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PYTHAGOREAN WORD PROBLEMS

PYTHAGOREAN WORD PROBLEMS. Word Problems Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

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Word Problem  We actually see the Pythagorean Theorem used in our daily lives. For example a television is measured by the diagonal dimension of its screen. For example, a 24-in television has a diagonal measure of 24 in.

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Page 1: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

PYTHAGOREANWORD PROBLEMS

Page 2: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems Today’s focus is going to be on the

dreaded word problems. More often problems that need to be solved using the Pythagorean Theorem appear as word problems. This is where everything we have been doing in this unit comes together.

Page 3: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problem We actually see the

Pythagorean Theorem used in our daily lives. For example a television is measured by the diagonal dimension of its screen. For example, a 24-in television has a diagonal measure of 24 in.

Page 4: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems The first thing you always want to do

when solving a word problem, is draw a picture. The pictures aids in making sure that you label the correct sides.

Page 5: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television screen is 16 in high and 22

in wide. What is its diagonal dimension to the nearest whole number?

Page 6: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television screen is 16 in high and 22

in wide. What is its diagonal dimension to the nearest whole number? Remember to first draw a picture

Page 7: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television screen is 16 in high and 22

in wide. What is its diagonal dimension to the nearest whole number? Remember to first draw a picture

16 in

Page 8: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television screen is 16 in high and 22

in wide. What is its diagonal dimension to the nearest whole number? Remember to first draw a picture

16 in

22 in

Page 9: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television

screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number? Remember to first

draw a picture

a2 + b2 = c2

16 in

22 in

Page 10: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television

screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number? Remember to first

draw a picture

a2 + b2 = c2

162 + 222 = c2

16 in

22 in

Page 11: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 1 A television

screen is 16 in high and 22 in wide. What is its diagonal dimension to the nearest whole number? Remember to first

draw a picture

a2 + b2 = c2

162 + 222 = c2

256 + 484 = c2

740 = c2

27.2 = c Therefore the

television has a diagonal measurement of 27.2 inches.

16 in

22 in

Page 12: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip from a camp

walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp? Remember to first draw a picture

Page 13: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip from a camp

walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp? Remember to first draw a picture

1.5 km

Page 14: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip from a camp

walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp? Remember to first draw a picture

1.5 km

1.7 km

Page 15: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip

from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?

a2 + b2 = c2

1.7 km

1.5 km

Page 16: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip

from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?

a2 + b2 = c2

1.72 + 1.52 = c2

1.7 km

1.5 km

Page 17: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip

from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?

a2 + b2 = c2

1.72 + 1.52 = c2

2.89 + 2.25 = c2

1.7 km

1.5 km

Page 18: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip

from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?

a2 + b2 = c2

1.72 + 1.52 = c2

2.89 + 2.25 = c2

5.14 = c2

1.7 km

1.5 km

Page 19: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 2 Two hikers start a trip

from a camp walking 1.5 km due east. They turn due north and walk 1.7 km to a waterfall. To the nearest tenth of kilometer, how far is the waterfall from the camp?

a2 + b2 = c2

1.72 + 1.52 = c2

2.89 + 2.25 = c2

5.14 = c2

2.3 = c Therefore the

waterfall is 2.3 km from camp.

1.7 km

1.5 km

Page 20: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is attaching a brace to the

back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace? Remember to first draw a picture

Page 21: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is attaching a brace to the

back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace? Remember to first draw a picture

40 in

30 in

Page 22: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is

attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?

a2 + b2 = c2

30 in

40 in

Page 23: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is

attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?

a2 + b2 = c2

302 + 402 = c2

30 in

40 in

Page 24: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is

attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?

a2 + b2 = c2

302 + 402 = c2

900 + 1600 = c2

30 in

40 in

Page 25: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is

attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?

a2 + b2 = c2

302 + 402 = c2

900 + 1600 = c2

2500 = c2

30 in

40 in

Page 26: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 3 A carpenter is

attaching a brace to the back of a frame that measures 30 in. by 40 in. What the is length, in inches, of the brace?

a2 + b2 = c2

302 + 402 = c2

900 + 1600 = c2

2500 = c2

50 = c Therefore the

brace is 50 inches long.

30 in

40 in

Page 27: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m under water to the

anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter. Remember to first draw a picture

Page 28: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m under water to the

anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter. Remember to first draw a picture

20 m

Page 29: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m under water to the

anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter. Remember to first draw a picture

20 m10 m

Page 30: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

20 m10 m

Page 31: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

102 + b2 = 202

20 m10 m

Page 32: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

102 + b2 = 202

100 + b2 = 400

20 m10 m

Page 33: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

102 + b2 = 202

100 + b2 = 400 b2 = 400 – 100

20 m10 m

Page 34: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

102 + b2 = 202

100 + b2 = 400 b2 = 400 – 100 b2 = 300

20 m10 m

Page 35: PYTHAGOREAN WORD PROBLEMS. Word Problems  Todays focus is going to be on the dreaded word problems. More often problems that need to be solved using

Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a buoy that is 10 m below the surface of the water. On the surface, how far is the buoy located from the place where the diver started? Round to the nearest meter.

a2 + b2 = c2

102 + b2 = 202

100 + b2 = 400 b2 = 400 – 100 b2 = 300 b = 17.3 m Therefore the buoy

is 17.3 meters from where the diver started.

20 m10 m