Transcript Applications of the Pythagorean Theorem
Applications of the Pythagorean Theorem
Lesson 3.3.3
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Lesson
3.3.3
Applications of the Pythagorean Theorem
California Standard:
Measurement and Geometry 3.3
Know and understand the Pythagorean theorem
and its converse
line segments and use it to find the length of the missing side of a right triangle and the lengths of other
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 2
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.
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Lesson
3.3.3
Applications of the Pythagorean Theorem Use the Pythagorean Theorem in Other Shapes Too
You can use the lots of
shapes Pythagorean theorem
— you just have to
split
to find lengths in them up into
right triangles
.
Here’s a reminder of the formula.
c
2 =
a
2 +
b
2 which rearranges to:
a
2 =
c
2 –
b
2 (
c
is the hypotenuse length, and
a
and
b
are the leg lengths.)
a b c
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Lesson
3.3.3
Applications of the Pythagorean Theorem Example 1
Find the area of rectangle ABCD, shown.
A 13 inches B Solution
The formula for the area of a rectangle is:
Area = length × width
.
D 12 inches
You know that the
length
of the rectangle is
12 inches
, but you don’t know the rectangle’s
width
,
BC
.
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
.
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C
Lesson
3.3.3
Example 1 Applications of the Pythagorean Theorem B
Find the area of rectangle ABCD, shown.
13 inches Solution (continued)
BC 2 = 13 2 – 12 2 BC 2 = 169 – 144
D
You can use the
Pythagorean theorem
to find the length of side
BC
.
BC 2 = BD 2 – CD 2
Write out the equation Substitute the values you know Simplify the equation 12 inches
BC 2 = 25
BC = = 5 inches C
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Solution continues…
Lesson
3.3.3
Applications of the Pythagorean Theorem Example 1
Find the area of rectangle ABCD, shown.
A 5 inches 13 inches Solution (continued) D 12 inches
BC is the width of the rectangle. Now you can find its area.
Area = length × width Area = 12 inches × 5 inches =
60 inches 2 B C
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Lesson
3.3.3
Example 2 Applications of the Pythagorean Theorem R
Find the area of isosceles triangle QRS.
15 cm Solution
The formula for the area of a
Q Q
triangle is:
1 Area = base × height
.
2
The
base
of the triangle is
18 cm
, but you don’t know its
height
,
MR .
S M 9 cm 18 cm
This is half the base of the original triangle.
Isosceles triangles
can be split up into two right triangles. 8
Lesson
3.3.3
Applications of the Pythagorean Theorem Example 2 R
Find the area of isosceles triangle QRS.
15 cm Solution (continued)
You can use the
Pythagorean theorem
to find the length of side
MR
.
MR 2 = RS 2 – MS 2
Write out the equation 12 cm M
MR 2 = 15 2 – 9 2 MR 2 = 225 – 81
Substitute the values you know Simplify the equation 9 cm S
MR 2 = 144
MR = = 12 cm
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Solution continues…
Lesson
3.3.3
Applications of the Pythagorean Theorem Example 2 R
Find the area of isosceles triangle QRS.
Solution (continued)
Now put the value of MR into the area formula: 1 Area = base × height 2
Q 15 cm 12 cm M 18 cm 15 cm S
1 Area = (18 cm) × 12 cm = 9 cm × 12 cm =
108 cm 2
2 10
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.
E F 2.
U 10 cm 6 cm 34 ft 34 ft H Area =
x
cm 2 GH 2 = 10 2 – 6 2 GH = 8 cm
x
= 8 • 6 = 48 = 64 G T V P UP 2 = 34 2 32 ft Area =
x
– (32 ÷ 2) ft 2 UP = 30 ft
x
= 0.5 • 32 • 30 = 480 2 = 900
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Solution follows…
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.
W 25 in X 4.
W X 17 in
x
in.
x
m 12 m Z Y 33 in
x
2 = 17
x
= 15 2 – (33 – 25) 2 = 225 Z Area = 108 m 2 ZY = 108 ÷ 12 = 9 m
x
2 = 12 2 + 9 2 = 225
x
= 15 Y
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Solution follows…
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. 13
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 Path Yard
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.
32 feet 24 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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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)
c
2 =
a
2 +
b
2
Write out the equation
c
2 = 32 2 + 24 2
c
2 = 1024 + 576
c
2 = 1600
Substitute the values you know Simplify the equation
c c
= = 40 feet
a
= 32 ft
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Solution continues…
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) Path
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 Yard 40 feet 32 feet 24 feet
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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.
Let
l l
2 = 8 2 = length of ladder: + 6 2 = 100
l
= 10 m
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
s
= distance of straight-line walk and
w s
2 = 0.5
= distance Sam walked: 2 + 1.2
2 = 1.69
s
= 1.3 miles and
w
= 1.2 + 0.5 = 1.7 miles Difference = 1.7 – 1.3 = 0.4 miles
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Solution follows…
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?
Let
s
= side length of the tablecloth Then 4 2 =
s
2 +
s
2
10 cm
16 = 2
s
2 So the area of the table cloth is:
s
2 = 8 feet 2
8 cm
8.
Megan is making the kite shown in the diagram on the right. The crosspieces are made of thin cane. What length of cane will she need in total?
17 cm
Let
l
Then = total length needed
l
= (2 × 8) + + = 16 + 6 + 15 = 37 cm
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Solution follows…
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.
B 2.
P Q
x
inches 15 inches 13 cm 13 cm A H 10 cm Area =
x
cm 2
x
= 60 C S R Area = 300 inches 2
x
= 25
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Solution follows…
Lesson
3.3.3
Applications of the Pythagorean Theorem
Independent Practice In Exercises 3 –4 use the Pythagorean theorem to find the missing value,
3.
E 32 m
x
.
F 4.
K 20 feet L 34 m
x
m 17 feet
x
feet H G N M 48 m
x
= 30 36 feet
x
= 15
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Solution follows…
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?
270 m 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 m 2 . How many cans should he buy?
5 cans
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Solution follows…
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 feet 2 5 feet 5 feet 11 feet 15 feet 20 feet
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Solution follows…
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 Pitcher’s plate 3rd base 1st base Home plate 23
Solution follows…
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 theorem
to find the
missing lengths of sides
in many different shapes — it just takes practice to be able to spot the right triangles.
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