Lesson 3.3.3

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Lesson 3.3.3Lesson 3.3.3

Applications of thePythagorean Theorem

Applications of thePythagorean Theorem

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Lesson

3.3.3Applications of the Pythagorean TheoremApplications of the Pythagorean Theorem

California Standard:Measurement and Geometry 3.3Know and understand the Pythagorean theorem and its converse and use it to find the length of the missing side of a right triangle and the lengths of other line segments and, 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

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Applications of the Pythagorean TheoremApplications of the Pythagorean TheoremLesson

3.3.3

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.

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Applications of the Pythagorean TheoremApplications of the Pythagorean Theorem

Use the Pythagorean Theorem in Other Shapes Too

Lesson

3.3.3

You can use the Pythagorean theorem to find lengths in lots of shapes — you just have to split them up into right triangles.

Here’s a reminder of the formula.

c2 = a2 + b2 which rearranges to: a2 = c2 – b2

(c is the hypotenuse length, and a and b are the leg lengths.)

a

b

c

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Example 1

Solution follows…

Lesson

3.3.3

Find the area of rectangle ABCD, shown.A B

D C

13 inches

12 inchesSolution

The formula for the area of a rectangle is: Area = length × width.

But you do know the length of the diagonal BD and since all the corners of a rectangle are 90° angles, you know that BCD is a right triangle.

You know that the length of the rectangle is 12 inches, but you don’t know the rectangle’s width, BC.

Solution continues…

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You can use the Pythagorean theorem to find the length of side BC.

Simplify the equation

Substitute the values you know


Example 1

Lesson

3.3.3

Find the area of rectangle ABCD, shown.B

D C

13 inches

12 inchesSolution (continued)

BC = = 5 inches

BC2 = BD2 – CD2

BC2 = 132 – 122

BC2 = 169 – 144

BC2 = 25

Write out the equation


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BC is the width of the rectangle. Now you can find its area.


Example 1

Lesson

3.3.3

Find the area of rectangle ABCD, shown.A B

D C

13 inches

12 inchesSolution (continued)

Area = 12 inches × 5 inches = 60 inches2

Area = length × width

5 inches

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15 cm

Q


Example 2

Solution follows…

Lesson

3.3.3

Find the area of isosceles triangle QRS.

Solution

Q

R

S

15 cm

18 cm

15 cm

The base of the triangle is 18 cm, but you don’t know its height, MR.

Isosceles triangles can be split up into two right triangles.

MThe formula for the area of a

triangle is: Area = base × height. 1

2

9 cm

This is half the base of the original triangle.


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You can use the Pythagorean theorem to find the length of side MR.


Example 2

Lesson

3.3.3


Solution (continued)

R

S

15 cm

M 9 cmMR2 = RS2 – MS2

MR2 = 144

MR2 = 225 – 81

MR2 = 152 – 92

MR = = 12 cm

12 cm





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Now put the value of MR into the area formula:


Example 2

Lesson

3.3.3



Q

R

S

15 cm

18 cm

15 cm

M

12 cm

Area = base × height1

2

Area = (18 cm) × 12 cm = 9 cm × 12 cm = 108 cm21

2

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Guided Practice

Solution follows…

Lesson

3.3.3

In Exercises 1–2 use the Pythagorean theorem to find the missing value, x.

1. 2.

10 cm6 cm

Area = x cm2

E F

H G

T

U

V

34 ft

32 ft

34 ft

P

Area = x ft2GH2 = 102 – 62 = 64GH = 8 cmx = 8 • 6 = 48 UP2 = 342 – (32 ÷ 2)2 = 900

UP = 30 ft x = 0.5 • 32 • 30 = 480

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Guided Practice

Solution follows…

Lesson

3.3.3


3. 4.

x m12 m

Area = 108 m2

W X

Z Y

x2 = 172 – (33 – 25)2 = 225x = 15 ZY = 108 ÷ 12 = 9 m

x2 = 122 + 92 = 225 x = 15

W X

Z Y

25 in

33 in

17 in x in.

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The Pythagorean Theorem Has Real-Life Applications

Lesson

3.3.3

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.

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Example 3

Solution follows…

Lesson

3.3.3

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.

Yard

Path

24 feet

32 feet

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.


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Example 3

Lesson

3.3.3



c2 = a2 + b2

c2 = 322 + 242

c2 = 1024 + 576

c2 = 1600

c = = 40 feet




a = 32 ft

b = 2

4 ft

c


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Example 3

Lesson

3.3.3



Yard

Path

24 feet

32 feet

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

40 feet

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Guided Practice

Solution follows…

Lesson

3.3.3

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

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Guided Practice

Solution follows…

Lesson

3.3.3

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?

17 cm

8 cm

10 cmLet s = side length of the tableclothThen 42 = s2 + s2

16 = 2s2 So the area of the table cloth is: s2 = 8 feet2

Let l = total length neededThen l = (2 × 8) + + = 16 + 6 + 15 = 37 cm

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Independent Practice

Solution follows…

Lesson

3.3.3


1. 2.

A

B

C

13 cm

10 cm

13 cm

H

Area = x cm2

15 inchesx inches

Area = 300 inches2

P Q

S R

x = 25

x = 60

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Solution follows…

Lesson

3.3.3

In Exercises 3–4 use the Pythagorean theorem to find the missing value, x.3. 4.

x feet

K L

N M

x = 30 x = 15

E F

H G48 m

34 m x m

32 m

36 feet

20 feet

17 feet

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Solution follows…

Lesson

3.3.3

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

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Solution follows…

Lesson

3.3.3

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

15 feet

11 feet

5 feet

5 feet

20 feet

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Solution follows…

Lesson

3.3.3

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

3rd base 1st base

2nd base

Home plate

Pitcher’s plate

90 feet

90 fe

et

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Round UpRound Up

Lesson

3.3.3

You can break up a lot of shapes into right triangles. This means you can use the Pythagorean theorem to find the missing lengths of sides in many different shapes — it just takes practice to be able to spot the right triangles.

Lesson 3.3.3

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