Squares and Square Roots — NCERT Solutions, AI Tutor & Practice

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

Here we have $8$ odd numbers, so $n = 8$.

$1 + 3 + 5 + \cdots + 15 = 8^2 = 64.$

A perfect square can only end in $0, 1, 4, 5, 6$ or $9$. It can never end in $2, 3, 7$ or $8$.

• $121 = 11^2$ → perfect square.
• $132$ ends in $2$ → not a perfect square.
• $196 = 14^2$ → perfect square.
• $250$ ends in a single $0$ (a perfect square ending in $0$ must end in an even number of zeros) → not a perfect square.

So $121$ and $196$ are perfect squares.

Write $45 = 40 + 5$, so $a = 40$ and $b = 5$.

$45^2 = (40 + 5)^2 = 40^2 + 2(40)(5) + 5^2.$

$= 1600 + 400 + 25 = 2025.$

Between $n^2$ and $(n+1)^2$ there are exactly $2n$ natural numbers.

Here $n = 15$, so the count is

$2n = 2 \times 15 = 30.$

(Check: $15^2 = 225$ and $16^2 = 256$; the numbers from $226$ to $255$ are $30$ in all.)

For any number $m > 1$, the following form a Pythagorean triplet:

$2m, \quad m^2 - 1, \quad m^2 + 1.$

We want the smallest member $2m = 8$, so $m = 4$.

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

The triplet is $(8, 15, 17)$.

Check: $8^2 + 15^2 = 64 + 225 = 289 = 17^2.$

Prime factorise $1764$:

$1764 = 2 \times 2 \times 3 \times 3 \times 7 \times 7.$

Pair the equal factors:

$1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7).$

Take one factor from each pair:

$\sqrt{1764} = 2 \times 3 \times 7 = 42.$

Prime factorise $1152$:

$1152 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3.$

Group into pairs:

$= (2\times2)(2\times2)(2\times2)(3\times3)\times 2.$

One $2$ is unpaired, so multiply by $2$:

$1152 \times 2 = 2304.$

Now $2304 = (2\times2)(2\times2)(2\times2)(2\times2)(3\times3)$.

$\sqrt{2304} = 2 \times 2 \times 2 \times 2 \times 3 = 48.$

Smallest multiplier $= 2$ and $\sqrt{2304} = 48$.

Prime factorise $2925$:

$2925 = 3 \times 3 \times 5 \times 5 \times 13.$

Group into pairs:

$= (3\times3)(5\times5)\times 13.$

The factor $13$ is unpaired, so divide by $13$:

$2925 \div 13 = 225.$

Now $225 = (3\times3)(5\times5)$, so

$\sqrt{225} = 3 \times 5 = 15.$

Smallest divisor $= 13$ and $\sqrt{225} = 15$.

Group the digits of $4096$ in pairs from the right: $\overline{40}\ \overline{96}$.

Step 1: Largest number whose square $\le 40$ is $6$ (since $6^2 = 36$). Write $6$ as the first quotient digit. Remainder $= 40 - 36 = 4$. Bring down $96$ to get $496$.

Step 2: Double the quotient: $6 \times 2 = 12$. We need a digit $x$ such that $\overline{12x} \times x \le 496$.

Try $x = 4$: $124 \times 4 = 496$. Exact.

So the quotient is $64$ and remainder is $0$.

$\sqrt{4096} = 64.$

For $6.25$, group the whole part in pairs from the decimal point leftward, and the decimal part in pairs rightward: $\overline{6}\ .\ \overline{25}$.

Step 1: Largest number whose square $\le 6$ is $2$ ($2^2 = 4$). Quotient digit $= 2$, remainder $= 6 - 4 = 2$. Place a decimal point in the quotient and bring down $25$ to get $225$.

Step 2: Double the quotient: $2 \times 2 = 4$. Find $x$ with $\overline{4x} \times x \le 225$.

Try $x = 5$: $45 \times 5 = 225$. Exact.

So the quotient is $2.5$.

$\sqrt{6.25} = 2.5.$

For a square, area $=$ side$^2$, so side $= \sqrt{\text{area}}$.

Find $\sqrt{5184}$ by prime factorisation:

$5184 = 2^6 \times 3^4 = (2^3)^2 \times (3^2)^2.$

$\sqrt{5184} = 2^3 \times 3^2 = 8 \times 9 = 72.$

So the side $= 72$ m.

Perimeter $= 4 \times \text{side} = 4 \times 72 = 288$ m.

By long division for $\sqrt{5000}$:

Group as $\overline{50}\ \overline{00}$.

$7^2 = 49 \le 50$, remainder $1$; bring down $00$ → $100$. Double quotient $= 14$; try $\overline{14x}\times x \le 100$: with $x = 0$, $140\times 0 = 0$ works but leaves large remainder, try higher.

Actually the integer part of $\sqrt{5000}$ is $70$ since $70^2 = 4900$ and $71^2 = 5041 > 5000$.

So the greatest perfect square not exceeding $5000$ is $70^2 = 4900$.

Least number to subtract $= 5000 - 4900 = 100.$

The resulting perfect square is $4900 = 70^2$.