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THE

PYTHAGOREAN. THE. THEOREM. 2500 YEARS AGO, A GREEK MATHEMATICAN NAMED PYTHAGORIS , SAID THAT IN A RIGHT TRIANGLE , THE SUMS OF THE SQUARES OF THE LENGTHS OF THE LEGS (A AND B) IS EQUAL TO THE SQUARE OF THE HYPOTENUSE. IDENTIFYING THE TRIANGLE. HYPOTENUSE. IS ACROSS FROM THE RIGHT ANGLE.

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  1. PYTHAGOREAN THE THEOREM

  2. 2500 YEARS AGO, A GREEK MATHEMATICAN NAMED PYTHAGORIS, SAID THAT IN A RIGHT TRIANGLE, THE SUMS OF THE SQUARES OF THE LENGTHS OF THE LEGS (A AND B) IS EQUAL TO THE SQUARE OF THE HYPOTENUSE.

  3. IDENTIFYING THE TRIANGLE HYPOTENUSE IS ACROSS FROM THE RIGHT ANGLE LEG A LEG B

  4. This is the formula a² + b² = c² ADD THE SQUARES OF A AND B TO GET THE SQUARE OF C ! c² a² b²

  5. This is the formula a² + b² = c² c² ADD THE SQUARES OF A AND B TO GET THE SQUARE OF C ! a² b²

  6. IT’S PROBLEM TIME! a² + b² = c² WRITE FORMULA 5² + 12² = c² PLUG IN NUMBERS 25 + 144 = C² SOLVE 169= C² 169 = C c 5m 13 = C REMEMBER THE WARMUP SLIDE? 12m

  7. NOW YOU TRY ONE, AT YOUR DESK! A IS 12, B IS 16, AND THEY ARE LOOKING FOR THE HYPOTENUSE! c 12 16

  8. DID YOU GET IT RIGHT? a² + b² = c² 12² + 16² = c² 144 + 256 = c² 400 = c² C 400 = c 12 20 = c 16 BE SURE TO WRITE THE FORMULA EACH TIME, PLUG IN THE NUMBERS AND SOLVE

  9. WRITE THE FORMULA, PLUG IN THE NUMBERS, SQUARE, THEN ADD, THEN FIND THE SQUARE ROOT OF THE NUMBER. a² + b² = c² 3² + 4² = c² C 3 9 + 16 = c² 25 = c² 4

  10. A 35 FOOT TELEPHONE POLE NEEDS A SUPPORT WIRE ATTACHED AT THE TOP OF THE POLE TO A POINT IN THE GROUND 20 FEET AWAY FROM THE BASE. ABOUT HOW LONG SHOULD THE WIRE BE? a² + b² = c² 35² + 20² = c² 1225 + 400 = c² 35 C 1625 = c² 40.3 or 40ft = c 20

  11. Pythagorean Theorem – Day 2 x x

  12. Try these on your own! x x

  13. WORD PROBLEMS A 20ft LADDER IS LEANED AGAINST A WALL. IF THE BASE IS 5ft FROM THE WALL, HOW HIGH CAN YOU CLIMB? a² + b² = c² b = 375 5² + b² = 20² 20 b = 19.3 25 + b² = 400 ? 19,20 b² = 400 - 25 b² = 375 5 AS TOM SWAM ACROSS A 60m RIVER, THE CURRENT MOVED HIM 20m DOWNSTREAM. HOW FAR DID HE SWIM? HELP! HELP! a² + b² = c² 60² + 20² = c² 3600 + 400² = c² 4000 = c² 60 20 BRODERICK SWIMMING ACROSS THE RIVER 4000 = c² 63.2 = c² 64,65

  14. HOW FAR IS THE HELICOPTER FROM IT’S STARTING POINT? a² + b² = c² 9mi 6² + 9² = c² 36 + 81 = c² 6mi 117 = c² 117 = c² 10.82 = c² A 35ft LADDER IS LEANED AGAINST A BLDG. HOW HIGH IS THE WINDOW LEDGE ABOVE THE GROUND? a² + b² = c² a² = 1000 b = 31.6 35ft a² + 15² = 35² a² = 1000 ? a + 225 = 1225 a = 31.62 a² = 1225 - 225 15ft 31² = 961 and 32² = 1024

  15. Missing Leg Wkst – EVENS ONLY!Use the scrambled answers to help you check your answers!

  16. Extra Practice with Pythagorean Theorem • A small rectangular cabinet door has a width of 12 inches and a diagonal of 28 inches. Find the length of the door. • What is the AREA and PERIMETER of the door?

  17. A BIG SCREEN T.V. IS 6ft WIDE AND 11ft LONG. HOW LONG IS THE DIAGONAL ACROSS THE SCREEN? 11 ? a² + b² = c² 157 = c² 6 6² + 11² = c² 157 = c 36 + 121 = c² 12.53 = c KERSTIN LEFT HER CAMPSITE AND HIKED 5mi EAST AND THEN 2mi NORTH. HOW FAR WAS SHE FROM THE CAMPSITE? a² + b² = c² 29 = c² ? 5² + 2² = c² 29 = c 2 5 25 + 4 = c² 5.39 = c EACH SIDE OF A EQUILATERAL TRIANGLE IS 10m. FIND THE HEIGHT OF THE TRIANGLE.

  18. PRACTICE EACH SIDE OF A EQUILATERAL TRIANGLE IS 10m. FIND THE HEIGHT OF THE TRIANGLE. a² + b² = c² a² + 5² = 10² 10 10 a + 25 = 100 a = 8.66 a² = 100 - 25 a = 75 5 10 5 A ROPE IS TIED TO THE TOP OF A 9ft TENT AND TIED TO A STAKE IN THE GROUND 10ft FROM THE BASE. HOW LONG IS THE ROPE? ? 9 10 a² + b² = c² c = 181 9² + 10² = c² c = 13.45 81 + 100 = c² 181 = c²

  19. FIND THE SQUARE OF EACH: FIND THE SQUARE ROOT OF EACH: 1. 5² = 2. 2² = 3. 11² = 4. 8² = 5. 12² = 6. 15² = 7. 20² = 5x5=25 1. √36 = 2. √168 = 3. √625 = 4. √288 = 5. √105 = 6. √49 = 7. √9 = 6 2x2=4 11x11=121 25 8x8=64 12x12=144 15x15=225 7 20x20=400 3 QUIZ ON FRIDAY OVER PYTH. THEOREM!

  20. WARM-UP 5/12/11 TRY THESE ON YOUR OWN, FIND THE HYPOTENUSE FOR EACH: 8 13 7 10 a² + b² = c² a² + b² = c² 7² + 10² = c² 8² + 13² = c² 49 + 100 = c² 64 + 169 = c² 149 = c² 233 = c² 149 = c 233 = c 12.2 = c 12, 13 15.2 = c 15, 16 QUIZ TOMORROW!!!!!!

  21. WARM-UP 5/17/11 A 35 FOOT TELEPHONE POLE NEEDS A SUPPORT WIRE ATTACHED AT THE TOP OF THE POLE TO A POINT IN THE GROUND 20 FEET AWAY FROM THE BASE. ABOUT HOW LONG SHOULD THE WIRE BE? a² + b² = c² 35² + 20² = c² 1225 + 400 = c² 35 C 1625 = c² 40.31 = c 20

  22. A 35 FOOT TELEPHONE POLE NEEDS A SUPPORT WIRE ATTACHED AT THE TOP OF THE POLE, TO A POINT IN THE GROUND 20 FEET AWAY FROM THE BASE. ABOUT HOW LONG SHOULD THE WIRE BE? a² + b² = c² 35² + 20² = c² 1225 + 400 = c² 35 1625 = c² C 40.31 = c 20

  23. A 20ft FLAGPOLE IS SECURED BY A 25ft WIRE TO A STAKE IN THE GROUND? HOW FAR FROM THE BASE IS THE STAKE? a² + b² = c² 20 c a² + 20² = 25² 25 a² + 400 = 625 b a² = 625 - 400 a a = √225 a = 15 12 IF A STAIRCASE IS TO FIT INTO A SPACE THAT IS 12ft HIGH AND 10ft LONG, HOW LONG IS THE RISER ? 10 a² + b² = c² 10² + 12² = c² 100 + 144 = c² 244 = c² √244 = c² 15.62 = c

  24. EXTRA PRACTICE A MAST ON A SAILBOAT IS 10m TALL. A ROPE IS STRETCHED 15m FROM THE TOP OF THE MAST TO A CLEAT ON THE DECK. HOW FAR IS THE CLEAT FROM THE BASE OF THE MAST? a² + b² = c² b = 125 10² + b² = 15² 15 b = 11.2 10 100 + b² = 225² b² = 225 - 100 b² = 125 ? TWO JETS LEAVE AN AIRPORT AT THE SAME TIME. ONE WEST AT 400mph AND THE OTHER NORTHWEST AT 500mph. HOW FAR APART WERE THE JETS AT THE END OF THE HOUR? a² + b² = c² 400² + b² = 500² 160,000 + b² = 250,000² 500 ? b² = 250000 - 160000 b = 300 b² = 90,000 400 b² = 90,000

  25. TELL IF IT IS A RIGHT TRIANGLE OR NOT!?!? a² + b² = c² ? 7 4² + 6² = 7² 4 ? 16 + 36 = 49 6 52 = 49

  26. PRACTICE TELL IF IT IS A RIGHT TRIANGLE OR NOT! 20 1. a² + b² = c² 2. a² + b² = c² YES 30 8² + 15² = 17² 10² + 20² = 30² 17 8 64 + 225 = 289 100 + 400 = 900 289 = 289 500 = 900 15 10 NO a² + b² = c² 3. 8 9 TEST FRIDAY ON PYTHAGOREAN THEOREM AND WARMUPS 8² + 9² = 15² 64 + 81 = 225 145 = 225 15 a² + b² = c² 4. 15 YEP 15² + 20² = 25² 25 225 + 400 = 625 625 = 625 20

  27. PROBLEM PROCESS ( SHOW WORK) SOLUTION Pythagorean Theorem A 35 FOOT TELEPHONE POLE NEEDS A SUPPORT WIRE ATTACHED AT THE TOP OF THE POLE TO A POINT IN THE GROUND 20 FEET AWAY FROM THE BASE. ABOUT HOW LONG SHOULD THE WIRE BE?

  28. WARMUP 5/19/11 A HIGHWAY BRIDGE IS ANCHORED TO THE GROUND BY A RIGHT TRIANGLE. HOW LONG IS THE EMBANKMENT LABELED x 14 a² + b² = c² √596 = c² 14² + 20² = c² 24.41 = c 196 + 400 = c² 24,25 X 20 596 = c² TEST TOMORROW!!!! FIND THE MISSING SIDE OF EACH RIGHT TRIANGLE a = 16 b = 20 EQUILATERAL SIDES OF 16 a² + b² = c² a² + b² = c² 16 16² + 20² = c² a² + 8² = 16² 256 + 400 = c² a² + 64 = 256 8 8 656 = c² a²= 256 - 64 √656 = c FIND THE HEIGHT a²= 192 25,26 25.61 = c a² = √192 a = 13.86 = height 13,14

  29. AGENDA 5/20/11 • COMPLETE UNIT TEST **Show all of your work!!! **Check your answers!!! When you are finished with your test and if you are all caught up with your class work, you may go to the computer lab to work in Study Island.

  30. WARMUPS GIVE THE SQUARE ABOVE AND BELOW EACH OF THE GIVEN SQUARE ROOTS. EXAMPLE IS GIVEN USE THE LIST YOU COPIED YESTERDAY

  31. SOLVE EACH

  32. NEW PYTHAG, SQUARES, AREA, ETC

  33. WARMUP If a ladder 39ft tall is leaned against a building and the base of the ladder is 15ft from the building, how tall is the building? a² + b² = c² 39 a² + 15² = 39² a c a a² + 225 = 1521 a² = 1521-225 a² = 1296 b a = 36 15 Cost %Disc $Disc S.P. %Tax $Tax F.C. Item $300 20% $60 $240 10% $24 $264 shoe You should take notes on these for the review for the test on Friday.

  34. WARMUP Fraction Decimal Percent 5 500% .2425 24¼% 24.25% Some of these will be on the Warmup quiz next week. I will give out all answers to review on Thursday.

  35. WARMUP SOLVE EACH: 30² = 900 12² = 144 1² = 1 5² = 25 18² = 324 13² = 169 17² = 289 7² = 49 8² = 64 14² = 196 19² = 361 10² = 100 COMBINE LIKE TERMS: 12 + 5 – 10 + 2x 3x + 10 – x + 5 - 3 2x + 7 2x + 12 14 +10c + 4c + 12c + 8c - 14 -22a + 79 34c 5 – 11a + 35 – 11a + 39 We are starting area and perimeter today Test Friday

  36. WARMUPS GIVE THE SQUARE ABOVE AND BELOW EACH OF THE GIVEN SQUARE ROOTS. EXAMPLE IS GIVEN USE THE LIST YOU COPIED

  37. WEDNESDAY A and P 9 The area of a rectangle is 45in². One side of it is 5in. What is the perimeter? A = l • w 45 = 5 • 9 5 45in² 5 = 9 + 5 + 9 + 5 = 28in. 9 20 The perimeter of rectangle is 60in. One side is 10in. What is the area? A = l • w 10 200in² 10 A = 10 • 20 20 60-20=40 A = 200in² The length of a rectangle is 9cm. The width is 7cm. What is the area? Perimeter? 9 A = l • w Perimeter = 9 + 7 + 9 + 7 = 32in. 7 7 A = 9 • 7 A = 63in² 9 QUIZ TOMORROW OVER THIS AND THE WKSHTS!!!!!

  38. WEDNESDAY AREA AND PERIMETER CIRCUMFERENCE FIND THE AREA AND PERIMETER OF EACH: CIRCUMFERENCE DIAMETER DISTANCE ACROSS THE MIDDLE 8m 8m RADIUS HALF OF THE DIAMETER  = 3.14 A =  • r² C =  • d A = 3.14 • 8² C = 3.14 • 16 = R A = 3.14 • 64 C = 50.24 A = 200.96 IT REALLY HELPS TO WRITE THE FORMULA AND WRITE NEATLY AND PLUG IN THE #’S = D

  39. WEDNESDAY FIND THE “A or C” A =  r² C =  d 5 A = 3.14 • 5² C = 3.14 • 8 8 A = 3.14 • 25 A = 25.12cm 3•8 = 24 A = 78.5cm 3•25 = 75 Estimate your answer by mult. “3 • the whole #”!! A =  r² C =  d A = 3.14 • 6² 12 9 C = 3.14 • 18 A = 3.14 • 36 A = 56.25cm A = 113.04cm 3•18 = 54 3•36 = 108 R=9 D=18 D=12 R=6 IT IS ALWAYS GOOD TO WRITE OUT FORMULA AND THE “D and R” AND THEN ESTIMATE YOUR ANSWER. QUIZ TOMORROW OVER THIS AND THE WKSHTS!!!!!

  40. THURSDAY WARMUP Put the following in order from least to greatest. Which problem situation can be solved with t = 60w? a. What is t, the # of ounces in w gallons? b. What is t, the volume of a cylinder with a radius of 60 and height of w? c. What is t, the # of hours in w minutes? d. What is t, the # of seconds in w minutes? QUIZ TODAY ON A and P

  41. THURSDAY WARMUP 15º What is the compliment of 15º? 75º 145º 35º What is the suppliment of 35º? Timmy wants to buy a bike for $47.95 with a 20% discount. About how much will he pay? Timmy wants to buy a bike for $47.95 with a 20% discount. About how much will he pay? DISC = $50 • .20 = $10 $50 – 10 = $40 Which company has the best price? .45 22.50 ÷ 50 .40 56 ÷ 140 .35 4.20 ÷ 12 .50 QUIZ TODAY ON A and P

  42. THURSDAY WARMUP Solve for Area Solve for Circumference A =  r² C =  d A = 3.14 • 7.5² 15 7 C = 3.14 • 14 A = 3.14 • 56.25 A = 43.96cm A = 176.625cm² D=15 R=7.5 3•56 = 168 R=7 D=14 3•14 = 42 Solve for Area 10 Solve for Perimeter 21 A = l • w A = l • w 8 P = 36 8 A = 7 • 21 7 A = 147m² 7 A = 8 • 10 P = 56 A = 80in² 10 21 36 – 16 = 20 We will check review and take quick questions, then check hmwk, then take quiz over this material. QUIZ TODAY ON A and P

  43. BE SEATED QUICKLY, HAVE REVIEW READY TO TURN IN, AND GET STARTED ON TEST. YOU NEED TO FINISH BEFORE THE BELL RINGS, IF YOU STUDIED YOU SHOULD DO FINE. WRITE NEATLY ON ANSWER KEY. STARING AT THE TEST WILL NOT GET IT DONE! NO TALKING, NO LOOKING AROUND, SIT UP STRAIGHT, KEEP TEST IN FRONT OF YOU, KEEP YOUR EYES ON YOUR PAPER, DURING TEST Scratch off the 2nd #57,

  44. TUESDAY WARMUP Which model represents Which rule can be used to determine the following? Simplify the following: 6² ÷ 3 • 6 – 2 + 20 m c 36 ÷ 3 • 6 – 2 + 20 12 • 6 – 2 + 20 72 – 2 + 20 70 + 20 90 We will finish the test the last 30 minutes of class. 2c + 1 4c - 3 4m - 3 2m + 1

  45. TUESDAY WARMUP Solve for Area Solve for Circumference A =  r² C =  d A = 3.14 • 5² 10 5 C = 3.14 • 10 A = 3.14 • 25 A = 31.40cm A = 78.50cm² D=10 R=5 3•25 = 75 R=5 D=10 3•10 = 30 FRACTION DECIMAL PERCENT .60 60% .245% 24½% 24.5% NOTES ON IRREGULAR SHAPES, THEN PRACTICE PROBLEMS, THEN SOME WILL FINISH THE TEST FROM FRIDAY AND THE REST WILL WORK ON WKSHT….WARMUP QUIZ TOMORROW!!!!!!

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