1 / 24

Lesson 3.3.3

Applications of the Pythagorean Theorem. Lesson 3.3.3. Lesson 3.3.3. Applications of the Pythagorean Theorem. California Standard: Measurement and Geometry 3.3

emery
Download Presentation

Lesson 3.3.3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Applications of the Pythagorean Theorem Lesson 3.3.3

  2. Lesson 3.3.3 Applications of the Pythagorean Theorem California Standard: Measurement and Geometry 3.3 Know and understand the Pythagorean theoremand its converseand use it to find the length of the missing side of a right triangle and the lengths of other line segmentsand, in some situations, empirically verify the Pythagorean theorem by direct measurement. What it means for you: You’ll see how the Pythagorean theorem can be used to find lengths in more complicated shapes and in real-life situations. • Key words: • Pythagorean theorem • right triangle • hypotenuse • legs • right angle

  3. Lesson 3.3.3 Applications of the Pythagorean Theorem In the last two Lessons you’ve seen what the Pythagorean theorem is, and how you can use it to find missing side lengths in right triangles. Now you’ll see how it can be used to help find missing lengths in other shapes too — by breaking them up into right triangles. It can help solve real-life measurement problems too.

  4. c2 = a2 + b2 a2 = c2 –b2 c a b Lesson 3.3.3 Applications of the Pythagorean Theorem Use the Pythagorean Theorem in Other Shapes Too You can use thePythagorean theoremto find lengths in lots of shapes— you just have to splitthem up into right triangles. Here’s a reminder of the formula. which rearranges to: (c is the hypotenuse length, and a and b are the leg lengths.)

  5. Lesson 3.3.3 Applications of the Pythagorean Theorem Example 1 A B Find the area of rectangle ABCD, shown. 13 inches Solution D C 12 inches The formula for the area of a rectangle is: Area = length × width. You know that the lengthof the rectangle is 12 inches, but you don’t know the rectangle’s width, BC. But you do know the length of thediagonal BD and since all the corners of a rectangle are 90° angles, you know that BCDis a right triangle. Solution continues… Solution follows…

  6. BC = = 5 inches Lesson 3.3.3 Applications of the Pythagorean Theorem Example 1 B Find the area of rectangle ABCD, shown. 13 inches Solution (continued) D C 12 inches You can use the Pythagorean theoremto find the length of side BC. BC2 = BD2 – CD2 Write out the equation BC2 = 132 – 122 Substitute the values you know BC2 = 169 – 144 Simplify the equation BC2 = 25 Solution continues…

  7. Lesson 3.3.3 Applications of the Pythagorean Theorem Example 1 A B Find the area of rectangle ABCD, shown. 13 inches 5 inches Solution (continued) D C 12 inches BC is the width of the rectangle. Now you can find its area. Area = length × width Area = 12 inches × 5 inches = 60 inches2

  8. 1 2 The formula for the area of a triangleis: Area = base × height. Q Lesson 3.3.3 Applications of the Pythagorean Theorem Example 2 R Find the area of isosceles triangle QRS. 15 cm 15 cm 15 cm Solution Q S M 9 cm 18 cm This is half the base of the original triangle. The baseof the triangle is 18 cm, but you don’t know its height, MR. Isosceles triangles can be split up into two right triangles. Solution continues… Solution follows…

  9. MR = = 12 cm Lesson 3.3.3 Applications of the Pythagorean Theorem Example 2 R Find the area of isosceles triangle QRS. 15 cm Solution (continued) 12 cm You can use the Pythagorean theoremto find the length of side MR. S M 9 cm MR2 = RS2 – MS2 Write out the equation MR2 = 152 – 92 Substitute the values you know MR2 = 225 – 81 Simplify the equation MR2 = 144 Solution continues…

  10. 1 1 2 2 Area = base × height Area = (18 cm) × 12 cm = 9 cm × 12 cm = 108 cm2 Lesson 3.3.3 Applications of the Pythagorean Theorem Example 2 R Find the area of isosceles triangle QRS. 15 cm 15 cm Solution (continued) 12 cm Now put the value of MR into the area formula: Q S M 18 cm

  11. Lesson 3.3.3 Applications of the Pythagorean Theorem Guided Practice In Exercises 1–2 use the Pythagorean theorem to find the missing value, x. 1.2. E F U 10 cm 6 cm 34 ft 34 ft H G T V Area = x cm2 P 32 ft GH2 = 102 – 62 = 64GH = 8 cmx = 8 • 6 = 48 Area = x ft2 UP2 = 342 – (32 ÷ 2)2 = 900UP = 30 ft x = 0.5 • 32 • 30 = 480 Solution follows…

  12. Lesson 3.3.3 Applications of the Pythagorean Theorem Guided Practice In Exercises 3–4 use the Pythagorean theorem to find the missing value, x. 3.4. 25 in W X W X x m 17 in 12 m x in. Z Y Z Y 33 in Area = 108 m2 x2 = 172 – (33 – 25)2 = 225x = 15 ZY = 108 ÷ 12 = 9 mx2 = 122 + 92 = 225 x = 15 Solution follows…

  13. Lesson 3.3.3 Applications of the Pythagorean Theorem The Pythagorean Theorem Has Real-Life Applications Because you can use the Pythagorean theorem to find lengths in many different shapes, it can be useful in lots of real-life situations too.

  14. Path Yard 24 feet 32 feet Lesson 3.3.3 Applications of the Pythagorean Theorem Example 3 Monique’s yard is a rectangle 24 feet long by 32 feet wide. She is laying a diagonal gravel path from one corner to the other. One sack of gravel will cover a 10-foot stretch of path. How many sacks will she need? Solution The first thing you need to work out is the length of the path. It’s a good idea to draw a diagram to help sort out the information. You can see from the diagram that the path is the hypotenuse of a right triangle. So you can use the Pythagorean theorem to work out its length. Solution continues… Solution follows…

  15. c b = 24 ft a = 32 ft c = = 40 feet Lesson 3.3.3 Applications of the Pythagorean Theorem Example 3 Monique’s yard is a rectangle 24 feet long by 32 feet wide. She is laying a diagonal gravel path from one corner to the other. One sack of gravel will cover a 10-foot stretch of path. How many sacks will she need? Solution (continued) c2 = a2 + b2 Write out the equation c2 = 322 + 242 Substitute the values you know c2 = 1024 + 576 Simplify the equation c2 = 1600 Solution continues…

  16. Path 40 feet Yard 24 feet 32 feet Lesson 3.3.3 Applications of the Pythagorean Theorem Example 3 Monique’s yard is a rectangle 24 feet long by 32 feet wide. She is laying a diagonal gravel path from one corner to the other. One sack of gravel will cover a 10-foot stretch of path. How many sacks will she need? Solution (continued) The question tells you that one sack of gravel will cover a 10-foot length of path. To work out how many are needed, divide the path length by 10. Sacks needed = 40 ÷ 10 = 4 sacks

  17. Lesson 3.3.3 Applications of the Pythagorean Theorem Guided Practice 5. Rob is washing his upstairs windows. He puts a straight ladder up against the wall. The top of the ladder is 8 m up the wall. The bottom of the ladder is 6 m out from the wall. How long is the ladder? 6. To get to Gabriela’s house, Sam walks 0.5 miles south and 1.2 miles east around the edge of a park. How much shorter would his walk be if he walked in a straight line across the park? Let l = length of ladder:l2 = 82 + 62 = 100l = 10 m Let s = distance of straight-line walkand w = distance Sam walked: s2 = 0.52 + 1.22 = 1.69s = 1.3 miles and w = 1.2 + 0.5 = 1.7 milesDifference = 1.7 – 1.3 = 0.4 miles Solution follows…

  18. 10 cm 8 cm 17cm Let l = total length neededThen l = (2 × 8) + + = 16 + 6 + 15 = 37 cm Lesson 3.3.3 Applications of the Pythagorean Theorem Guided Practice 7. The diagonal of Akil’s square tablecloth is 4 feet long. What is the area of the tablecloth? 8. Megan is making the kite shown in thediagram on the right. The crosspieces are made of thin cane. What length of cane will she need in total? Let s = side length of the tableclothThen 42 = s2 + s216 = 2s2So the area of the table cloth is: s2 =8 feet2 Solution follows…

  19. Lesson 3.3.3 Applications of the Pythagorean Theorem Independent Practice In Exercises 1–2 use the Pythagorean theorem to find the missing value, x. 1. 2. B P Q x inches 15 inches 13 cm 13 cm S R A C Area = 300 inches2 H 10 cm x = 25 Area = x cm2 x = 60 Solution follows…

  20. Lesson 3.3.3 Applications of the Pythagorean Theorem Independent Practice In Exercises 3–4 use the Pythagorean theorem to find the missing value, x. 3. 4. 32 m 20 feet E F K L 34 m x m 17 feet x feet H G N M 48 m 36 feet x = 30 x = 15 Solution follows…

  21. Lesson 3.3.3 Applications of the Pythagorean Theorem Independent Practice 5. A local radio station is getting a new radio mast that is 360 m tall. It has guy wires attached to the top to hold it steady. Each wire is 450 m long. Given that the mast is to be put on flat ground, how far out from the base of the mast will the wires need to be anchored? 6. Luis is going to paint the end wall of his attic room, which is an isosceles triangle. The attic is 7 m tall, and the length of each sloping part of the roof is 15 m. One can of paint covers a wall area of 20 m2. How many cans should he buy? 270 m 5 cans Solution follows…

  22. 5 feet 5 feet 11 feet 15 feet 20 feet Lesson 3.3.3 Applications of the Pythagorean Theorem Independent Practice 7. Maria is carpeting her living room, shown in the diagram on the left. It is rectangular, but has a bay window. She has taken the measurements shown on the diagram. What area of carpet will she need? 332 feet2 Solution follows…

  23. Lesson 3.3.3 Applications of the Pythagorean Theorem Independent Practice 8. The diagram below shows a baseball diamond. The catcher throws a ball from home plate to second base. What distance does the ball travel? 127 feet 2nd base 90 feet Pitcher’s plate 3rd base 1st base 90 feet Home plate Solution follows…

  24. Lesson 3.3.3 Applications of the Pythagorean Theorem Round Up You can break up a lot of shapes into right triangles. This means you can use the Pythagorean theoremto find the missing lengths of sides in many different shapes — it just takes practice to be able to spot the right triangles.

More Related