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A b c. Pythagorean Theorem Essential Questions a 2 + b 2 = c 2 The Pythagorean Theorem a 2 + b 2 = c 2 “For any right triangle, the sum of the areas.

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Presentation on theme: "A b c. Pythagorean Theorem Essential Questions a 2 + b 2 = c 2 The Pythagorean Theorem a 2 + b 2 = c 2 “For any right triangle, the sum of the areas."— Presentation transcript:

1 a b c

2 Pythagorean Theorem Essential Questions

3 a 2 + b 2 = c 2 The Pythagorean Theorem a 2 + b 2 = c 2 “For any right triangle, the sum of the areas of the two small squares is equal to the area of the larger.”

4 Proof

5 a a a2a2 b b c c b2b2 c2c2 Let’s look at it this way…

6 The little square 90 o in the angle tells you it is a right angle.

7 Pythagorus realized that if you have a right triangle, 3 4 5

8 and you square the lengths of the two sides that make up the right angle, 3 4 5

9 and add them together, 3 4 5

10 you get the same number you would get by squaring the other side. 3 4 5

11 Is that correct? ? ?

12 The two sides which come together in a right angle are called

13

14

15 The lengths of the legs are usually called a and b. a b

16 The side across from the right angle a b is called the

17 And the length of the hypotenuse is usually labeled c. a b c

18 The relationship Pythagorus discovered is now called The Pythagorean Theorem: a b c

19 The Pythagorean Theorem says, given the right triangle with legs a and b and hypotenuse c, a b c

20 then a b c

21 Find the length of a rectangle given its diagonal and side: 15" ?" 17

22 Find the length of a leg of the right triangle: 15" ?" 17 b = ? a = 15 c

23 b = ? a = 15 C=17

24 Side length of the rectangle is 8”: 15" 8" 17

25 Practice using The Pythagorean Theorem to solve these right triangles:

26 10 b 26

27 10 b 26 = 24 (a)(a) (c)(c)

28 Using a 2 + b 2 = c 2 Looking for length of a leg a 2 + b 2 = c 2 6 2 + b 2 = 10 2 36 + b 2 = 100 b 2 =100 -36 b 2 = 64 b = 8

29 5 ? 13 TRY THIS

30 5 12 13

31 Check It Out! Example 2 A rectangular field has a length of 100 yards and a width of 33 yards. About how far is it from one corner of the field to the opposite corner of the field? Round your answer to the nearest tenth.

32 Check It Out! Example 2 Continued 1 Understand the Problem Rewrite the question as a statement. Find the distance from one corner of the field to the opposite corner of the field. The segment between the two corners is the hypotenuse. The sides of the fields are legs, and they are 33 yards long and 100 yards long. List the important information: Drawing a segment from one corner of the field to the opposite corner of the field divides the field into two right triangles.

33 Check It Out! Example 2 Continued 2 Make a Plan You can use the Pythagorean Theorem to write an equation.

34 Check It Out! Example 2 Continued Solve 3 a 2 + b 2 = c 2 33 2 + 100 2 = c 2 1089 + 10,000 = c 2 11,089 = c 2 105.304  c The distance from one corner of the field to the opposite corner is about 105.3 yards. Use the Pythagorean Theorem. Substitute for the known variables. Evaluate the powers. Add. Take the square roots of both sides. 105.3  c Round.

35 Baseball Problem A baseball “diamond” is really a square. You can use the Pythagorean theorem to find distances around a baseball diamond.

36 Baseball Problem The distance between consecutive bases is 90 feet. How far does a catcher have to throw the ball from home plate to second base?

37 Baseball Problem To use the Pythagorean theorem to solve for x, find the right angle. Which side is the hypotenuse? Which sides are the legs? a 2 + b 2 = c 2 Now use: a 2 + b 2 = c 2

38 Baseball Problem Solution The hypotenuse is the distance from home to second, or side x in the picture. The legs are from home to first and from first to second. Solution: x 2 = 90 2 + 90 2 = 16,200 x = 127.28 ft

39 Ladder Problem A ladder leans against a second-story window of a house. If the ladder is 25 meters long, and the base of the ladder is 7 meters from the house, how high is the window?

40 Ladder Problem Solution First draw a diagram that shows the sides of the right triangle. Label the sides: –Ladder is 25 m –Distance from house is 7 m Use a 2 + b 2 = c 2 to solve for the missing side. Distance from house: 7 meters

41 Ladder Problem Solution 7 2 + b 2 = 25 2 49 + b 2 = 625 b 2 = 576 b = 24 m How did you do?


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