Pythagoras Word Problems

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PYTHAGORAS WORD PROBLEMS

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Pythagoras Word Problems. Introduction/Course Description. Introduction Introductory notes Introductory notes Introductory notes. Agenda. Math Minute Discuss Agenda Applying the Pythagorean Theorem to Word Problems. Word Problems. - PowerPoint PPT Presentation

Transcript of Pythagoras Word Problems

Page 1: Pythagoras Word Problems

PYTHAGORASWORD PROBLEMS

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Introduction/Course Description

Introduction Introductory notes

Introductory notes

Introductory notes

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Agenda Math Minute Discuss Agenda Applying the Pythagorean Theorem to

Word Problems

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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.

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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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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.

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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?

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

anchor of a bout 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

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

anchor of a bout 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

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

anchor of a bout 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

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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 Therefore the

brace is 50 inches long.

20 m10 m

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Word Problems – Example 4 A diver swims 20 m

under water to the anchor of a bout 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

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Word Problems As a team, work on the worksheet. Your

exit slip will be one of the problems from the worksheet.

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Exit Slip Your exit slip is problem number

Group 6: Problem #1 Group 7: Problem #2 Group 8: Problem #3 Group 9 and 10: Problem #4