Class 8 Maths Square and Square Roots Worksheet - Free Printable

Problem Set Solution

#### 1. Fill in the blanks

(i) The digit at the unit's place of square of 239 = _______

- To find the unit digit of \(239^2\), we only need to consider the unit digit of 239, which is 9.
- The unit digit of \(9^2\) is \(81\), so the unit digit is 1.
- Answer: \(1\)

(ii) \(29 \times 31 = \_\_\_\_\_\_ - 1\)

- Notice that \(29 \times 31\) can be written as \((30 - 1)(30 + 1)\), which is a difference of squares:
\[
29 \times 31 = 30^2 - 1^2 = 900 - 1 = 899
\]
- Therefore, \(29 \times 31 = 900 - 1\).
- Answer: \(900\)

(iii) \(17 \times 23 = \_\_\_\_\_\_ - 3^2\)

- Notice that \(17 \times 23\) can be written as \((20 - 3)(20 + 3)\), which is a difference of squares:
\[
17 \times 23 = 20^2 - 3^2 = 400 - 9 = 391
\]
- Therefore, \(17 \times 23 = 400 - 9\).
- Answer: \(400\)

(iv) \((151)^2 - (150)^2 = \_\_\_\_\_\_

- Using the difference of squares formula \(a^2 - b^2 = (a - b)(a + b)\):
\[
(151)^2 - (150)^2 = (151 - 150)(151 + 150) = 1 \times 301 = 301
\]
- Answer: \(301\)

(v) The sum of the first five odd numbers = \_\_\_\_\_\_

- The first five odd numbers are: \(1, 3, 5, 7, 9\).
- The sum of the first \(n\) odd numbers is given by \(n^2\). Here, \(n = 5\):
\[
1 + 3 + 5 + 7 + 9 = 5^2 = 25
\]
- Answer: \(25\)

(vi) If \(6^x = 1296\), then \(x = \_\_\_\_\_\_

- We need to express 1296 as a power of 6:
\[
1296 = 6^4
\]
- Therefore, \(x = 4\).
- Answer: \(4\)

---

#### 2. Write the perfect square numbers between 100 and 150.

- The perfect squares between 100 and 150 are:
\[
11^2 = 121 \quad \text{and} \quad 12^2 = 144
\]
- Answer: \(121, 144\)

---

#### 3. Write 17 as the sum of two consecutive integers.

- Let the two consecutive integers be \(n\) and \(n+1\). Their sum is:
\[
n + (n + 1) = 17 \implies 2n + 1 = 17 \implies 2n = 16 \implies n = 8
\]
- The two consecutive integers are \(8\) and \(9\).
- Answer: \(8 + 9\)

---

#### 4. Find the Pythagorean triplet whose smallest number is 10.

- A Pythagorean triplet can be generated using the formula \( (m^2 - n^2, 2mn, m^2 + n^2) \) where \(m > n\).
- Given the smallest number is 10, we set \(m^2 - n^2 = 10\).
- Trying small values for \(m\) and \(n\):
- Let \(m = 6\) and \(n = 4\):
\[
m^2 - n^2 = 6^2 - 4^2 = 36 - 16 = 10
\]
\[
2mn = 2 \times 6 \times 4 = 48
\]
\[
m^2 + n^2 = 6^2 + 4^2 = 36 + 16 = 50
\]
- The triplet is \(10, 48, 50\).
- Answer: \(10, 48, 50\)

---

#### 5. Find the smallest number by which 192 must be multiplied to make the product a perfect square.

- Prime factorize 192:
\[
192 = 2^6 \times 3
\]
- For the product to be a perfect square, all prime factors must have even exponents. Currently, the exponent of 3 is 1 (odd). We need to multiply by another 3 to make it even.
- Therefore, the smallest number to multiply by is \(3\).
- Answer: \(3\)

---

#### 6. Find the square root of the following:

(i) \(10609\)

- Prime factorize \(10609\):
\[
10609 = 103^2
\]
- Therefore, \(\sqrt{10609} = 103\).

(ii) \(33.64\)

- Rewrite \(33.64\) as \(3364 \div 100\):
\[
\sqrt{33.64} = \sqrt{\frac{3364}{100}} = \frac{\sqrt{3364}}{\sqrt{100}} = \frac{58}{10} = 5.8
\]

(iii) \(0.4489\)

- Rewrite \(0.4489\) as \(4489 \div 10000\):
\[
\sqrt{0.4489} = \sqrt{\frac{4489}{10000}} = \frac{\sqrt{4489}}{\sqrt{10000}} = \frac{67}{100} = 0.67
\]

(iv) \(\frac{289}{361}\)

- Take the square root of the numerator and denominator separately:
\[
\sqrt{\frac{289}{361}} = \frac{\sqrt{289}}{\sqrt{361}} = \frac{17}{19}
\]

(v) \(1 \frac{7}{9}\)

- Convert the mixed fraction to an improper fraction:
\[
1 \frac{7}{9} = \frac{16}{9}
\]
- Take the square root:
\[
\sqrt{\frac{16}{9}} = \frac{\sqrt{16}}{\sqrt{9}} = \frac{4}{3}
\]

- Answers:
\[
\boxed{103, 5.8, 0.67, \frac{17}{19}, \frac{4}{3}}
\]

---

#### 7. Find:

(i) \(\sqrt{55} \times \sqrt{220}\)

- Use the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\):
\[
\sqrt{55} \times \sqrt{220} = \sqrt{55 \times 220} = \sqrt{12100} = 110
\]

(ii) \(\sqrt{0.25} \times \sqrt{0.09}\)

- Use the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\):
\[
\sqrt{0.25} \times \sqrt{0.09} = \sqrt{0.25 \times 0.09} = \sqrt{0.0225} = 0.15
\]

- Answers:
\[
\boxed{110, 0.15}
\]

---

#### 8. Find the smallest number of four digits which is a perfect square.

- The smallest four-digit number is 1000. We need to find the smallest perfect square greater than or equal to 1000.
- The square root of 1000 is approximately 31.62. The next whole number is 32.
- \(32^2 = 1024\).
- Answer: \(1024\)

---

#### 9. Find the smallest square number which is divisible by the numbers 4, 12, and 16.

- First, find the least common multiple (LCM) of 4, 12, and 16:
\[
4 = 2^2, \quad 12 = 2^2 \times 3, \quad 16 = 2^4
\]
\[
\text{LCM}(4, 12, 16) = 2^4 \times 3 = 48
\]
- Now, we need the smallest square number that is a multiple of 48. Factorize 48:
\[
48 = 2^4 \times 3
\]
- To make it a perfect square, we need an even power of 3. Multiply by 3:
\[
48 \times 3 = 144
\]
- Check if 144 is a perfect square:
\[
144 = 12^2
\]
- Answer: \(144\)

---

#### 10. Find \(x\) if \(\frac{x}{16} = \sqrt{\frac{9}{16}}\).

- Simplify the right-hand side:
\[
\sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}
\]
- So the equation becomes:
\[
\frac{x}{16} = \frac{3}{4}
\]
- Cross-multiply to solve for \(x\):
\[
x = 16 \times \frac{3}{4} = 12
\]
- Answer: \(12\)

---

#### 11. The area of a square field is \(4225 \, \text{m}^2\). If a girl cycles along its boundary, how much distance she will be able to cover if she covers the boundary twice.

- The area of the square field is \(4225 \, \text{m}^2\). Let the side length be \(s\):
\[
s^2 = 4225 \implies s = \sqrt{4225} = 65 \, \text{m}
\]
- The perimeter of the square is:
\[
4s = 4 \times 65 = 260 \, \text{m}
\]
- If the girl cycles along the boundary twice, the total distance covered is:
\[
2 \times 260 = 520 \, \text{m}
\]
- Answer: \(520 \, \text{m}\)

---

#### 12. For the international yoga day, a yoga teacher has to arrange 1024 students in such a way that each row has as many students as the number of rows. Find the number of students in each row.

- Let the number of rows be \(n\). Then the number of students in each row is also \(n\).
- The total number of students is:
\[
n \times n = n^2 = 1024
\]
- Solve for \(n\):
\[
n = \sqrt{1024} = 32
\]
- Answer: \(32\)

---

Final Answers:

\[
\boxed{1, 900, 400, 301, 25, 4, 121, 144, 8 + 9, 10, 48, 50, 3, 103, 5.8, 0.67, \frac{17}{19}, \frac{4}{3}, 110, 0.15, 1024, 144, 12, 520, 32}
\]