Download Class 8 Square and Square Roots

ID : in-8-Square-and-Square-Roots [1]
Class 8
Square and Square Roots
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Answer t he quest ions
(1)
Which is the smallest number that can be multiplied to 10400 to give a perf ect square?
(2)
What is the value of
(3)
T he total population of a village is a perf ect square. T he number of men in the village is 39808,
and the number of women in the village is 51012.
If we know that the number of children in the village is also a perf ect square and is more than
20730, then what is the smallest possible number of children in the village?
?
(4) What is the smallest number that must be added to 295 to get a perf ect square?
(5)
Find the smallest perf ect square number that is larger than the smallest 4 digit number?
(6) What is the square root of 0.0361?
(7) In a school auditorium, the number of chairs in a row is the same as the number of rows of
chairs. T here are 3948 students in the school. If we know that less than
22 chairs were empty, how many chairs were empty af ter all the students sat down?
(8)
Students f rom grade 3 class decided to collect money f or earthquake relief . T he amount (in Rs.)
collected by each students equals to number of students in the class, If they collected a total of
Rs.3364, how many students were there in the class?
(9) What is the expansion of (a - b)2?
Choose correct answer(s) f rom given choice
(10) Which of the f ollowing statements is always true:
a. A square number cannot be the sum of two
b. A number that has 5 in the units place can
other square numbers
never be a perf ect square
c. A number that leaves a remainder of 3 when d. A number that has 7 in the units place may
divided by 4 cannot be a perf ect square
be a perf ect square
(11) Which of the f ollowing numbers cannot be the area of a square that has an integer value side?
a. 774
b. 361
c. 1521
d. 784
(12) What is the smallest number that 200 must be multiplied with to get a perf ect square?
a. 0
b. 4
c. 2
d. 5
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ID : in-8-Square-and-Square-Roots [2]
(13) For any given number x, which of the f ollowing can be pythagorean triplets?
a. 2m+1, m2-1,m2+1
b. 2m, m2-1,m2+1
c. 2m-1, m2-1,m2
d. m, m2,m2+1
Fill in t he blanks
(14) T here are two squares, and the ratio of their perimeters is 6:17
If the area of the f irst square is 144 m2, then the area of the second is
(15)
m2.
T he smallest even number whose square is larger than 2363 is
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ID : in-8-Square-and-Square-Roots [3]
Answers
(1)
26
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 10400 = 2 × 2 × 2 × 2 × 2 × 5 × 5 × 13
Step 3
Lets' create pairs of prime f actors,
10400 = (2 × 2) × (2 × 2) × (5 × 5) × 2 × 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. 2 × 13 = 26), all prime f actors will be paired and
number will be perf ect square.
Step 5
T heref ore, 26 is the smallest number that can be multiplied to 10400 to give the perf ect
square.
(2)
25
49
Step 1
We have been asked to f ind the value of
.
Step 2
Now,
=
5×5
7 ×7
=
25
49
Step 3
T heref ore, the value of
is
25
.
49
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ID : in-8-Square-and-Square-Roots [4]
(3)
20736
Step 1
T otal population of the village is a perf ect square.
Number of men in the village = 39808
Number of women in the village = 51012
Step 2
It is given that the number of children in the village is also a perf ect square, and is more
than 20730. Let us f irst f ind the values of perf ect squares more than 20730.
Step 3
Square root of 20730 = 143.97916515941 and the perf ect squares more than 20730 will be
the squares of all integers more than 143.97916515941, ie 144 2, 1452, 1462, etc.
Step 4
T he population of children will be the smallest value out of 144 2, 1452, 1462, etc, f or which
the total population of the village is a perf ect square.
Step 5
Let us see if the population of the children can be 144 2 (= 20736). In this case, the total
population of the village = 39808 + 51012 + 144 2
= 39808 + 51012 + 20736
= 111556
Since the square root of 111556 is 334, which is an integer, the total population of the
village in this case is indeed a perf ect square.
Step 6
T heref ore, the smallest possible number of children in the village is 144 2 = 20736.
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ID : in-8-Square-and-Square-Roots [5]
(4) 29
Step 1
Let's f irst f ind the square root of 295 using division method,
17.1
1 2 95 . 00
+1 1
27
195
+7
189
341
600
+1
341
259
T his shows that square root of 295 is between 17 and 18
Step 2
Now, square of 18,
182 = 18 × 18 = 324
Step 3
T heref ore, number that must be added to 295, to get perf ect square,
= 324 - 295
= 29
(5)
1024
Step 1
T he smallest 4 digit number is 1000. We need to f ind the smallest perf ect square number
that is larger than 1000.
Step 2
Let us f irst look at the square root of 1000. T he square root of 1000 = 31.62278
T he smallest perf ect square number larger than 1000 must be the square of the integer just
larger than 31.62278.
Step 3
T he integer just larger than 31.62278 is 32 and the square of 32 = (32)2
= 1024
Step 4
T heref ore, the smallest perf ect square number that is larger than the smallest 4 digit
number is 1024.
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ID : in-8-Square-and-Square-Roots [6]
(6) ±0.19
Step 1
361
Let's write 0.0361 as:
10000
Step 2
Let's f ind the square root of
361
10000
=
√ 361
√ 10000
=
√ 19 × 19
√ 100 × 100
=
√(19)2
√(100)2
=
±19
±100
= ±0.19
Step 3
T hus the square root of 0.0361 is ±0.19.
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ID : in-8-Square-and-Square-Roots [7]
(7) 21
Step 1
We have been told that the number of chairs in a row is the same as the number of rows of
chairs.
It means if there are n chairs in a row, then the number of rows also is n. T hat is, the total
number of chairs makes a perf ect square.
Step 2
T otal number of students in the school = 3948
Square root of 3948 = 62.83311228962
Since the number of chairs is a perf ect square, the number of chairs in a row is an integer
more than 62.83311228962.
Step 3
T he integers more than 62.83311228962 are 63, 64, 65 and so on. How do we tell how
many chairs are there in a row out of 63, 64, 65 and so on? In order to f ind that, let us
make use of the f act that there were less than 22 chairs empty af ter all students were
seated.
Step 4
T he number 22 is less than 62.83311228962, the square root of 3948. T his means, the
number of chairs in a row should be the integer just greater than 62.83311228962. T his
means, the number of chairs in a row is 63 and the total number of chairs is 632 = 3969.
Step 5
T otal number of chairs = 3969
Number of students = 3948
Number of chairs lef t = T otal number of chairs - Number of students
= 3969 - 3948
= 21
Step 6
T heref ore, there were 21 chairs lef t in the auditorium of the school af ter all the students
sat down.
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ID : in-8-Square-and-Square-Roots [8]
(8)
58
Step 1
Let us assume the number of students is x.
Step 2
According to the question, if there are x students and each student collected Rs.x.
So, total collected money = x2.
Step 3
Collected money by the students = Rs.3364
x2 = 3364
⇒ x = √3364
⇒ x = 58 (ignore negative since number of students cannot be negative)
Step 4
T heref ore, there were 58 students in the class.
(9) 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 .
(10) c. A number that leaves a remainder of 3 when divided by 4 cannot be a perf ect square
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ID : in-8-Square-and-Square-Roots [9]
(11) a. 774
Step 1
We know the area of square = side2
Since it is given that side has an integer value, the area must be a perf ect square.
Let's check if given areas are perf ect square or not
Step 2
Let's check if 1st area of 774 is perf ect square or not using division method,
27.8
2 7 74 . 00
+2 4
47
374
+7
329
548
4500
+8
4384
116
774 is not a perf ect square, theref ore it cannot be the are of given square
Step 3
Let's check if 2nd area of 361 is perf ect square or not using division method,
19
1 3 61
+1 1
29
261
+9
261
0
361 is a perf ect square, theref ore it could be the are of given square
Step 4
Let's check if 3rd area of 1521 is perf ect square or not using division method,
39
3 15 21
+3 9
69
621
+9
621
0
1521 is a perf ect square, theref ore it could be the are of given square
Step 5
Let's check if 4 th area of 784 is perf ect square or not using division method,
28
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ID : in-8-Square-and-Square-Roots [10]
2 7 84
+2 4
48
384
+8
384
0
784 is a perf ect square, theref ore it could be the are of given square
Step 6
T heref ore, 774 cannot be the area of given square.
(12) c. 2
Step 1
We have been asked to f ind the smallest number that 200 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 200 are 2 × 2 × 2 × 5 × 5
Step 4
Lets identif y pairs of f actors,
(2 × 2) × (5 × 5) × 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
200 × 2 = 400
[Prime f actor of 400 are 2 × 2 × 2 × 2 × 5 × 5, theref ore it is the perf ect square.]
Step 7
T heref ore, 2 is the smallest number that 200 must be multiplied with to get the perf ect
square.
(13) b. 2m, m2-1,m2+1
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ID : in-8-Square-and-Square-Roots [11]
(14)
1156
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 = 6:17
Step 2
Area of the f irst square, a2 = 144
⇒ a2 = 122
⇒ a = 12
Step 3
Now the ratio of perimeter of the squares,
4a
4b
⇒
6
=
17
12
6
=
b
17
⇒ 12 × 17 = 6 × b
⇒
204
=b
6
⇒ b = 34 m
Step 4
Area of the second square, b2 = (34)2
= 34 × 34
= 1156 m2
Step 5
T heref ore, the area of the second square is 1156.vm2
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ID : in-8-Square-and-Square-Roots [12]
(15)
50
Step 1
We have been asked to f ind the smallest even number whose square is larger than 2363.
Step 2
First of all we have to f ind the square root of 2363.
48.6
4 23 63 . 00
+ 4 16
88
763
+8
704
966
5900
+6
5796
104
Step 3
We can notice that square root of 2363 is greater than 48 and less than 49.
Step 4
T heref ore, the smallest even number whose square is larger than 2363, will be an even
number which is larger than 48. T hus, answer is 50.
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