Lesson 3.2
Perfect Squares, Perfect Cubes, and Their Roots Exercises (pages 146–147)
A
4. Use a calculator to write each number as a product of its prime factors, then arrange the factors in 2
equal groups. The product of the factors in one group is the square root.
a) 196 = 2 · 2 · 7 · 7
= (2 · 7)(2 · 7)
= 14 · 14
196 = 14
b) 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
= (2 · 2 · 2 · 2)(2 · 2 · 2 · 2)
= 16 · 16
256 = 16
c) 361 = 19 · 19
361 = 19
d) 289 = 17 · 17
289 = 17
e) 441 = 3 · 3 · 7 · 7
= (3 · 7)(3 · 7)
= 21 · 21
441 = 21
5. Use a calculator to write each number as a product of its prime factors, then arrange the factors in 3
equal groups. The product of the factors in one group is the cube root.
a) 343 = 7 · 7 · 7
3
343 = 7
b) 512 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
= (2 · 2 · 2)(2 · 2 · 2)(2 · 2 · 2)
=8·8·8
3
512 = 8
c) 1000 = 2 · 2 · 2 · 5 · 5 · 5
= (2 · 5)(2 · 5)(2 · 5)
= 10 · 10 · 10
3
1000 = 10
d) 1331 = 11 · 11 · 11
3
1331 = 11
e) 3375 = 3 · 3 · 3 · 5 · 5 · 5
= (3 · 5)(3 · 5)(3 · 5)
= 15 · 15 · 15
3
3375 = 15
B
6. Use a calculator to write each number as a product of its prime factors. If the factors can be arranged
in 2 equal groups, then the number is a perfect square. If the factors can be arranged in 3 equal
groups, then the number is a perfect cube.
a) 225 = 3 · 3 · 5 · 5
= (3 · 5)(3 · 5)
The factors can be arranged in 2 equal groups, so 225 is a perfect square.
b) 729 = 3 · 3 · 3 · 3 · 3 · 3
= (3 · 3)(3 · 3)(3 · 3)
= (3 · 3 · 3)(3 · 3 · 3)
The factors can be arranged in 2 equal groups and in 3 equal groups, so 729 is both a perfect
square and a perfect cube.
c) 1944 = 2 · 2 · 2 · 3 · 3 · 3 · 3 · 3
The factors cannot be arranged in 2 or 3 equal groups, so 1944 is neither a perfect square nor a
perfect cube.
d) 1444 = 2 · 2 · 19 · 19
= (2 · 19)(2 · 19)
The factors can be arranged in 2 equal groups, so 1444 is a perfect square.
e) 4096 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2
= (2 · 2 · 2 · 2 · 2 · 2)(2 · 2 · 2 · 2 · 2 · 2)
= (2 · 2 · 2 · 2)(2 · 2 · 2 · 2)(2 · 2 · 2 · 2)
The factors can be arranged in 2 equal groups and 3 equal groups, so 4096 is both a perfect square
and a perfect cube.
f) 13 824 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 · 3 · 3 · 3
= (2 · 2 · 2 · 3)(2 · 2 · 2 · 3)(2 · 2 · 2 · 3)
The factors can be arranged 3 equal groups, so 13 824 is a perfect cube.
7. The side length of each square is the square root of its area. Use a calculator to write each number as
a product of its prime factors, then arrange the factors in 2 equal groups. The product of the factors in
one group is the square root.
a) 484 = 2 · 2 · 11 · 11
= (2 · 11)(2 · 11)
= 22 · 22
484 = 22
The side length of the square is 22 mm.
b) 1764 = 2 · 2 · 3 · 3 · 7 · 7
= (2 · 3 · 7)(2 · 3 · 7)
= 42 · 42
1764 = 42
The side length of the square is 42 yd.
8. The edge length of each cube is the cube root of its volume. Use a calculator to write each number as
a product of its prime factors, then arrange the factors in 3 equal groups. The product of the factors in
one group is the cube root.
a) 5832 = 2 · 2 · 2 · 3 · 3 · 3 · 3 · 3 · 3
= (2 · 3 · 3)(2 · 3 · 3)(2 · 3 · 3)
= 18 · 18 · 18
3
5832 = 18
The edge length of the cube is 18 in.
b) 15 625 = 5 · 5 · 5 · 5 · 5 · 5
= (5 · 5)(5 · 5)(5 · 5)
= 25 · 25 · 25
3 15 625 = 25
The edge length of the cube is 25 ft.
9. The volume of the cube is 64 cubic feet.
The edge length of the cube, in feet, is: 3 64 = 4
The surface area of the cube, in square feet, is 6 times the area of one face: 6(42) = 96
The surface area of the cube was 96 square feet.
10. The surface area of the cube is 6534 square feet.
6534
The area of one face, in square feet, is:
= 1089
6
The side length of a face, in feet, is 1089 .
1089 = 3 · 3 · 11 · 11
= (3 · 11)(3 · 11)
= 33 · 33
1089 = 33
The volume of the cube, in cubic feet, is: 333 = 35 937
The volume of the cube is 35 937 cubic feet.
11. If 2000 is a perfect cube, then a cube could be constructed with 2000 interlocking cubes.
Determine the factors of 2000.
2000 = 2 · 2 · 2 · 2 · 5 · 5 · 5
Since these factors cannot be arranged in 3 equal groups, 2000 is not a perfect cube, and a cube could
not be constructed with 2000 interlocking cubes.
12. Use estimation or guess and check to determine the perfect square and perfect cube closest to the first
number in each given pair, then calculate the squares and cubes of all whole numbers until the second
number in each pair is reached or exceeded.
a) 315 – 390
172 = 289
182 = 324
192 = 361
202 = 400
3
3
3
6 = 216
7 = 343
8 = 512
Between 315 and 390, the perfect squares are 324 and 361; and the perfect cube is 343.
b) 650 – 750
252 = 625
262 = 676
272 = 729
282 = 784
3
3
3
8 = 512
9 = 729
10 = 1000
Between 650 and 750, the perfect squares are 676 and 729; and the perfect cube is 729.
c) 800 – 925
282 = 784
292 = 841
302 = 900
312 = 961
3
3
9 = 729
10 = 1000
Between 800 and 925, the perfect squares are 841 and 900; there are no perfect cubes.
d) 1200 – 1350
342 = 1156
352 = 1225
362 = 1296
372 = 1369
3
3
3
10 = 1000
11 = 1331
12 = 1728
Between 1200 and 1350, the perfect squares are 1225 and 1296; and the perfect cube is
1331.
13. For a number to be a perfect square and a perfect cube, its prime factors must be arranged in 2 equal
groups and 3 equal groups; that is, each factor occurs 2(3), or 6 times, or a multiple of 6 times.
The first number, after 0 and 1, that is a perfect square and perfect cube is:
2 · 2 · 2 · 2 · 2 · 2 = 64
Another number that is a perfect square and a perfect cube is: 3 · 3 · 3 · 3 · 3 · 3 = 729
Another number that is a perfect square and a perfect cube is: 5 · 5 · 5 · 5 · 5 · 5 = 15 625
14. Since the rectangular prism has a square cross-section, the prism is a square prism. Its volume is 1440
cubic feet.
1440
Its height is 10 ft., so its base area, in square feet, is:
= 144
10
The side length of the square base, in feet, is: 144 = 12
So, the length and width of the base are 12 ft.
C
15. a) The tent has 4 congruent square faces; 2 congruent rectangular faces, and 2 congruent
triangular faces.
The area, in square feet, of each square face is: (x)(x) = x2
2
 5x  5x
The area, in square feet, of each rectangular face is: (x)   =
8
 8 
To determine the area of each triangle, first determine its height, h.
Use the Pythagorean Theorem in ΔABC.
2
 5x 
x
h2 =   –  
2
 8 
2
25 x
x2
h2 =
–
64
4
2
25 x 2
16 x 2
–
64
64
2
9x
h2 =
64
h2 =
9 x2
64
3x
h=
8
h=
The area, in square feet, of each triangular face is:
3x 2
1
3x
 x    =
16
2
 8 
So, the surface area of the tent is:
 5x2 
 3x 2 
10 x 2
3x 2
2
4(x2) + 2 
+
 + 2
 = 4x +
8
8
 8 
 16 
2
2
32 x
13 x
=
+
8
8
45 x 2
=
8
2
45 x
The surface area of the tent is
square feet.
8
b) Write an equation.
45 x 2
= 90
Multiply each side by 8.
8
45x2 = 720
Divide each side by 45.
x2 = 16
x = 16
x=4
16. Let the edge length of the cube be x.
Then the area of one face is x2.
And its surface area is 6x2.
The volume of the cube is x3.
The volume and surface area are equal, so:
x3 = 6x2
Since x is not equal to 0, divide each side by x2.
x=6
The dimensions of a cube that has its surface area numerically the same as its volume are 6 units by 6
units by 6 units.
17. a) The side length of a square with area 121x4y2 is: 121x 4 y 2
Factor 121x4y2: 11 · 11 · x · x · x · x · y · y
Rearrange the factors in 2 equal groups: (11 · x · x · y)(11 · x · x · y)
So, 121x 4 y 2 = 11x2y
The side length of the square is 11x2y.
b) The edge length of a cube with volume 64x6y3 is: 3 64x 6 y 3
Factor 64x6y3: 2 · 2 · 2 · 2 · 2 · 2 · x · x · x · x · x · x · y · y · y
Rearrange the factors in 3 equal groups:
(2 · 2 · x · x · y)(2 · 2 · x · x · y)(2 · 2 · x · x · y)
So, 3 64x 6 y 3 = 4x2y
The edge length of the cube is 4x2y.
18. List all the perfect cubes up to a number close to 1729:
1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728
Use guess and check.
The 1st and 12th perfect cubes have the sum: 1 + 1728 = 1729
The 9th and 10th perfect cubes have the sum: 729 + 1000 = 1729
No other perfect cubes have the sum 1729.