Right Triangles and Quadrilaterals

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1 CHATER. RIGHT TRIANGLE AND UADRILATERAL Choose always the way that seems the best, however rough it may be; custom will soon render it easy and agreeable. ythagoras CHATER Right Triangles and uadrilaterals.1 The ythagorean Theorem In a right triangle, the side of the triangle opposite the right angle is called the hypotenuse and the other two sides are called the legs of the triangle. We also often use the terms legs and hypotenuse to refer the lengths of the legs and hypotenuse of a right triangle. leg In this section, we explore one of the most famous math theorems, the ythagorean Theorem, which is a powerful relationship among the sides of a right triangle. We ll start by walking through one of the many proofs of the ythagorean Theorem. ( ythagorean is pronounced puh-thag-uh-ree-uhn. ) leg hypotenuse roblems roblem.1: Four identical right triangles with legs of lengths and are attached to the sides of square W as shown, such that W = = R = = and = = R = W =. (a) Explain why \W = 180, and why R is a square. (b) What is the area of R? (c) Find the area of W. W (d) Find W. R

2 .1. THE THAGOREAN THEOREM roblem.: In this problem, we follow in the steps of the previous problem to prove the ythagorean Theorem. We start again with four copies of a right triangle, attached to the sides of square W as shown at the right. Let the lengths of the legs of each triangle be a and b, as shown, and let the hypotenuse of each right triangle have length c. a W b c (a) Find the area of W in terms of c. (b) Find the area of R in terms of a and b. R (c) Find the area of W in terms of a and b. (d) how that a + b = c. roblem.: Find the missing side lengths in each of the three triangles shown below. B C 5 A T 5 8 U p 15 roblem.: Must the hypotenuse of a right triangle be the longest side of the triangle? Why or why not? roblem.5: In roblems.1 and., we have seen two right triangles in which all three side lengths are integers. Can you find any more right triangles in which all three side lengths are integers? Hints: roblem.6: (a) Find the hypotenuse of a right triangle whose legs are and. (b) Find the hypotenuse of a right triangle whose legs are 5 and 5. (c) Find the hypotenuse of a right triangle whose legs are 011 and 011. (d) Find the hypotenuse of a right triangle whose legs are and roblem.: The length of one leg of a right triangle is 10 and the triangle s hypotenuse has length 50. What is the length of the other leg?

3 CHATER. RIGHT TRIANGLE AND UADRILATERAL roblem.1: Four identical right triangles with legs of lengths and are attached to the sides of square W as shown, such that W = = R = = and = = R = W =. (a) Explain why \W = 180, and why R is a square. (b) What is the area of R? (c) Find the area of W. W (d) Find W. R olution for roblem.1: (a) Back on page 0, we learned that the acute angles of a right triangle add to 90. Therefore, in right triangle W we have \W + \W = 90. ince triangles W and W are identical, we have \W = \W. ubstituting this into the equation above gives \W + \W = 90. We are told that W is a square, so \W = 90, and we have \W = \W + \W + \W = \W \W = 90 + (\W + \W) = = 180. Therefore, \W is a straight angle. This means that W is on. imilarly, each vertex of W is on one of the sides of quadrilateral R. Each side of R has length + =, and each angle of R is the right angle of one of the triangles. o, all the sides of R are congruent, and all the angles of R are congruent, which means R is a square. (b) ince R is a square with side length, its area is = 9. (c) Each right triangle has area ()()/ = 6 square units. Removing the four right triangles from R leaves W, so we have [W] = [R] (6) = 9 = 5. (d) The area of W is the square of its side length. Because the area of W is 5, its side length must be p 5, which equals 5.

4 .1. THE THAGOREAN THEOREM roblem.: In this problem, we follow in the steps of the previous problem to prove the ythagorean Theorem. We start again with four copies of a right triangle, attached to the sides of square W as shown at the right. Let the lengths of the legs of each triangle be a and b, as shown, and let the hypotenuse of each right triangle have length c. a W b c (a) Find the area of W in terms of c. (b) Find the area of R in terms of a and b. R (c) Find the area of W in terms of a and b. (d) how that a + b = c. olution for roblem.: (a) ince W is a square with side length c, its area is c. (b) As in the previous problem, R is a square, and the vertices of W are on the sides of R. Each side of R has length a + b, so the area of R is (a + b). We can expand (a + b) with the distributive property: [R] = (a + b) = (a + b)(a + b) = a(a + b) + b(a + b) = a + ab + ba + b = a + ab + ab + b = a + ab + b. (c) The area of each of the right triangles is ab/, so the four right triangles together have area (ab/) = ab. We can find the area of W in terms of a and b by subtracting the areas of the triangles from the area of R: [W] = [R] (ab/) = a + ab + b ab = a + b. (d) In part (a), we found that [W] = c, and in part (c), we found that [W] = a + b. Equating our expressions for [W], we have a + b = c. 5

5 CHATER. RIGHT TRIANGLE AND UADRILATERAL Important: The ythagorean Theorem tells us that in any right triangle, the sum of the squares of the legs equals the square of the hypotenuse. o, in the diagram to the right, we have a + b = c. A b C a c B Our work in roblem. is the same as the work we did in roblem.1, except that we replaced the numbers in roblem.1 with variables a, b, and c in roblem.. Concept: pecific examples can sometimes be used as guides to discover proofs. The ythagorean Theorem also works in reverse. By this, we mean that if the side lengths of a triangle satisfy the ythagorean Theorem, then the triangle must be a right triangle. o, for example, if we have a triangle with side lengths,, and 5, then we know that the triangle must be a right triangle because + = 5. Let s get a little practice using the ythagorean Theorem. roblem.: Find the missing side lengths in each of the three triangles shown below. B C 5 A T 5 8 U p 15 olution for roblem.: What s wrong with this solution: Bogus olution: Applying the ythagorean Theorem to ABC gives + 5 = BC. Therefore, we find BC = = 6. Taking the square root gives us BC = p 6. 6

6 .1. THE THAGOREAN THEOREM This solution is incorrect because it applies the ythagorean Theorem incorrectly. ide BC is a leg, not the hypotenuse. Applying the ythagorean Theorem to ABC correctly gives AC + BC = AB. B 5 ubstituting AC = and AB = 5 gives us + BC = 5, so 9 + BC = 65. ubtracting 9 from both sides gives BC = 56. Taking the square root of 56 gives BC =. (Note that ( ) = 56 too, but we can t have a negative length, so BC cannot be.) C A WARNING!! j Be careful when applying the ythagorean Theorem. Make sure you correctly identify which sides are the legs and which is the hypotenuse. Applying the ythagorean Theorem to TU gives T + TU = U, so we have = U from the side lengths given in the problem. Therefore, we have U = = 89. Taking the square root gives us U = p 89. In, the ythagorean Theorem gives us T 5 8 U + =, so + ( p 15) =. This gives us =, so = 6 and = 8. p 15 WARNING!! j A common mistake when using the ythagorean Theorem to find the hypotenuse length of a right triangle is forgetting that the hypotenuse is squared in the equation, too. One quick way to avoid this error is to consider the three side lengths after finding the hypotenuse. For example, suppose a right triangle has legs of lengths and. The hypotenuse clearly can t be + = 5, because the lengths,, and 5 don t satisfy the Triangle Inequality. Taking the square root of 5, we see that the hypotenuse should be 5, not 5. roblem.: Must the hypotenuse of a right triangle be the longest side of the triangle? Why or why not?

7 CHATER. RIGHT TRIANGLE AND UADRILATERAL olution for roblem.: es. The square of the hypotenuse equals the sum of the squares of the legs. The sum of any two positive numbers is greater than both of the numbers being added. o, the square of the hypotenuse must be greater than the square of each leg. Therefore, the hypotenuse must be longer than each leg. roblem.5: In roblems.1 and., we have seen two right triangles in which all three side lengths are integers. Can you find any more right triangles in which all three side lengths are integers? olution for roblem.5: There are lots and lots of right triangles in which all three side lengths are integers! To search for some, we can list the first 0 positive perfect squares: 1,, 9, 16, 5, 6, 9, 6, 81, 100, 1, 1, 169, 196, 5, 56, 89,, 61, 00. Then, we look for pairs of squares that add up to another square. We immediately see 9+16 = 5, which is + = 5. We already saw this example in roblem.1. We also see = 169, which is 5 + = 1. o, a right triangle with legs of lengths 5 and has a hypotenuse with length 1. We also find = 89, which is = 1. This gives us a right triangle with 8 and 15 as the legs and 1 as the hypotenuse. A ythagorean triple is a group of three positive integers that satisfy the equation a +b = c. o, for example, {,, 5} is a ythagorean triple, as are {5,, 1} and {8, 15, 1}. There are lots of interesting patterns in ythagorean triples. ee if you can find more ythagorean triples, and look for patterns that you can use to find more ythagorean triples. We can find one such important pattern by looking back at our list of squares: 1,, 9, 16, 5, 6, 9, 6, 81, 100, 1, 1, 169, 196, 5, 56, 89,, 61, 00. We find that = 100, which is = 10. Here, the side lengths are double those of the first triangle we saw with sides of lengths,, and 5. We might wonder if tripling these three side lengths also gives us another right triangle. Indeed, we see that 9 + = 15, since = 5. Let s investigate further. roblem.6: (a) Find the hypotenuse of a right triangle whose legs are and. (b) Find the hypotenuse of a right triangle whose legs are 5 and 5. (c) Find the hypotenuse of a right triangle whose legs are 011 and 011. (d) Find the hypotenuse of a right triangle whose legs are and olution for roblem.6: (a) The legs have lengths and 16. Letting the hypotenuse be c, the ythagorean Theorem gives us c = + 16 = = 00. 8

8 .1. THE THAGOREAN THEOREM Taking the square root gives us c = 0. Notice that 0 = 5. (b) The legs have lengths 15 and 0. Letting the hypotenuse be c, the ythagorean Theorem gives us c = = = 65. Taking the square root gives us c = 5. Notice that 5 = 5 5. (c) The legs have lengths 60 and 80. Um, squaring those doesn t look like much fun. Let s see if we can find a more clever way to solve this problem. We know that a right triangle with legs and has hypotenuse 5. In part (a), we saw that if the legs of a right triangle are and, then the hypotenuse is 5. In part (b), we saw that if the legs of a right triangle are 5 and 5, then the hypotenuse is 5 5. It looks like there s a pattern! Concept: earching for patterns is a powerful problem-solving strategy. It appears that if the legs of a right triangle are x and x, then the hypotenuse is 5x. We can test this guess with the ythagorean Theorem. uppose the legs of a right triangle are x and x. Then, the sum of the squares of the legs is ince (x) + (x) = x + x = 9x + 16x = 5x. (5x) = 5 x = 5x, we have (x) + (x) = (5x), which means that the length of the hypotenuse is indeed 5x. This means that we don t have to square 60 and 80! A right triangle with legs of lengths 011 and 011 has a hypotenuse with length = (d) There s nothing in our explanation in part (c) that requires x to be a whole number; it 1 can be a fraction, too! o, in a right triangle with legs of length and 1, the hypotenuse has length 5 = Our work in roblem.6 is an example of why knowing common ythagorean triples is useful. Any time we have a right triangle in which the legs have ratio :, then we know that all three sides of the triangle are in the ratio : : 5. As we saw in the final two parts of roblem.6, this can allow us to find the hypotenuse quickly without using the ythagorean Theorem directly. We can also sometimes use this approach to quickly find the length of a leg when we know the lengths of the other leg and the hypotenuse. roblem.: The length of one leg of a right triangle is 10 and the triangle s hypotenuse has length 50. What is the length of the other leg? 9

9 CHATER. RIGHT TRIANGLE AND UADRILATERAL olution for roblem.: We find the ratio of the given leg length to the hypotenuse length, hoping it will match the corresponding ratio in one of the ythagorean triples we know. We have 10 : 50 = : = : 5, so the ratio of the given leg to the hypotenuse is : This reminds us of the {,, 5} ythagorean triple that we saw in roblem.. ince the ratio of the known leg to the hypotenuse is : 5, we know that all three sides are in the ratio : : 5. The first leg is 0 and the hypotenuse is 5 0, so the other leg of the right triangle is 0 = 0. Important: If we multiply all three side lengths of a right triangle by the same positive number, then the three new side lengths also satisfy the ythagorean Theorem. In other words, if side lengths a, b, and c satisfy a + b = c, then (na) + (nb) = (nc), for any positive number n. Exercises.1.1 Find the missing side lengths below: 15 U R T V.1. Bill walks 1 mile south, then mile east, and finally 1 mile south. How many miles is he, in a direct line, from his starting point? (ource: AMC 8).1. Find a formula for the length of a diagonal of a rectangle with length l and width w..1. The bases of a 9-foot pole and a 15-foot pole are 5 feet apart, and both poles are perpendicular to the ground. The ground is flat between the two poles. What is the length of the shortest rope that can be used to connect the tops of the two poles?.1.5 A square, a rectangle, a right triangle, and a semicircle are combined to form the figure at the right. Find the area of the whole figure in square units..1.6? Find the hypotenuse of a right triangle whose legs have lengths and

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