Download Grade 8 Square and Square Roots

ID : ru-8-Square-and-Square-Roots [1]
Grade 8
Square and Square Roots
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Answer t he quest ions
(1)
If you subtract a number x f rom 41 times that number, and then take the square of the
dif f erence, what will the result be?
(2)
What is the value of
(3)
What is the smallest number that 72 must be multiplied with to get a perf ect square?
.
Choose correct answer(s) f rom given choice
(4) Which is the smallest number that can be multiplied to 20592 to give a perf ect square?
(5)
a. 141
b. 39
c. 33
d. 143
Which of the f ollowing numbers is a perf ect square?
a. 1575
b. 1225
c. 980
d. 700
(6) What is the square root of 0.0225?
a. ±0.015
b. ±0.45
c. ±0.15
d. ±3.15
(7) For any given number x, which of the f ollowing can be pythagorean triplets?
(8)
a. 2m, m2-1,m2+1
b. 2m+1, m2-1,m2+1
c. 2m, m2,m2+1
d. 2m, m2-1,m2
What is the expansion of (a - b)2?
a. a2 + 2ab + b 2
b. a2 + 2ab + b
c. a2 - 2ab + b 2
d. a2 - ab + b 2
Fill in t he blanks
(9)
(10)
T he smallest perf ect square that can be divided by 77 and 14 is
T he sum of the f irst 43 odd numbers is
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ID : ru-8-Square-and-Square-Roots [2]
(11)
T he smallest f raction that must be added to the square root of 8
8
so that the resulting
49
number is a positive whole number is
(12) T here are two squares, and the ratio of their perimeters is 7:10
If the area of the f irst square is 1225 m2, then the area of the second is
m2.
(13) T here are two numbers such that the dif f erence of their squares is 155 and sum of the
numbers is 31, their dif f erence =
(14)
(15)
.
If the square root of 2 is 1.41, then the square root of 200 is
T he square root of 32 correct to one place of decimal is
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ID : ru-8-Square-and-Square-Roots [3]
Answers
(1)
1600x2
Step 1
According to the question,
f irst number is x and
second number is 41 x.
Step 2
Dif f erence of two numbers = (41 x - x)
Square of the dif f erence = (41 x - x)2
= (40 x)2
= 1600 x2
Step 3
T heref ore the result will be 1600x2.
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ID : ru-8-Square-and-Square-Roots [4]
(2)
±11
7
Step 1
Let's f irst f ind prime f actors of 242.
242 = 11 × 11 × 2
Step 2
Similarly f ind prime f actors of 98.
98 = 7 × 7 × 2
Step 3
Now,
=
√(11 × 11 × 2)
√(7 × 7 × 2)
=
√(11 × 11 × 2)
√(7 × 7 × 2)
=
√(11)2
√(7)2
=
±11
7
Step 4
T heref ore, the value of
is
±11
.
7
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ID : ru-8-Square-and-Square-Roots [5]
(3)
2
Step 1
We have been asked to f ind the smallest number that 72 must be multiplied with to get a
perf ect square.
Step 2
Perf ect square is a number made by squaring a whole number.Steps to f ind the perf ect
square are as f ollows.
i. Resolve the given number into prime f actors.
ii. Make pairs of prime f actors such that both the f actors in each pair are equal.
iii. If there is nothing lef t with the pair of prime numbers so the number is perf ect square
otherwise not.
Step 3
Prime f actor of 72 are 2 × 2 × 2 × 3 × 3
Step 4
Lets identif y pairs of f actors,
(2 × 2) × (3 × 3) × 2
Step 5
T his number is not the perf ect square.T here is a prime f actor (2) lef t with the pair of prime
f actors.T o obtain the perf ect square we have to multiply the given number with the 2.
Step 6
72 × 2 = 144
[Prime f actor of 144 are 2 × 2 × 2 × 2 × 3 × 3, theref ore it is the perf ect square.]
Step 7
T heref ore, 2 is the smallest number that 72 must be multiplied with to get the perf ect
square.
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ID : ru-8-Square-and-Square-Roots [6]
(4) d. 143
Step 1
Perf ect square is a number made by squaring a whole number. Steps to check if a number is
perf ect square are as f ollows.
i. Resolve the given number into prime f actors.
ii. Make pairs of prime f actors such that both the f actors in each pair are equal.
iii. If there is nothing lef t without the pair of prime numbers so the number is perf ect
square otherwise not.
Step 2
Prime f actors of 20592 = 2 × 2 × 2 × 2 × 3 × 3 × 11 × 13
Step 3
Lets' create pairs of prime f actors,
20592 = (2 × 2) × (2 × 2) × (3 × 3) × 11 × 13
Step 4
T his number is not the perf ect square, since all f actors are not paired. If we multiply this
number with remaining prime f actors (i.e. 11 × 13 = 143), all prime f actors will be paired and
number will be perf ect square.
Step 5
T heref ore, 143 is the smallest number that can be multiplied to 20592 to give the perf ect
square.
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ID : ru-8-Square-and-Square-Roots [7]
(5)
b. 1225
Step 1
Perf ect square is a number made by squaring a whole number.
Steps to f ind the perf ect square are as f ollows.
i. Resolve the given number into prime f actors.
ii. Make pairs of prime f actors such that both the f actors in each pair are equal.
iii. If there is nothing lef t without the pair of prime numbers so the number is perf ect
square otherwise not.
Let's check all given numbers
Step 2
Prime f actors of 1575,
1575 = 5 × 5 × 3 × 3 × 7
Now create pairs,
1575 = (5 × 5) × (3 × 3) × 7
Since all f actors are not paired, this is not a perf ect square.
Step 3
Prime f actors of 1225,
1225 = 5 × 5 × 7 × 7
Now create pairs,
1225 = (5 × 5) × (7 × 7)
Since all f actors are paired, this is a perf ect square.
Step 4
Prime f actors of 980,
980 = 2 × 2 × 7 × 7 × 5
Now create pairs,
980 = (2 × 2) × (7 × 7) × 5
Since all f actors are not paired, this is not a perf ect square.
Step 5
Prime f actors of 700,
700 = 5 × 5 × 2 × 2 × 7
Now create pairs,
700 = (5 × 5) × (2 × 2) × 7
Since all f actors are not paired, this is not a perf ect square.
Step 6
T heref ore correct answer is 1225.
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ID : ru-8-Square-and-Square-Roots [8]
(6) c. ±0.15
Step 1
225
Let's write 0.0225 as:
10000
Step 2
Let's f ind the square root of
225
10000
=
√ 225
√ 10000
=
√ 15 × 15
√ 100 × 100
=
√(15)2
√(100)2
=
±15
±100
= ±0.15
Step 3
T hus the square root of 0.0225 is ±0.15.
(7) a. 2m, m2-1,m2+1
(8)
c. a2 - 2ab + b 2
Step 1
If you look at the question caref ully, you will notice that you have to f ind out the expansion
of (a - b)2 .
Step 2
Now,
(a - b)2 = (a - b)(a - b)
= (a - b)a - (a - b)b
= a2 -ab - ab - (-b)b
= a2 - 2ab + (b)b
= a2 - 2ab + b 2
Step 3
T heref ore the expansion of (a - b)2 is a2 - 2ab + b 2 .
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ID : ru-8-Square-and-Square-Roots [9]
(9)
23716
Step 1
According to the question, we have to f ind the smallest perf ect square that can be divided
by 77 and 14.
Step 2
Perf ect square is a number made by squaring a whole number.Steps to f ind the perf ect
square are as f ollows.
i. Resolve the given number into prime f actors.
ii. Make pairs of prime f actors such that both the f actors in each pair are equal.
iii. If there is nothing lef t with the pair of prime f actors so the number is perf ect square
otherwise not.
Step 3
Prime f actors of 77 and 14 are 7 × 11 × 2.
T o make perf ect square the prime f actors must f ollow the condition of perf ect square.
Prime f actors of smallest perf ect square = 7 × 11 × 2 × 7 × 11 × 2
= 11 × 11 × 7 × 7 × 2 × 2
= 23716
Step 4
T heref ore, the smallest perf ect square that can be divided by 77 and 14 is 23716.
(10)
1849
Step 1
If you look at the question caref ully, you will notice that you have to f ind out the value of
the sum of the f irst 43 odd numbers.
Step 2
Sum of f irst n odd numbers = n 2
(where, n = 43)
= (43)2
= 1849
Step 3
T heref ore the sum of f irst 43 odd numbers is 1849.
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(11)
1
7
Step 1
Let's f irst convert mixed f raction to improper f raction
8
8
=
49
400
49
Step 2
Now,
=
√400
√49
⇒ =
±20
±7
⇒ =
20
.......... [ Use only positive number since question is about positive whole
7
number]
⇒ =2+
6
7
Step 3
Now f or above number to be a whole number, we just need to f ind a f raction that must me
added to 6/7, such that sum if 1. T heref ore required f raction to be added,
=1-
6
7
=
1
7
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ID : ru-8-Square-and-Square-Roots [11]
(12)
2500
Step 1
Let us assume the side of f irst square is a.
Let us assume the side of second square is b.
Perimeter of f irst square = 4a
Perimeter of second square = 4b
Ratio of the perimeter = 7:10
Step 2
Area of the f irst square, a2 = 1225
⇒ a2 = 352
⇒ a = 35
Step 3
Now the ratio of perimeter of the squares,
4a
4b
⇒
7
=
10
35
=
b
7
10
⇒ 35 × 10 = 7 × b
⇒
350
=b
7
⇒ b = 50 m
Step 4
Area of the second square, b2 = (50)2
= 50 × 50
= 2500 m2
Step 5
T heref ore, the area of the second square is 2500.vm2
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ID : ru-8-Square-and-Square-Roots [12]
(13)
5
Step 1
Let us assume x and y are the two numbers.
Step 2
It is given that, the dif f erence of their squares is 155, that is, x2 - y2 = 155 ----(1)
Step 3
It is also given that sum of the numbers is 31, that is, x + y = 31 ----(2)
Step 4
We know that,
x2 - y2 = (x + y)(x - y)
Step 5
On substituting values f rom Eq. (1) and (2),
155 = 31 × (x - y)
x - y = 155/31
x- y=5
Step 6
T heref ore, their dif f erence is 5.
(14)
14.1
Step 1
We have been asked to f ind the square root of 200 if square root of 2 is 1.41.
Step 2
Now
√(200) = √(100 × 2)
=√((10)2 × 2)
= 10 × √(2) = 10 × 1.41
= 14.1
Step 3
T heref ore, square root of 200 is 14.1 if square root of 2 is 1.41.
(15)
5.7
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