1. A. 2 Square Units, 2 Square Units, 4 Square Units

Answers | Investigation 3

Applications

1. a. 2 square units, 2 square units,
4 square units

b. The side lengths are units,
units, and 2 units, and
(that is, 2 + 2 = 4),
so the side lengths satisfy the
Pythagorean Theorem.

2. The sides have lengths units,
units, and units and, because
(that is, 5 + 5 =
10), the triangle satisfies the Pythagorean
Theorem.

Note: This is a convenient place to remind
students that ≠ even
though .

3. a. 52 + 122 = 169; 169 in.2

b. 13 in.

4. 32 + 62 = 45, so c = cm, or about
6.7 cm

5. 42 + 32 = 25, so h = in. = 5 in.

6. 32 + 82 = 73, so k = cm ≈ 8.5 cm

7. WX is the hypotenuse of a right triangle
with legs of length 4 units and 1 unit.
Because 42 + 12 = 17, the length of
segment WX is units. Therefore,
W and X are units apart.

8. YZ is the hypotenuse of a right triangle
with legs of length 4 units and 2 units.
Because 42 + 22 = 20, the length of
segment YZ is units. Therefore, the
distance between Y and Z is units.

9. Because 42 + 32 = 25, the distance is
5 blocks.

10. Because 62 + 52 = 61, the distance is
blocks ≈ 7.8 blocks.

11. The distance is 4 blocks.

12. Because 42 + 42 = 32, the distance is
blocks ≈ 5.7 blocks.

13. D

14. F

15. This is a right triangle.

16. This is not a right triangle. 92 + 162 ≠ 252

Note: In fact, these side lengths will not
form a triangle of any kind. As in Exercise
16, watch for students who incorrectly
write that

Connections

17. a. 6.5 cm

b. You do not need to know the value of
a to find the volume, but it is needed
to find the surface area. To find the
volume, you multiply 4 by the area of
the triangular face, which you can find
using only the given base and height.
To find the surface area, you need to
find the areas of the rectangular faces.
For one of these faces, you need to
know the value of a.

c. 30 cm3; 0.5(6 × 2.5) × 4 = 30

d. 75 cm2; (2.5 × 4) + 2 [0.5(6 × 2.5)] +
(6 × 4) + (6.5 × 4) = 10 + 15 + 24 +
26 = 75

Answers | Investigation 3

e. Possible sketch:

18. B 19. H 20. B 21. H

22. No. This can be checked using the
Pythagorean Theorem. The two sides
squared do not equal the diagonal
squared, so the sides do not form a right
angle, so it cannot be a rectangle.

The question we need to ask is, does
162 + 202 equal 252, or 625? Since
256 + 400 = 656 ≠ 625, we know these
cannot be measures of a right triangle, so
this quadrilateral does not have a right angle
between the sides measuring 16 and 20
units, which means it cannot be a rectangle.

23. The diagonal must be 15 feet. For
a rectangle, the angle created by
the two sides must be a right angle,
so you can use the Pythagorean
Theorem to find the diagonal. This will
guarantee the base is a rectangle. Since
122 + 92 = 144 + 81 = 225, and the square
root of 225 is 15, the diagonal must be
15 feet long.

24. For the barn wall to be vertical, the pole
must be the hypotenuse of a right triangle.
The hypotenuse is 10 feet and one leg is
6 feet, so the other leg must be 8 feet.
This triangle has the same angles as a
3–4–5 right triangle, but it is twice a big.

25. a. 4 blocks

b. blocks. Find the length of the
segment connecting the points. It is the
hypotenuse of a right triangle with leg
lengths 1 and 3. Then 32 + 12 = 10, so
the distance is blocks.

26. Points A and B are 5 units apart. Point F is
also 5 units from point A.

Extensions

27. 72 – 42 = 33, so x = m ≈ 5.7 m
212 – 42 = 425, so y = m ≈ 20.6 m

28. a. Possible answer: Draw a right
triangle as shown below, and use the
Pythagorean Theorem to find the
hypotenuse, which is the radius.

b. 5 units

Note: You can give additional extension
problems to interested students. For
example, you might ask students to find
the length of a diagonal of a square with

side length a. Or, you could ask them
to draw a square of side a inscribed in
a circle and then to find the radius and
area of the circle in terms of a.

29. a. The radii of the circles are 1.5, 2, and
2.5 units, so the areas are 1.125π, 2π
and 3.125π square units.

b. When the areas of the half circles on
both legs are added together, you
get the area of the half circle on the
hypotenuse.

30. a. Using the Pythagorean Theorem,
you can calculate the heights of the
triangles as approximately 2.59, 3.46,
and 4.33 units, so the areas of the
triangles are approximately 3.89, 6.92,
and 10.83 square units.

b. When the areas of the triangles on both
legs are added together, you get the
area of the triangle on the hypotenuse.

Answers | Investigation 3

31. a. The areas of the hexagons are 3.89,
6.92, and 10.83 square units (6 times the
areas of the triangles in Exercise 30).

b. When the areas of the regular
hexagons on both legs are added
together, you get the area of the
regular hexagon on the hypotenuse.
(i.e., 23.4 + 41.6 = 65.0)

+ / 1 / 4 / 9 / 16 / 25 / 36 / 49 / 64
1 / 2 / 5 / 10 / 17 / 26 / 37 / 50 / 65
4 / 5 / 8 / 13 / 20 / 29 / 40 / 53 / 68
9 / 10 / 13 / 18 / 25 / 34 / 45 / 58 / 73
16 / 17 / 20 / 25 / 32 / 41 / 52 / 65 / 80
25 / 26 / 29 / 34 / 41 / 50 / 61 / 74 / 89
36 / 37 / 40 / 45 / 52 / 61 / 72 / 85 / 100
49 / 50 / 53 / 58 / 65 / 74 / 85 / 98 / 113
64 / 65 / 68 / 73 / 80 / 89 / 100 / 113 / 128

32. a.

b. 1 and 9

c. 9 and 16

d. 25 and 64

e. 1 + 25 = 26, so a triangle with leg
lengths 1 unit and 5 units has a
hypotenuse of length units.

f. 36 + 64 = 100, so a triangle with
leg lengths 6 units and 8 units has a
hypotenuse of length 10 units.

g. 9 + 1 = 10, so a triangle with leg
lengths 3 units and 1 unit has a
hypotenuse of length units.

h. 49 + 1 = 50, so a triangle with leg
lengths 7 units and 1 unit has a
hypotenuse of length units.

33. Yes. units is the length of the
hypotenuse of a right triangle with leg
lengths of 1 unit.

34. No. 3 is not the sum of two square
numbers.

35. Yes. , so just draw a horizontal or
vertical segment with length 2 units.

36. Yes. units is the length of the
hypotenuse of a right triangle with leg
lengths of 2 units and 1 unit.

37. No. 6 is not the sum of two square numbers.

38. No. 7 is not the sum of two square numbers.

39. a. J(1, 1); K(4, 7)

b. About 6.7 units. You can draw a right
triangle with hypotenuse JK. The length
of one leg is the positive difference
of the x-coordinates, which is 4 – 1,
or 3. The length of the other leg is the
positive difference of the y-coordinates,
which is 7 – 1 = 6. So, the length of JK
is , or approximately
6.7 units.

Note: In high school, students will see
the distance formula,
or
d2 = (x2 – x1)2 + (y2 – y1)2. The
distance formula follows directly from
the Pythagorean Theorem. If you use
the segment between two points as
the hypotenuse of a right triangle,
the length of the horizontal leg will
be x2 – x1 and the length of the
vertical side will be y2 – y1, so the
distance between the points, which is
the length of the hypotenuse, will be
.

c. 2.8 units.
, or about 2.8.