leg will be 20, our hypotenuse will be 5 more which is 25. And our shorter leg will be five less which is 15.

Dimensions of a Rectangle

Here's a different type of problem that we can solve using the Pythagorean Theorem. A rectangle is three inches longer than it is wide. If the diagonal of the rectangle is 15 inches long, what are the rectangle's length and width? Try labeling this diagram with the appropriate expressions and then see if you can solve for the length and the width. When you're entering in the length, the length will be the longer side or the longer dimension. And the width will be the shorter one. Good luck here.

Dimensions of a Rectangle

The length of the rectangle is 12 units long and the width is 9 units. Great work if you found these two numbers. The key to this problem is that the diagonal is 15 inches long. The diagonal rectangle cuts from one corner to an opposite corner. So, we can draw in this line and consider this to be one of the diagonals. This line would be the other diagonal. But we only need 1 triangle in order to figure out the side links of our rectangle. We know the diagonal is 15 inches long so we can label this with 15. We'll let the width of the rectangle be x. Since we know the length is 3 inches longer than the width. This means that the length must be x plus 3. Here again is another. We're using algebra to relate two things. We have one unknown which is the width, and relating to the length. Since the length is three inches longer than the width. Now we have a right triangle and we're ready to use the pythagoreum theorem to solve for x. Whole squared e to the legs which is this side and this side and we know if we sum those squares together we'll get the squared hypotenuse or 15 squared. We'll combine these like terms together to get 2x squared plus 6x plus expression said equal to 0. These terms are all even, so we can divide our equation through by 2. This gives us x squared plus 3 x minus 108 equals 0. We find factors of negative 108 that sum to positive 3. Those factors are negative equal to 0 to solve for x. So, X can equal positive nine, or X can equal negative 12. A negative side length wouldn't make sense in this case, so we'll let X equal 9 here. So, if the width is 9, then we know that the length must be thinking if you got these two numbers.

Ladder Problem Diagram

Here's another challenging problem that uses the Pythagorean theorem. This time we'll have a ladder that 17 feet long, that leans against a building. The distance from the bottom of the ladder to the building is 7 feet less than the distance from the top of the ladder to the ground. What you need to find is the distance from the top of the ladder to the base of the building. Now, we'll walk through this problem just like one before. First, I want you to tell me what are the expressions for each of these sides? Try to label the parts of the diagram that you know, and then when you think that you have numbers or expressions for each of these. Enter them in the corresponding box. If you use a variable, make sure that you use the variable x. Good luck here.

Ladder Problem Diagram

Well, we know the ladder is 17 feet long. So we'll have 17 for this distance of the ladder. We know that the distance from the bottom of the ladder to the house or the building is 7 feet less, than the distance from the top of the ladder to the ground. So we know that this distance is 7 units less than this distance. Since we don't know this distance, we can use the variable x. And then we can write the expression x minus 7 for the distance from the base of the ladder to the base of the building. This is how we get this number and these two expressions. Great work if you found all 3.

Ladder Problem Solved

Now that we've set up a right triangle, I want you to find the distance from the top of the ladder to the base of the building. Solve for X.

Ladder Problem Solved

When we use the Pythagorean theorem to solve, we'll find that this distance, from the top of the ladder to the base of the building, is 15 feet. Great solving if you found it. But when you use the Pythagorean Theorem, we'll have x squared plus x minus 7 squared equals 17 squared. We'll square these legs and take the sum of the squares to get the square of the hypotenuse. Squaring these expressions, we get this equation. We'll add the like terms together here to get 2x squared minus 14x plus 49 equals 289. Then, we'll subtract 289 from both sides to get our quadratic expression set equal to 0. Each of these terms is divisible by 2, so we can divide our equation by 2. We take our quadratic expression, and we factor it by finding the factors of negative 120 that sum to negative 7. Those factors are negative 15 and positive 8. We set each of these factors equal to 0, and then we solve for values of x. So x can equal positive feet, since distance is always positive. So instead, x is going to equal 15 feet. This is the distance from the top of the ladder to the base of the building.

Pythagorean Theorem Practice 1

Now let's see if you can use the Pythagorean theorem to solve some practice problems on your own. First try to find the length of a hypotenuse of a right triangle with side lengths 18 inches and 24 inches. When you think you have the answer, type it in here.

Pythagorean Theorem Practice 1

Here, the hypotenuse is 30 inches. Great work if you found this number. We'll start with a right triangle, with side lengths 18 inches and 24 inches. We'll use x for the length of the hypotenuse since that's what we're finding. Now we just plug into the Pythagorean theorem, squaring the legs, and squaring our unknown hypotenuse length. When we square 18, we get 324. And when we square of both sides to solve for x. Here, we'll keep the positive root only, since we're talking about distance. The length that this hypotenuse is 30 inches. This triangle is actually a special right triangle with side lengths in the ratio of 3 to 4 to 5. So, we can actually use this to help us solve for x without using the Pythagorean Theorem. If we multiply 3 by 6, we'll get 18. And if we multiple 4 by 6 we'll get 24. This means to find this unknown hypotneuse we multiple the hypotneuse by 5 by our same multipler 6. We multiple 5 times 6 we'll get 30. Now not every right triangle will be so special. In this case it was since these side lengths were in the ratio of 3 to 4 to 5. This is why we looked at those Pythagorean triples earlier. Sometimes they can help us problem solve more quickly. But either way we solve we'll still get a hypotenuse of 30.

Pythagorean Theorem Practice 2

Let's see if you can solve the second problem. Here we'll have a right traingle with a side of 36 units and a hypotenuse of 45 units. I want you to find the missing side length.

Pythagorean Theorem Practice 2

The missing side length is 27 units. Great work if you found that number. Again, we'll start with the right triangle and this time we'll have one side of across from the 90 degree angle. This means that this missing side link must be x. We'll square the legs and sum those together to get the square of the hypotenuse. So we'll have x squared plus 1,296 equals 2,025. We'll subtract take the square root of both sides, keeping only the positive root, so x is equal to 27. The other way to solve this problem is to use the special triangle we would get 45. This must mean that 9 has to be the multiplier for each of the side lengths. So to find this missing piece, we simply multiply 3 by 9 to get If you know a couple of them it might make your problem solving easier.

Pythagorean Theorem Practice 3

Here's our 3rd practice problem for the Pythagorean Theorem. I want you to try to find the side lengths of a triangle whose hypotenuse is 4x and who's legs are 3x plus 1 and 2x plus 2. When you think you've found the numbers for each of them, enter them here.

Pythagorean Theorem Practice 3

The shorter leg is 12, the longer leg is 16, and the hypotenuse is 20. Great problem solving if you found those numbers. Our setup will be very similar to the problems before. We'll have the hypotenuse of 4x, and legs of 3x plus 1, and 2x plus 2. We square the first leg to get 9x squared, plus 6x plus 1. We square the second leg to get 4x squared plus 8x plus 4 and we square the hypotenuse to get 16x squared. Next we'll combine the like terms of x squared, the like terms for x and the constant terms to get this left hand expression. right hand side of the equation by subtracting them. We subtract each of these, so that way we can get an x squared term that's positive and we can get 0 on the left hand side. Next we'll use factoring by grouping, to get the factors of factor equal to 0. Solving for x, we'll have x can equal negative one third, or x can equal positive 5. Now, we won't let x equal negative one third since this side length will end up being negative four thirds. That wouldn't make sense. Instead, we'll let x equal 5, and we'll substitute it in for each of these expressions. This is how we get the shortest leg to be 12, the longer leg to be

Pythagorean Theorem Practice 4

Here's our fourth problem with the Pythagorean Theorem. See if you can solve it, and find each of the side legs for our right triangle. Think carefully as you set up the expressions, because this one's tough. Take your time, and be sure to check your answers in the end. Good luck.

Pythagorean Theorem Practice 4

The shorter leg is 20 meters long, the longer leg is 21 meters, and the hypotenuse is 29 meters. Great work if you found all three. Now, the setup might have been tough, so if you got stuck there, watch part of the solution, and then pause it and see if you can continue solving. To start setting up this problem, we'll let x be the shortest side of our right triangle. We do this since we know the longer leg of the right triangle is 1 meter more than the shorter leg. So, we know this longer leg has to be X plus 1. And now here's where we want to be careful. We know the hypotenuse is 8 meters longer than the longer leg. So, we'll take the length of the longer leg, X plus 1, and we'll add on 8 meters. So, the entire length of our hypotenuse Is x plus 1 plus 8, or just x plus 9. Now we have expressions for each side of our right triangle, and we can use the Pythagorean Theorem to solve. We square x, which is one of the legs, and x plus 1, the other leg. This gives us these two expressions and then we square by hypotenuse x plus 9, to get this expression on the right hand side of our equation. We combine the light terms here to get 2x squared and the rest of our equation remains the same. Next we subtract each of these terms from both sides of the equation. So we have 0 on the right, and x squared minus 16 x minus 80 on the left. Next, we factor this expression to get x minus 20, times x plus 4, equals 0. We set each factor equal to 0, which means the solutions for x could be positive 20 or negative 4, the opposite sign of the factors in the parentheses. We know a negative side length for x wouldn't make sense in this case, because we would have negative 4 here. So, this solution is not valid. When we plug in 20 for x in each of these expressions, we'll get our side links of 20, 21, and 29. With the hypotenuse being the longest one of 29 units. And since our original side links were reported in meters, we report each of these in meters.

Pythagorean Theorem Practice 5

Here's your fifth and final problem on the Pythagorean Theorem. Try setting up a diagram to model the situation and then when you think you have the solution, enter it here.

Pythagorean Theorem Practice 5

The distance from the base of the building to the foot of the ladder, or the bottom of the ladder, is 18ft. Great solving for getting this one right. This one was tough. To start, we should begin with a diagram. We'll have a building and then we'll have a 30ft ladder stretching from the ground up to reach the top, or near the top of the building. We need to place the foot of the ladder, this part, so that it's distance away from the building, is 6ft less than the distance up the building. So if we let this distance be x, we know this distance must be x minus 6. Now, this might look confusing. We see a distance that's longer here, than it is here. Well, my drawings not to scale. Just because you draw a diagram, it doesn't mean it needs to be perfect. We're just doing a sketch of the situation so we can solve our problem. So, now that we have our expressions for this side and this side and, we have a length of the hypotenuse. We're ready to use the Pythagorean theorem. So, we'll square the legs, and then we'll square the hypotenuse. This gives us x squared plus x squared minus 12x plus 36 equals 900. Combining the like terms here we will get to get this equation. Each of these terms is even, so we can divide our equation by a factor of two. Now we want a factor of this quadratic. We find factors of negative 432 that sum to negative 6. Those two factors are negative values of x. Positive 24 and negative 18. We know that this distance needs to be positive, so this answer is out. This must mean that the distance from the bottom of the building to the top of the ladder is 24ft. And the distance from the base of the building to the base of the ladder is 18ft. Great solving if you got all the way to the end. [NOISE]