NCERT Solutions for Class 8 Maths Chapter 5: Squares and Square Roots — Complete Guide

Property 1: A perfect square never ends in $2, 3, 7,$ or $8$.
- Possible ending digits: $0, 1, 4, 5, 6, 9$.
- Example: $144$ ends in $4$ (possible), $123$ ends in $3$ (cannot be a perfect square).

Property 2: A perfect square ending in $0$ has an even number of zeros at the end.
- $100 = 10^2$ (two zeros) ✓, $400 = 20^2$ (two zeros) ✓.
- $1000$ has three zeros — not a perfect square.

Property 3: The square of an even number is even; the square of an odd number is odd.
- $12^2 = 144$ (even), $13^2 = 169$ (odd).

Property 4: A perfect square, when expressed in prime factorization, has even powers of every prime factor.
- $36 = 2^2 \times 3^2$ (all even powers) ✓.
- $48 = 2^4 \times 3^1$ (power of $3$ is odd) — not a perfect square.

Property 5: Between two consecutive perfect squares $n^2$ and $(n+1)^2$, there are exactly $2n$ non-perfect-square numbers.
- Between $9$ ($3^2$) and $16$ ($4^2$): $10, 11, 12, 13, 14, 15$ — that is $6 = 2 \times 3$ numbers.

Problem: Without calculating, determine whether the following are perfect squares: (a) $1057$ (b) $6241$ (c) $5765$ (d) $1024$.

Solution:

(a) $1057$ ends in $7$. Since perfect squares never end in $7$, not a perfect square.

(b) $6241$ ends in $1$. This is possible. Need further checking. $78^2 = 6084$, $79^2 = 6241$. Yes, **$6241 = 79^2$** is a perfect square.

(c) $5765$ ends in $5$. If a perfect square ends in $5$, it must end in $25$. Since $5765$ ends in $65$, not a perfect square.

(d) $1024$ ends in $4$. $32^2 = 1024$. Yes, **$1024 = 32^2$** is a perfect square.

The sum of the first $n$ odd natural numbers equals $n^2$:

In general: $1 + 3 + 5 + \cdots + (2n-1) = n^2$.

Converse: If a number can be expressed as a sum of $n$ consecutive odd numbers starting from $1$, then it is the perfect square $n^2$.

The difference between two consecutive perfect squares is always an odd number:

Examples: $4^2 - 3^2 = 16 - 9 = 7 = 2(3) + 1$. $10^2 - 9^2 = 100 - 81 = 19 = 2(9) + 1$.

This means: If we know $n^2$, we can quickly find $(n+1)^2$ by adding $2n + 1$.
- $25^2 = 625$, so $26^2 = 625 + 2(25) + 1 = 625 + 51 = 676$.

For a number ending in $5$, say $n5$ (where $n$ represents the digits before $5$):

Examples:
- $25^2 = 2 \times 3 \times 100 + 25 = 625$
- $35^2 = 3 \times 4 \times 100 + 25 = 1225$
- $65^2 = 6 \times 7 \times 100 + 25 = 4225$
- $115^2 = 11 \times 12 \times 100 + 25 = 13225$

This pattern is a fantastic shortcut for mental calculations!

A Pythagorean triplet is a set of three natural numbers $(a, b, c)$ such that $a^2 + b^2 = c^2$.

For any natural number $m > 1$, the set $(2m, m^2 - 1, m^2 + 1)$ is a Pythagorean triplet.

**Verification for $(3, 4, 5)$:** $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓

Problem: Find $\sqrt{49}$ by repeated subtraction.

Solution:
$49 - 1 = 48$, $48 - 3 = 45$, $45 - 5 = 40$, $40 - 7 = 33$, $33 - 9 = 24$, $24 - 11 = 13$, $13 - 13 = 0$.

$7$ subtractions. $\sqrt{49} = 7$ ✓.

Problem: Find $\sqrt{441}$.

Solution:

All prime factors appear in pairs:

Answer: $\sqrt{441} = 21$.

Problem: Find the smallest number by which $1008$ must be multiplied to get a perfect square. Also find the square root of the resulting number.

Solution:

The prime factor $7$ appears an odd number of times ($1$). To make it even, multiply by $7$:

Answer: Multiply by $7$. Square root $= 84$.

Problem: Find the smallest number by which $2028$ must be divided to get a perfect square.

Solution:

The prime factor $3$ appears an odd number of times. Divide by $3$:

Answer: Divide by $3$. Square root $= 26$.

Problem: Find $\sqrt{\dfrac{169}{289}}$.

Solution:

Since $13^2 = 169$ and $17^2 = 289$.

Answer: $\sqrt{\dfrac{169}{289}} = \dfrac{13}{17}$.

Problem: Find $\sqrt{7921}$ using the long division method.

Solution:

Group as $\overline{79}\;\overline{21}$.

Step 1: $8^2 = 64 \leq 79$ and $9^2 = 81 > 79$. First digit $= 8$. Subtract: $79 - 64 = 15$.

Step 2: Bring down $21$. New dividend $= 1521$. Double the root: $2 \times 8 = 16$. We need $16d \times d \leq 1521$.
Try $d = 9$: $169 \times 9 = 1521$. ✓

Remainder $= 0$.

Problem: Find $\sqrt{2.56}$.

Solution:

Group as $\overline{2}.\overline{56}$.

Step 1: $1^2 = 1 \leq 2$. First digit $= 1$. Subtract: $2 - 1 = 1$.

Step 2: Bring down $56$. New dividend $= 156$. Double root: $2 \times 1 = 2$. Place decimal point.
Try $d = 6$: $26 \times 6 = 156$. ✓

Verification: $1.6^2 = 2.56$ ✓

Problem: Find $\sqrt{3}$ correct to $2$ decimal places.

Solution:

$3$ is not a perfect square. Write as $3.000000$ and group: $\overline{3}.\overline{00}\;\overline{00}\;\overline{00}$.

Step 1: $1^2 = 1 \leq 3$. Root so far $= 1$. Remainder $= 2$.

Step 2: Bring down $00$. Dividend $= 200$. Double: $2 \times 1 = 2$. Try $d = 7$: $27 \times 7 = 189 \leq 200$. Root $= 1.7$. Remainder $= 11$.

Step 3: Bring down $00$. Dividend $= 1100$. Double: $2 \times 17 = 34$. Try $d = 3$: $343 \times 3 = 1029 \leq 1100$. Root $= 1.73$. Remainder $= 71$.

Step 4: Bring down $00$. Dividend $= 7100$. Double: $2 \times 173 = 346$. Try $d = 2$: $3462 \times 2 = 6924 \leq 7100$. Root $= 1.732$.

Problem: Find the least $4$-digit perfect square and the greatest $3$-digit perfect square.

Solution:

Least 4-digit perfect square: The smallest $4$-digit number is $1000$. $\sqrt{1000} \approx 31.6$. So $32^2 = 1024$ is the least $4$-digit perfect square.

Greatest 3-digit perfect square: The largest $3$-digit number is $999$. $\sqrt{999} \approx 31.6$. So $31^2 = 961$ is the greatest $3$-digit perfect square.

Answer: Least $4$-digit perfect square $= 1024$. Greatest $3$-digit perfect square $= 961$.

Problem: Express $49$ as the sum of consecutive odd numbers starting from $1$.

Solution:

$49 = 7^2$, so it is the sum of the first $7$ odd numbers:

Verification: $1 + 3 = 4$, $4 + 5 = 9$, $9 + 7 = 16$, $16 + 9 = 25$, $25 + 11 = 36$, $36 + 13 = 49$ ✓

Problem: If $31^2 = 961$, find $32^2$ without direct multiplication.

Solution:

Using the pattern $(n+1)^2 = n^2 + 2n + 1$:

Answer: $32^2 = 1024$.

Problem: Write a Pythagorean triplet whose smallest member is $8$.

Solution:

Using the formula $(2m, m^2 - 1, m^2 + 1)$. We need $2m = 8$, so $m = 4$.

$m^2 - 1 = 16 - 1 = 15$. $m^2 + 1 = 16 + 1 = 17$.

Triplet: $(8, 15, 17)$.

Verification: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓

Problem: Find $75^2$ using the shortcut for numbers ending in $5$.

Solution:

$(75)^2 = 7 \times 8 \times 100 + 25 = 5600 + 25 = 5625$.

Verification by direct calculation: $75 \times 75 = 75 \times 70 + 75 \times 5 = 5250 + 375 = 5625$ ✓

Problem: Find $\sqrt{5184}$ by prime factorization.

Solution:

Answer: $\sqrt{5184} = 72$.

Problem: Find $\sqrt{324 \times 576}$ without computing the product.

Solution:

Since $18^2 = 324$ and $24^2 = 576$.

Answer: $\sqrt{324 \times 576} = 432$.

Problem: Find $\sqrt{54756}$ by the long division method.

Solution:

Group: $\overline{5}\;\overline{47}\;\overline{56}$.

Step 1: $2^2 = 4 \leq 5$. Root $= 2$. Remainder $= 1$.

Step 2: Bring down $47$. Dividend $= 147$. Double: $2 \times 2 = 4$. Try $d = 3$: $43 \times 3 = 129 \leq 147$. Root $= 23$. Remainder $= 18$.

Step 3: Bring down $56$. Dividend $= 1856$. Double: $2 \times 23 = 46$. Try $d = 4$: $464 \times 4 = 1856$. Root $= 234$.

Problem: Find $\sqrt{51.84}$.

Solution:

Group: $\overline{51}.\overline{84}$.

Step 1: $7^2 = 49 \leq 51$. Root $= 7$. Remainder $= 2$.

Step 2: Bring down $84$. Dividend $= 284$. Double: $14$. Place decimal. Try $d = 2$: $142 \times 2 = 284$. Root $= 7.2$.

Verification: $7.2^2 = 51.84$ ✓

Problem: Estimate $\sqrt{500}$ without a calculator.

Solution:

$22^2 = 484$ and $23^2 = 529$.

Since $484 < 500 < 529$, we know $22 < \sqrt{500} < 23$.

Using linear interpolation:

Answer: $\sqrt{500} \approx 22.4$. (Actual value: $22.3607...$)

Problem: The area of a square garden is $3481$ m$^2$. Find the length of each side.

Solution:

Side of a square $= \sqrt{\text{Area}} = \sqrt{3481}$.

Using long division on $\overline{34}\;\overline{81}$:
- $5^2 = 25 \leq 34$. Root $= 5$. Remainder $= 9$.
- Bring down $81$. Dividend $= 981$. Double: $10$. Try $d = 9$: $109 \times 9 = 981$. Root $= 59$.

Verification: $59^2 = 3481$ ✓

Problem: How many perfect squares lie between $100$ and $400$?

Solution:

$\sqrt{100} = 10$ and $\sqrt{400} = 20$.

Perfect squares between $100$ and $400$ (exclusive): $11^2, 12^2, \ldots, 19^2$.

Count $= 19 - 11 + 1 = 9$.

Answer: $9$ perfect squares.

Problem: Find $\sqrt{148996}$ by the long division method.

Solution:

Group: $\overline{14}\;\overline{89}\;\overline{96}$.

• $3^2 = 9 \leq 14$. Root $= 3$. Remainder $= 5$.
  - Bring down $89$. Dividend $= 589$. Double: $6$. Try $d = 8$: $68 \times 8 = 544$. Root $= 38$. Remainder $= 45$.
  - Bring down $96$. Dividend $= 4596$. Double: $76$. Try $d = 6$: $766 \times 6 = 4596$. Root $= 386$.