1. Applications of the Pythagorean Theorem
Lesson 3.3.3

2. 3.3.3 Applications of the Pythagorean Theorem California Standard:
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 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

3. 3.3.3 Applications of the Pythagorean Theorem
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.

4. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Use the Pythagorean Theorem in Other Shapes Too
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
a2 = c2 – b2
which rearranges to:
(c is the hypotenuse length, and a and b are the leg lengths.) a b c

5. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 1 A B
Find the area of rectangle ABCD, shown.
13 inches
Solution D
12 inches C
The formula for the area of a rectangle is: Area = length × width.
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.
Solution continues…
Solution follows…

6. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 1 B
Find the area of rectangle ABCD, shown.
13 inches
Solution (continued) D
12 inches C
You can use the Pythagorean theorem to find the length of side BC.
BC2 = BD2 – CD2
Write out the equation
BC2 = 132 – 122
Substitute the values you know
BC2 = 169 – 144
Simplify the equation
BC2 = 25
BC = = 5 inches
Solution continues…

7. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 1 A B
Find the area of rectangle ABCD, shown.
13 inches
5 inches
Solution (continued) D
12 inches C
BC is the width of the rectangle. Now you can find its area.
Area = length × width
Area = 12 inches × 5 inches = 60 inches2

8. This is half the base of the original triangle.
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 2 R
Find the area of isosceles triangle QRS. Q
15 cm
15 cm
15 cm
Solution
The formula for the area of a triangle is: Area = base × height. 1 2 Q S M
9 cm
18 cm
This is half the base of the original triangle.
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.
Solution continues…
Solution follows…

9. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 2 R
Find the area of isosceles triangle QRS.
15 cm
Solution (continued)
12 cm
You can use the Pythagorean theorem to find the length of side MR. S M
9 cm
MR2 = RS2 – MS2
Write out the equation
MR2 = 152 – 92
Substitute the values you know
MR2 = 225 – 81
Simplify the equation
MR2 = 144
MR = = 12 cm
Solution continues…

10. 3.3.3 Applications of the Pythagorean Theorem
Lesson
3.3.3
Applications of the Pythagorean Theorem
Example 2 R
Find the area of isosceles triangle QRS.
15 cm
15 cm
Solution (continued)
12 cm
Now put the value of MR into the area formula: Q S M
18 cm
Area = base × height 1 2
Area = (18 cm) × 12 cm = 9 cm × 12 cm = 108 cm2 1 2

11. 3.3.3 Applications of the Pythagorean Theorem Guided Practice
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. E F U
10 cm
6 cm
34 ft
34 ft H G T V
Area = x cm2 P
32 ft
GH2 = 102 – 62 = 64 GH = 8 cm x = 8 • 6 = 48
Area = x ft2
UP2 = 342 – (32 ÷ 2)2 = 900 UP = 30 ft x = 0.5 • 32 • 30 = 480
Solution follows…

12. 3.3.3 Applications of the Pythagorean Theorem Guided Practice
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.
25 in W X W X
17 in
x m
x in.
12 m Z Y Z Y
33 in
Area = 108 m2
x2 = 172 – (33 – 25)2 = 225 x = 15
ZY = 108 ÷ 12 = 9 m x2 = = x = 15
Solution follows…

13. 3.3.3 Applications of the Pythagorean Theorem
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.

14. 3.3.3 Applications of the Pythagorean Theorem
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
Yard
Path
24 feet
32 feet
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.
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.
Solution continues…
Solution follows…

15. 3.3.3 Applications of the Pythagorean Theorem
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)
c2 = a2 + b2
Write out the equation
c2 =
Substitute the values you know
a = 32 ft
b = 24 ft c
c2 =
Simplify the equation
c2 = 1600
c = = 40 feet
Solution continues…

16. 3.3.3 Applications of the Pythagorean Theorem
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)
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.
40 feet
Sacks needed = 40 ÷ 10 = 4 sacks

17. 3.3.3 Applications of the Pythagorean Theorem Guided Practice
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. 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 = = 100 l = 10 m
Let s = distance of straight-line walk and w = distance Sam walked: s2 = = 1.69 s = 1.3 miles and w = = 1.7 miles Difference = 1.7 – 1.3 = 0.4 miles
Solution follows…

18. 3.3.3 Applications of the Pythagorean Theorem Guided Practice
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?
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?
Let s = side length of the tablecloth Then 42 = s2 + s2 16 = 2s2 So the area of the table cloth is: s2 = 8 feet2
17 cm
8 cm
10 cm
Let l = total length needed Then l = (2 × 8) = = 37 cm
Solution follows…

19. 3.3.3 Applications of the Pythagorean Theorem Independent Practice
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. B P Q
x inches
15 inches
13 cm
13 cm S R A C
Area = 300 inches2 H
10 cm
x = 25
Area = x cm2
x = 60
Solution follows…

20. 3.3.3 Applications of the Pythagorean Theorem Independent Practice
Lesson
3.3.3
Applications of the Pythagorean Theorem
Independent Practice
In Exercises 3–4 use the Pythagorean theorem to find the missing value, x.
32 m
20 feet E F K L
34 m
x m
17 feet
x feet H G N M
48 m
36 feet
x = 30
x = 15
Solution follows…

21. 3.3.3 Applications of the Pythagorean Theorem Independent Practice
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?
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
Solution follows…

22. 3.3.3 Applications of the Pythagorean Theorem Independent Practice
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 feet2
15 feet
11 feet
5 feet
20 feet
Solution follows…

23. 3.3.3 Applications of the Pythagorean Theorem Independent Practice
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
90 feet
Pitcher’s plate
3rd base
1st base
90 feet
Home plate
Solution follows…

24. 3.3.3 Applications of the Pythagorean Theorem Round Up
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.