The properties of perfect squares are explained here in each property with examples.
Numbers ending in 2, 3, 7 or 8 is never a perfect square but on the other hand, all the numbers ending in 1, 4, 5, 6, 9, 0 are not square numbers.
For example:
The numbers 10, 82, 93, 187, 248 end in 0, 2, 3, 7, 8 respectively.
So, none of them is a perfect square.
A number ending in an odd number of zeros is never a perfect square.
For example:
The numbers 160, 4000, 900000 end in one zero, three zeros and five zeros respectively.
So, none of them is a perfect square.
The square of an even number is always even.
For example:
2² = 4, 4² = 16, 6² = 36, 8² = 64, etc.
The square of an odd number is always odd.
For example:
1² = 1, 3² = 9, 5² = 25, 7² = 49, 9² = 81, etc.
The square of a proper fraction is smaller than the fraction.
For example:
(2/3)² = (2/3 × 2/3) = 4/9 and 4/9 < 2/3, since (4 × 3) < (9 × 2).
For every natural number n, we have
(n + 1)²  n² = (n + 1 + n)(n + 1  n) = {(n + 1) + n}.
Therefore, {(n + 1)²  n²} = {(n + 1) + n}.
For example:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8²
For every natural number n, we have
sum of the first n odd numbers = n²
For example:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8²
Three natural numbers m, n, p are said to form a Pythagorean triplet (m, n, p) if (m² + n²) = p².
Note:
For every natural number m > 1, we have (2m, m² – 1, m² + 1) as a Pythagorean triplet.
For example:
(i) Putting m = 4 in (2m, m² – 1, m² + 1) we get (8, 15, 17) as a Pythagorean triplet.
(ii) Putting m = 5 in (2m, m² – 1, m² + 1) we get (10, 24, 26) as a Pythagorean triplet.
1. Without adding, find the sum (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17).
Solution:
(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17) = sum of first 9 odd numbers = 9² = 81
2. Express 49 as the sum of seven odd numbers.
Solution:
49 = 7² = sum of first seven odd numbers
= (1 + 3 + 5 + 7 + 9 + 11 + 13).
3. Find the Pythagorean triplet whose smallest member is 12.
Solution:
For every natural number m > 1. (2m, m² – 1, m² + 1) is a Pythagorean triplet.
Putting 2m = 12, i.e., m = 6, we get the triplet (12, 35, 37).
● Square
Perfect Square or Square Number
● Square  Worksheets
8th Grade Math Practice
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