Pythagorean Theorem, Pythagoras, Right triangles, hypotenuse, adjacent side, opposite side, 90 degrees angle, angle, right angle

1. PYTHAGOREAN
THEOREM

2. Right Triangle
90°
legs

3. c2
b2
a2
a2
+b2
= c2
b
a
c
In a right-angled triangle,
the square on the
hypotenuse is equal to the
sum of the squares on the
other two legs.
Hypotenuse
Pythagorean Theorem

4. b
a
Cut the squares
away from the right
angle triangle and cut
up the segments
of square ‘a’
Draw line segment
xy, parallel with the
hypotenuse of the
triangle
x
y
q
p
Draw line segment
pq, at right angles to
Line segment xy.
To show how this works:

5. Now rearrange them
to look like this.
You can see that they
make a square with
length of side ‘c’.
This demonstrates that
the areas of squares
a and b
add up to be the
area of square c
a2
+b2
= c2

6. Write an equation representing the relationship among
the lengths of all sides of the following right triangle
where the variables and numbers are units of length.
1)
𝑝
𝑞
𝑟
2)
8𝑛
𝑚
3)
𝑎
𝑏
13
𝑟2
+ 𝑝2
= 𝑞2
82 + 𝑛2 = 𝑚2
𝑎2 + 𝑏2 = 132

7. Write an equation representing the relationship among
the lengths of all sides of the following right triangle
where the variables and numbers are units of length.
4)
𝑧
𝑥
𝑦
5)
6𝑒
14
7)
𝑡
𝑠
𝑟
8) 12
𝑎
7
6)
15
𝑏9
9)
16
𝑐
14
𝑥2 + 𝑦2 = 𝑧2
𝑒2
+ 62
= 142
92 + 𝑏2 = 152
𝑠2 + 𝑡2 = 𝑟2
72 + 122 = 𝑎2
𝑐2
+ 142
= 162

8. Find the lengths of the missing side of the right triangle
in each of the following figures.
1)
20
21
3)
30
34
6.5
6
2)
212
+ 202
= 𝑐2
441 + 400 = 𝑐2
841 = 𝑐2
𝑐 = 29
𝑎2 + 302 = 342
𝑎2
= 342
− 302
𝑎2 = 1156 − 900
𝑎2
= 256
𝑎 = 16
𝑎2 + 62 = 6.52
𝑎2
= 6.52
− 62
𝑎2 = 42.25 − 36
𝑎2
= 6.25
𝑎 = 2.5

9. Find the lengths of the missing side of the right triangle
in each of the following figures.
4)
13
12
5)
11
21
6)
15
17
7)
12.5
3.5 8)
7.5
4.5
9) 37
35
5
8 5
8
12 6
12

10. The numbers in each of the following items are the
lengths of the legs of the triangle. Find the length of the
hypotenuse.
1) 6, 8 2) 7, 24
3) 18, 24 4) 4.5, 6
5) 2, 4.8 6) 1.5, 2
10 25
30 7.5
5.2 2.5

11. Pythagorean Theorem
In a right-angled triangle, the square on the hypotenuse is equal to the
sum of the squares on the other two legs.
Reverse of the Pythagorean Theorem
In a triangle, if the square of the longest side is equal to the sum of the
squares of the other sides, then it is a right-angled triangle.
Example:
Let ∆𝐴𝐵𝐶 have 𝐴𝐵, 𝐴𝐶, and 𝐵𝐶, with 18, 24, and 30 centimeters long,
respectively. Determine whether the ∆𝐴𝐵𝐶 is a right triangle.
𝐴
𝐵 𝐶
18 24
30
𝑎2 = 182 = 324
𝑏2 = 242 = 576
𝑐2 = 302 = 900
𝑐2 = 𝑎2 + 𝑏2
900 = 576 + 324
900 = 900

12. Example:
Given ∆𝑃𝑄𝑅 where 𝑃𝑆 ⊥ 𝑄𝑅, and 𝑃𝑆, 𝑄𝑆, and 𝑆𝑅 have the lengths of
36, 27, and 48 units, respectively as shown in the figure. Show that
∆𝑃𝑄𝑅 is a right triangle.
𝑃
𝑄 𝑆 𝑅
36
27 48
𝑃𝑄2 = 362 + 272
𝑃𝑄2 = 1296 + 729
𝑃𝑄2 = 2025
𝑃𝑅2
= 362
+ 482
𝑃𝑄2 = 1296 + 2304
𝑃𝑄2 = 3600
𝑃𝑄2
+ 𝑃𝑅2
= 2025 + 3600
𝑃𝑄2
= 5625
𝑄𝑅 = 𝑄𝑆 + 𝑆𝑅
𝑄𝑅 = 27 + 48
𝑄𝑅 = 75
𝑄𝑅2 = 5625
𝑄𝑅2 = 𝑃𝑄2 + 𝑃𝑅2

13. Given the lengths of all the sides of triangles, determine
whether each of the following triangles is a right triangle.
1) 14, 48, 50 2) 15, 24, 26
3) 24, 70, 74 4) 33, 56,65
5) 55, 47, 76 6) 10, 10.5, 14.5
7) 16.5, 90, 91.5 8) 4.2, 5.6, 8.4
9) 4.7, 5.5, 7.2 10) 7.5, 10, 12.5
right not right
right right
right right
right not right
not right right

14. Determine whether each of the following triangles is a right
triangle.
1)
𝐴
𝐵 𝐶
30
12.5 72
2) 𝐴
𝐵 𝐶
30
32𝐷
24
right
right

15. Pythagorean
Theorem
Applications

16. From the given figure, how many units is 𝑆𝑅 longer than 𝑄𝑆?
𝑃
𝑅
52
𝑄 𝑆
50
𝑃𝑄2 = 𝑃𝑆2 + 𝑄𝑆2
𝑄𝑆2 = 𝑃𝑄2 − 𝑃𝑆2
𝑄𝑆2 = 502 − 482
𝑄𝑆2 = 2500 − 2304
𝑄𝑆2 = 196
𝑄𝑆 = 1414
48
𝑃
𝑅
52
𝑆
48
𝑃𝑅2 = 𝑃𝑆2 + 𝑆𝑅2
𝑆𝑅2 = 𝑃𝑅2 − 𝑃𝑆2
𝑆𝑅2 = 522 − 482
𝑄𝑆2 = 2704 − 2304
𝑄𝑆2
= 400
𝑆𝑅 = 20
20
20 − 14 = 6 units
𝑄 𝑆
50
𝑃

17. 𝐷
𝐵
𝐶
𝐷
𝐴 𝐵
Given the figure of ∆𝐴𝐵𝐷 with an area of 24 square centimeters, if 𝐴𝐵
is 6 centimeters, how long is 𝐶𝐷?
𝐷
𝐴 𝐵
𝐶
6
Area of ∆𝐴𝐵𝐷 =
1
2
𝐴𝐵 𝐴𝐷
24 =
1
2
6 𝐴𝐷
𝐴𝐷 = 88
𝐵𝐷2 = 𝐴𝐷2 + 𝐴𝐵2
𝐵𝐷2 = 82 + 62
𝐵𝐷 = 10
10
10
𝐷
𝐵
𝐶
10
10
𝐶𝐷2
= 𝐵𝐷2
+ 𝐵𝐶2
𝐶𝐷2 = 102 + 102
𝐶𝐷2 = 200
𝐶𝐷 = 10 2
10 2

18. Given that 𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻 is a rectangular box as shown, the length of
𝐴𝐵 is 12 centimeters, 𝐵𝐶 is 9 centimeters, and 𝐴𝐹 is 8 centimeters.
Find the length of 𝐴𝐻.
𝐴
𝐵
𝐶
12 9
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = 122 + 92
𝐴𝐶 = 15
15
𝐻
𝐶
𝐴𝐴
𝐵
𝐶
𝐷
𝐸
𝐹
𝐻
12 9
8 𝐺
15
8
𝐴𝐻2
= 𝐴𝐶2
+ 𝐶𝐻2
𝐴𝐻2 = 152 + 82
𝐴𝐻2
= 289
𝐴𝐻 = 17
𝐴𝐶2
= 225
17

19. A ladder 13 meters in length rests with its top against the edge of the
window. The foot of the ladder is placed 0.5 meters away from the
building. How high is the edge of the window above the ground?
𝐴
𝐵 𝐶
13 𝑚
0.5 𝑚
𝐴𝐵2 = 𝐴𝐶2 − 𝐵𝐶2
𝐴𝐵2 = 132 −
1
2
2
𝐴𝐵2 = 169 −
1
4
𝐴𝐵2 =
675
4
𝐴𝐵 =
675
4
𝐴𝐵 =
15 3
2
15 3
2
𝑚

20. A ship travels to the north 20 miles, and to the west 2 miles before
stopping. Then, it travels to the north 20 miles and moves to the east
11 miles. How many miles is this ship away from its origin?
𝐴
𝐵𝐶
𝐷 𝐸
20
2
20
11
𝐹
9
40
𝐴 𝐹
𝐸
40
9
𝐴𝐸2 = 𝐴𝐹2 + 𝐸𝐹2
= 92 + 402
= 81 + 1600
= 1681
𝐴𝐸 = 41
41

21. QUIZ
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