Examples for Square root of a perfect square by factorization
Example 1: Find the square root of 5929.
Solution:
Thus, 7 x 7 x 11 x 11 = 5929
Therefore, [group it]
(7 x 11) x (7 x 11) = 77 x 77 = 772 = 5929.
Example 2: Find the square root of 6724.
Solution:
Thus, 6724 = 2 x 2 x 41 x 41
6724 = (2 x 41) x (2 x 41)
= 82 x 82
Example 3: Find the smallest positive integer with which one has to divide 336 to get a perfect square.
Solution:
We observe that 336 = 2 x 2 x 2 x 2 x 3 x 7. Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.
We divide 336 by 3 and 7
42.
The required least number is 21.
Square root of a perfect square by factorization – Exercise 1.2.5
- Find the square roots of the following numbers by factorization:
i) 196
Solution:
196 = 2 x 2 x 7 x 7
= (2 x 7) x (2 x 7)
= 14 x 14
196 = 142
ii) 256
Solution:
256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
= (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2)
= 16 x 16
= 162
iii) 10404
Solution:
10404 = 2 x 2 x 3 x 3 x 17 x 17
= (2 x 3 x 17) x (2 x 3 x 17)
= (102) x (102)
= 1022
iv) 1156
Solution:
1156 = 2 x 2 x 17 x 17
= (2 x 17) x (2 x 17)
= 34 x 34
= 342
v) 13225
Solution:
13225 = 5 x 5 x 23 x 23
= (5 x 23) x (5 x 23)
= 115 x 115
= 1152
2. Simplify:
i) √100 + √36
Solution:
√100 + √36 = 10 + 6
= 16
ii) √(1360 + 9)
Solution:
√(1360 + 9) = √1369 = √(37 x 37) = 37
iii) √2704 + √144 + √289
Solution:
√2704 + √144 + √289 = √(52 x 52) + √(12 x 12) + √(17 x 17)
= 52 + 12 + 17
= 81
iv) √225 – √25
Solution:
√225 – √25 = √(15 x 15) – √(5 x 5)
= 15 – 5
= 10.
V) √1764 – √1444
Solution:
√1764 – √1444 = √(42 x 42) – √(38 x 38)
= 42 – 38
= 4
vi) √169 x √361
√169 x √361 = √(13 x 13) x √(19 x 19)
= 13 x 19
= 247
- A square yard has area 1764m2. From a corner of this yard, other square part of area 784m2 is taken out for public utility. The remaining portion is divided in to 5 equal parts. What is the perimeter of each of these equal parts?
Solution:
Area of square = a2
Area of given square yard = 1764 m2
Area used for public utility = 784 m2
∴ Area of remaining portion = 1764 – 784 = 980 m2
If the area is divided into 5 parts
Then, the area of each Square = 980/5 = 196 m2
Length of each side = √ 196, a = 14m
Perimeter of square = 4a = 4 x 14 = 56m
4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:
i) 847
Solution:
847 = 11 x 11 x 7
Here 7 occur only once. Hence we have to multiply by 7 them to get a perfect square.
ii) 450
Solution:
450 = 5 x 5 x 3 x 3 x 2
Here 2 occur only once. So we have multiply by 2 to get a perfect square.
iii) 1445
Solution:
1445 = 17 x 17 x 5
Here 5 occur only once. So we have multiply by 5 to get a perfect square.
iv) 1352
Solution:
1352 = 2 x 2 x 13 x 13 x 2
Here 2 occur only three times. So we have multiply by 2 to get a perfect square.
5. Find the largest perfect square factor of each of the following numbers:
i) 48
Solution:
48 = 2 x 2 x 2 x 2 x 3 = (2 x 2) x (2 x 2) x 3
= 4 x 4 x 3 = 16 x 3 = 42 + 3
Therefore, 16 is the largest perfect square factor of 48.
ii) 11280
Solution:
11280 = 2 x 2 x 2 x 2 x 3 x 47
Therefore, 16 is the largest perfect square factor of 11280
iii) 729
Solution:
729 = 3 x 3 x 3 x 3 x 3 x 3
= (27) x (27)
= 272
Therefore, 729 is the largest perfect square factor of 729.
iv) 1352
Solution:
1352 = 2 x 2 x 2 x 13 x 13
= (2 x 13) x (2 x 13) x 2
= 26 x 26 x 2
Therefore, 676 is the largest perfect square of 1352.
- Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?
Solution:
The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….
1) 48 + 1 = 49 = 72
2 ) 192 + 4 = 196 = 142
3)288 + 1 = 289 = 172
4 ) 96 + 4 = 100 = 102
5)96 + 48 = 144 = 122
6) 48 + 16 = 64 = 82
7)240 + 16 = 256 = 162
Yes, we can prove that there are infinitely many such fairs.
Squares, Square roots, cube and cube roots
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