Name: ________________________ Class: ___________________ Date: __________
Unit 1 Square Roots and Pythagrean Theorem
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. What is
a.
b.
c.
d.
36 ?
5
18
6
9
____
2. Between which pair of whole numbers is
a. 7 and 8
b. 49 and 64
c. 36 and 64
d. 6 and 7
____
3. What is the length of PQ ?
a.
b.
12 cm
25.6 cm
61 located?
c.
d.
1
144 cm
656 cm
ID: A
Name: ________________________
____
4. The areas of the squares on the two shorter sides of a triangle are given. What is the area of the shaded
square, in cm2?
a.
b.
c.
d.
____
ID: A
18
58
18
58
5. What is c in metres to one decimal place?
a.
b.
c.
d.
13.7
7.5
19.0
13.6
2
Name: ________________________
ID: A
Short Answer
6. The fence around a square field has a perimeter of 350 m.
What is the area of the field?
7. The string of a kite is tied to a stake in the ground. How long is the string? Round your
1 decimal place. [2 Marks]
3
answer to
Name: ________________________
ID: A
8. Determine the length of AD to one decimal place. [3 Marks]
4
ID: A
Unit 1 Square Roots and Pythagrean Theorem
Answer Section
MULTIPLE CHOICE
1. ANS: C
6 × 6 = 36 so
36 = 6
Feedback
A
B
C
D
Calculated 25 = 5.
Divided 36 by 2.
Correct!
Divided 36 by 4.
PTS: 1
TOP: 1.2
2. ANS: A
DIF: 2M
REF: CG p 38
OBJ: N1
MSC: Saquare Roots and the Pythagorean Theorem
61 is between the perfect squares 49 and 64. Since
49 = 7 and
64 = 8 then
whole numbers 7 and 8.
Feedback
C
Correct!
These are the perfect squares (square numbers)
These are perfect squares and 36 isn’t the closest!
D
6 is
A
B
PTS: 1
TOP: 1.4
36 ,
49 is closer.
DIF: 2A
REF: CG p.40
OBJ: N2
MSC: Square Roots and the Pythagorean Theorem
1
61 is between the
ID: A
3. ANS: A
Use the Pythagorean Theorem a 2 + b 2 = c 2 . c 2 , the area of the square on the hypotenuse is given:
400cm2 . b 2 , the area of the square on one of the legs is given : 256cm2
Use, therefore: c 2 − b 2 = a 2
c2 − b2 = a2
400 − 256 = a 2
144 = a 2
a=
144
a = 12cm
Feedback
A
B
C
D
Correct!
Did not switch the formula, used a 2 + b 2 = c 2
Did not take the square root.
Added the areas of the squares and did not take the square root.
PTS: 1
TOP: 1.5
4. ANS: D
DIF: 2A
REF: CG p.44
OBJ: SS1
MSC: Square Roots and the Pythagorean Theorem
The area of the square on the hypotenuse of a right triangle is equal to the sum of the areas of the
squares on the other two sides.
Feedback
A
B
C
D
You must add, not subtract!
This would give the side length, not the area.
This is the side length if you subtracted instead of added.
Correct!
PTS: 1
TOP: 1.5
DIF: 2M
REF: CG p 44-47 OBJ: 8SS1
MSC: Square Roots and the Pythagorean Theorem
2
ID: A
5. ANS: D
a2 + b2 = c2
11 2 + 8 2 = c 2
121 + 64 = c 2 '
185 = c 2
c=
185 = 13.601. . . ≈ 13.6
Feedback
A
B
C
D
Rounded 13.601 up to 13.7 when it should round down.
Subtracted the squares on the legs instead of adding.
Never use Pythagorean Theorem, just added the two legs.
Correct!
PTS: 1
TOP: 1.5
DIF: 2A
REF: CG 48
OBJ: SS1
MSC: Square Roots and the Pythagorean Theorem
SHORT ANSWER
6. ANS:
7656.3 m2
The perimeter of a square is 4 × sides. Therefore, if we divide 350 by 4 we will determine the length of each
side 350 ÷ 4 = 87.5m.
Next, to find area, use the formula A = s 2
A = 87.5 2 = 7656.3m2
The area is 7656.3 m2
PTS: 2
TOP: 1.1
7. ANS:
c2 = a2 + b2
DIF: 2A
REF: CG p.30
OBJ: N1
MSC: Square Roots and the Pythagorean Theorem
c 2 = 39 2 + 26 2
c 2 = 1521 + 676
c 2 = 2197
c = 46.9
PTS: 2
TOP: 1.7
DIF: 2A
REF: CGp48
OBJ: SS1
MSC: Square Roots and the Pythagorean Theorem
3
ID: A
8. ANS:
In order to calculate AD we need to first calculate AC using the triangle on the left.
2
2
AC + BC = AB
2
2
AC + 2 2 = 6 2
2
AC + 4 = 36
2
AC = 36 − 4
2
AC = 32
AC =
32 = 5.7 m
We can now calculate AD
2
2
AC + CD = AD
2
(5.7) + 8 2 = AD
32 + 64 = AD
2
2
2
2
AD = 96
AD =
96 ≈ 9.8 m
PTS: 3
TOP: 1.7
DIF: 3
REF: CG p 48
OBJ: SS1
MSC: Square Roots and the Pythagorean Theorem
4