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
From Properties of Perfect Squares to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 13, 24 10:24 AM
Nov 13, 24 09:23 AM
Nov 12, 24 01:36 PM
Nov 12, 24 12:07 PM
Nov 11, 24 02:08 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.