Squares, Square Roots, & BEDMAS Quiz Review
Part 1: Squares & Square Roots
1. Find the area of a square with each side length:
a) 8 cm- 8 × 8 = 64 cm2
b) 10 cm- 10 × 10 = 100 cm2
c) 3 cm- 3 × 3 = 9 cm2
2.
a)
b)
c)
Find the side length of a square with each area:
100 cm2- √100 = 10 cm
64 cm2- √64 = 8 cm
4 cm2- √4 = 2 cm
3. You want to carpet a square room that has an area of 144 metres squared. What is the length of one side
of the room? What is the perimeter of the room?
Given: Area = 144 m2
Required: One side length and room perimeter
To find side length: √144 = 12 m
The length of one side of the room is 12 m.
To find perimeter: Square has 4 sides. I know the length of one side is 12 m. I have to add up 12 4 times or
multiply 12 by 4 to find the perimeter (all 4 sides).
12 × 4 = 48 m.
The perimeter of the room is 48 m.
4. You have a square blanket that has an area of 1600 in 2. What is the length of 1 side of the blanket? What
is the perimeter of the blanket?
Given: Area = 1600 in2
Required: One side length and blanket perimeter
To find side length: √1600 = 40 in
The length of one side of the blanket is 40 in.
To find perimeter: Square has 4 sides. I know the length of one side is 40 in. I have to add up 40 4 times or
multiply 40 by 4 to find the perimeter (all 4 sides).
40 × 4 = 160 in.
The perimeter of the blanket is 160 in.
5. A square place mat has an area of 400 cm2. Aniesa decides to decorate the place mat by putting fringe
around the edges. How many centimetres of fringe will Aniesa need to buy?
Given: Area = 400 cm2
Required: perimeter of the place mat (centimetres around the
edges)
To find perimeter: First, I need to know the length of one side, so that I can add the 4 sides up to find the
perimeter of the place mat. To find the length of one side, I will find the square root of 400 cm 2.
Side length: √400 = 20 cm
Perimeter: 20 × 4 = 80 cm
Aniesa will need to buy 80 cm of fringe to put around the edges of the place mat.
Part 2: BEDMAS
1. 26 ÷ 2
= 64 ÷ 2
= 32
4. 32 + 4 ÷ 2
=9+4÷2
=9+2
= 11
2. (4 + 9) × 1
= 13 × 1
= 13
5. (5 – 8 ÷ 8) × 4
= (5 – 1) × 4
=4×4
= 16
7. 6 + 2 × 5
= 6 + 10
= 16
10. 8 + 12 ÷ 4 – 2
=8+3–2
= 11 – 2
=9
8. (6 + 2) × 5
=8×5
= 40
11. (8 + 12) ÷ (4 – 2)
= 20 ÷ 2
= 10
13.
(4 + 3)2 × 2 – 1
= 72 × 2 – 1
= 49 × 2 – 1
= 98 – 1
= 97
14.
32 + 4 × 2
=9+4×2
=9+8
= 17
3. 5 ÷ 1 + 7
=5+7
= 12
6. 8 ÷ (4 – 2) ÷ 4 × 2
=8÷2÷4×2
=4÷4×2
=1×2
=2
9. 14 – 6 ÷ 2
= 14 – 3
= 11
12. 4 + 32 × 2 – 1
=4+9×2–1
= 4 + 18 – 1
= 22 – 1
= 21
15. (54 – 72)2
= (54 – 49)2
= 52
= 25
16. Debbie bought 3 pairs of earrings on sale. The normal price is $6 a pair but they are on sale for $1 off.
Write an expression to find the final cost of her purchase in dollars. Evaluate your expression to determine
the final cost of her purchase.
Given: 3 pairs of earrings. $6 = normal price. On sale for $1 off.
Required: expression to find
cost and calculate final cost
Expression: 3 × (6 – 1)
Evaluate: 3 × (6 – 1)
=3×5
= 15
The final cost of the 3 pairs of earrings was $15.
17. Tristan orders two hamburgers and a can of pop for lunch. A hamburger is $4 and a can of pop is $2. Write
an expression to show how much he paid for lunch. Then, find the value of the expression.
Given: 2 hamburgers, 1 can of pop. $4 for a hamburgers, $2 for a can of pop.
Required: expression to
show cost of lunch and solve/calculate final cost.
Expression: 2 × 4 + 1 × 2
Evaluate: 2 × 4 + 1 × 2
=8+1×2
=8+2
= 10
The cost for two hamburgers and a can of pop for lunch was $10.