Math Notes For Class 8 Chapter 1 Ex 1.2 Solved & Quiz

Question 1 (Part i): Find the square root of $784$ using the prime factorization method.

Solution:
Step 1: Perform prime factorization of $784$:
$$784 = 2 \times 2 \times 2 \times 2 \times 7 \times 7$$ Step 2: Group prime factors into square pairs:
$$784 = (2 \times 2) \times (2 \times 2) \times (7 \times 7) = 2^2 \times 2^2 \times 7^2$$ Step 3: Take the square root on both sides:
$$\sqrt{784} = \sqrt{2^2 \times 2^2 \times 7^2} = 2 \times 2 \times 7$$ $$\sqrt{784} = 28$$
Answer: $$\sqrt{784} = 28$$

Question 1 (Part ii): Find the square root of $1296$ using the prime factorization method.

Solution:
Step 1: Perform prime factorization of $1296$:
$$1296 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$ Step 2: Group the prime factors into pairs:
$$1296 = (2^2) \times (2^2) \times (3^2) \times (3^2)$$ Step 3: Apply the square root:
$$\sqrt{1296} = \sqrt{2^2 \times 2^2 \times 3^2 \times 3^2} = 2 \times 2 \times 3 \times 3 = 36$$
Answer: $$\sqrt{1296} = 36$$

Question 2 (Part i): Find the square root of $529$ using the long division method.

Solution:
Step 1: Pair the digits from right to left: $\overline{5}\ \overline{29}$.
Step 2: Find the largest integer whose square is $\le 5$. That integer is $2$ because $2^2 = 4$.
Subtract $4$ from $5$, remainder is $1$. Bring down the next pair $29$, yielding $129$.
Step 3: Double the quotient $2$ to get new divisor prefix $4$.
Find digit $x$ such that $4x \times x = 129$. For $x = 3$:
$$43 \times 3 = 129$$ Subtracting gives remainder $0$. Quotient is $23$.
Answer: $$\sqrt{529} = 23$$

Question 2 (Part ii): Find the square root of $4096$ using the long division method.

Solution:
Step 1: Pair the digits from right to left: $\overline{40}\ \overline{96}$.
Step 2: Find the largest integer whose square is $\le 40$. That integer is $6$ ($6^2 = 36$).
Subtract $36$ from $40$, remainder is $4$. Bring down $96$ to get $496$.
Step 3: Double the quotient $6$ to get $12$.
Find digit $x$ such that $12x \times x = 496$. Testing $x = 4$:
$$124 \times 4 = 496$$ Subtracting gives remainder $0$. Final quotient is $64$.
Answer: $$\sqrt{4096} = 64$$

Question 3 (Part i): Find the cube root of $216$ using prime factorization.

Solution:
Step 1: Express $216$ as a product of prime factors:
$$216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$$ Step 2: Group factors in sets of three (triplets):
$$216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) = 2^3 \times 3^3$$ Step 3: Take the cube root on both sides:
$$\sqrt[3]{216} = \sqrt[3]{2^3 \times 3^3} = 2 \times 3 = 6$$
Answer: $$\sqrt[3]{216} = 6$$

Question 3 (Part ii): Find the cube root of $3375$ using prime factorization.

Solution:
Step 1: Perform prime factorization of $3375$:
$$3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5$$ Step 2: Group factors into triplets:
$$3375 = (3^3) \times (5^3)$$ Step 3: Apply the cube root operator:
$$\sqrt[3]{3375} = \sqrt[3]{3^3 \times 5^3} = 3 \times 5 = 15$$
Answer: $$\sqrt[3]{3375} = 15$$

Question 4 (Part i): Simplify the radical expression: $\sqrt{180} + \sqrt{45} - \sqrt{80}$.

Solution:
Step 1: Simplify each term into its radical components:
$$\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$$ $$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$$ $$\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$$ Step 2: Combine like radical terms:
$$\text{Expression} = 6\sqrt{5} + 3\sqrt{5} - 4\sqrt{5}$$ $$\text{Expression} = (6 + 3 - 4)\sqrt{5} = 5\sqrt{5}$$
Answer: $$5\sqrt{5}$$

Question 4 (Part ii): Simplify the expression: $\frac{\sqrt{32}}{\sqrt{8}} \times \sqrt{9}$.

Solution:
Step 1: Apply quotient rule of radicals to the fraction:
$$\frac{\sqrt{32}}{\sqrt{8}} = \sqrt{\frac{32}{8}} = \sqrt{4} = 2$$ Step 2: Evaluate the square root of $9$:
$$\sqrt{9} = 3$$ Step 3: Multiply the simplified terms:
$$\text{Result} = 2 \times 3 = 6$$
Answer: $$6$$

Question 5 (Part i): Classify $\sqrt{50}$ as Rational or Irrational and approximate its decimal value up to two decimal places.

Solution:
Step 1: Express $\sqrt{50}$ in simplified surd form:
$$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$$ Since $\sqrt{2}$ is a non-terminating, non-repeating decimal, $5\sqrt{2}$ is an **Irrational Number** ($\mathbb{Q'}$).
Step 2: Approximate the numerical value using $\sqrt{2} \approx 1.4142$:
$$5\sqrt{2} \approx 5 \times 1.4142 = 7.071$$ Rounding off to 2 decimal places yields $7.07$.
Answer: $$\text{Irrational Number, } \approx 7.07$$

Question 5 (Part ii): Classify $\frac{22}{7}$ as Rational or Irrational and compute its value to two decimal places.

Solution:
Step 1: Check quotient representation $\frac{p}{q}$ where $p=22, q=7 \neq 0$. Both $22$ and $7$ are integers. Thus, $\frac{22}{7}$ is a **Rational Number** ($\mathbb{Q}$).
Step 2: Divide $22$ by $7$ to convert to decimal:
$$\frac{22}{7} = 3.142857...$$ Rounding to two decimal places gives $3.14$.
Answer: $$\text{Rational Number, } \approx 3.14$$

Question 6 (Part i): A square garden has an area of $1225\text{ m}^2$. Find the length of each side and the total cost of fencing it at Rs. $150$ per meter.

Solution:
Step 1: Find side length $s$ using $A = s^2$:
$$s = \sqrt{1225}$$ By prime factorization: $1225 = 5^2 \times 7^2 \implies s = 5 \times 7 = 35\text{ m}$.
Step 2: Calculate the perimeter of the square garden ($P$):
$$P = 4 \times s = 4 \times 35 = 140\text{ m}$$ Step 3: Calculate total fencing cost:
$$\text{Total Cost} = 140\text{ m} \times \text{Rs. } 150/\text{m} = \text{Rs. } 21,000$$
Answer: $$\text{Side } = 35\text{ m}, \text{ Cost } = \text{Rs. } 21,000$$

Question 6 (Part ii): A cubic box has a volume of $729\text{ cm}^3$. Find the length of one edge of the box and its total surface area.

Solution:
Step 1: Find edge length $a$ using volume formula $V = a^3$:
$$a = \sqrt[3]{729}$$ Prime factorization: $729 = 3^6 = (3^2)^3 = 9^3 \implies a = 9\text{ cm}$.
Step 2: Compute total surface area $A$ using $A = 6a^2$:
$$A = 6 \times (9)^2 = 6 \times 81 = 486\text{ cm}^2$$
Answer: $$\text{Edge } = 9\text{ cm}, \text{ Total Surface Area } = 486\text{ cm}^2$$