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Squares and Square Roots - Properties of Square Numbers

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A square number, also called a perfect square, is obtained by multiplying a number by itself, denoted as n2=n×nn^2 = n \times n. Visually, this can be represented as an n×nn \times n grid of dots where the total number of dots equals the square number.

The ending digit (unit's place) of a perfect square follows a specific pattern: it must end in 0,1,4,5,6,0, 1, 4, 5, 6, or 99. Numbers ending in 2,3,7,2, 3, 7, or 88 are never perfect squares. Furthermore, if a number ends in 11 or 99, its square ends in 11; if it ends in 44 or 66, its square ends in 66.

Regarding zeros at the end of a number, if a number ends in kk zeros, its square will end in 2k2k zeros. This implies that a perfect square always contains an even number of zeros at the end (e.g., 102=10010^2 = 100, 1002=10000100^2 = 10000).

Between the squares of two consecutive natural numbers n2n^2 and (n+1)2(n+1)^2, there are exactly 2n2n non-perfect square numbers. For example, between 22=42^2=4 and 32=93^2=9, there are 2×2=42 \times 2 = 4 non-square numbers (5,6,7,85, 6, 7, 8).

The sum of the first nn odd natural numbers is equal to n2n^2. This can be visualized as 'adding layers' to a square: 11 (a 1×11 \times 1 square), 1+3=41+3=4 (a 2×22 \times 2 square), 1+3+5=91+3+5=9 (a 3×33 \times 3 square).

For any natural number m>1m > 1, we can generate a Pythagorean Triplet using the values 2m,m21,2m, m^2-1, and m2+1m^2+1. These three numbers satisfy the property that the sum of the squares of the first two equals the square of the third, representing the sides of a right-angled triangle.

A special property for numbers ending in 55: the square of a number (a5)(a5) can be calculated by multiplying aa by its successor (a+1)(a+1) and suffixing 2525 to the result. Visually, this simplifies calculating areas of squares with side lengths like 15,25,35,15, 25, 35, \dots.

📐Formulae

n2=n×nn^2 = n \times n

Number of non-square numbers between n2 and (n+1)2=2n\text{Number of non-square numbers between } n^2 \text{ and } (n+1)^2 = 2n

1+3+5++(2n1)=n21 + 3 + 5 + \dots + (2n-1) = n^2

Pythagorean Triplet: (2m)2+(m21)2=(m2+1)2\text{Pythagorean Triplet: } (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2

(a5)2=[a(a+1)]100+25(a5)^2 = [a(a+1)]100 + 25

💡Examples

Problem 1:

How many non-perfect square numbers lie between 15215^2 and 16216^2?

Solution:

Step 1: Identify the value of nn. Here, n=15n = 15 and n+1=16n+1 = 16. Step 2: Apply the formula for non-perfect square numbers between n2n^2 and (n+1)2(n+1)^2, which is 2n2n. Step 3: Calculate 2×15=302 \times 15 = 30.

Explanation:

The formula 2n2n gives the count of numbers that are greater than n2n^2 and less than (n+1)2(n+1)^2 without including the squares themselves.

Problem 2:

Find a Pythagorean triplet whose smallest member is 1212.

Solution:

Step 1: Let 2m=122m = 12. Solving for mm, we get m=122=6m = \frac{12}{2} = 6. Step 2: Find the second member using m21m^2 - 1. So, 621=361=356^2 - 1 = 36 - 1 = 35. Step 3: Find the third member using m2+1m^2 + 1. So, 62+1=36+1=376^2 + 1 = 36 + 1 = 37. Step 4: The triplet is (12,35,37)(12, 35, 37). Check: 122+352=144+1225=136912^2 + 35^2 = 144 + 1225 = 1369, and 372=136937^2 = 1369.

Explanation:

We use the general form (2m,m21,m2+1)(2m, m^2-1, m^2+1) to find the three integers that satisfy the Pythagorean theorem.