Class 8 : Maths – Lesson 6. We Distribute, Yet Things Multiply
EXPLANATION AND ANALYSIS
🌍 INTRODUCTION — WHEN SHARING INCREASES EFFECT
🍞 In daily life, we often distribute things—food, money, land, work, or time.
🧠 In mathematics, distribution is not only about sharing equally; it also reveals how numbers interact and grow.
📘 This lesson explains how:
distributing quantities involves multiplication
fractions and whole numbers work together
scaling changes size but keeps relationships intact
Understanding this lesson helps us see why multiplication appears naturally in sharing situations.
🔢 IDEA OF DISTRIBUTION
📦 Distribution means sharing a quantity into parts based on a rule or ratio.
🧠 When we distribute:
we divide a whole into parts
each part depends on the size of the group
multiplication helps calculate the total
📌 Distribution is closely connected with multiplication, not just division.
✖️ WHY DISTRIBUTION LEADS TO MULTIPLICATION
🧠 Suppose one group receives a certain amount.
📈 When the number of groups increases, the total amount increases.
📌 Key idea:
amount per group × number of groups = total amount
🔵 This is why, even though we distribute, things multiply.
🔄 DISTRIBUTION USING FRACTIONS
📐 Fractions are commonly used while distributing things.
📘 Examples of fractional distribution:
half of a quantity
one-third of a number
three-fourths of an amount
🧠 To find a fraction of a number, we multiply the number by the fraction.
📌 Fraction of a quantity = quantity × fraction
🔢 MULTIPLICATION WITH FRACTIONS
📘 Multiplying fractions follows a simple rule.
✨ Key idea:
multiply numerators
multiply denominators
🧠 Fraction multiplication helps find:
part of a whole
scaled values
distributed shares
Multiplication makes distribution clear and accurate.
📊 SCALING AND ITS EFFECT
📐 Scaling means increasing or decreasing a quantity by a factor.
📌 If a value is scaled by:
a factor greater than 1 → value increases
a factor between 0 and 1 → value decreases
🧠 Even when distributing, scaling can increase the total effect.
🔄 DISTRIBUTION IN RATIOS
⚖️ Distribution is often done in ratios.
📘 A ratio shows how a quantity is shared among different parts.
🧠 To distribute using ratios:
add all parts of the ratio
find the value of one part
multiply to get each share
📌 Multiplication plays a key role at every step.
🧮 DISTRIBUTIVE IDEA IN NUMBERS
🧠 Numbers follow a distributive pattern.
📘 This idea shows that:
multiplication spreads over addition
distribution simplifies calculations
📌 Instead of calculating step by step, distribution allows smart multiplication.
🔍 REAL-LIFE EXAMPLES OF DISTRIBUTION
🌍 Distribution appears in many real-life situations:
🔵 sharing money among people
🟡 dividing land into plots
🔴 distributing food equally
🟣 allocating time for tasks
🟠 dividing work among workers
🧠 In all these cases, multiplication helps find totals and shares.
⚠️ COMMON MISTAKES TO AVOID
🚫 confusing division with distribution
🚫 forgetting to multiply after finding one share
🚫 mishandling fractions
🚫 ignoring ratio totals
✔️ Always:
understand what is being distributed
identify the multiplier
check calculations carefully
🌟 IMPORTANCE OF THIS LESSON
🏆 builds understanding of fractions
🧠 improves reasoning with ratios
⚡ strengthens multiplication skills
📘 prepares for algebraic thinking
🌱 connects maths with real-life sharing
This lesson shows that distribution is not subtraction of value, but a structured multiplication process.
🧾 SUMMARY
🔵 distribution means sharing quantities
🟡 multiplication helps find total effect
🔴 fractions show parts of a whole
🟣 scaling changes size proportionally
🟠 ratios guide fair distribution
🟢 multiplication appears naturally in sharing
🔁 QUICK RECAP
🔵 distribution involves multiplication
🟡 fraction of a quantity uses multiplication
🟣 scaling can increase or decrease values
🟠 ratios help divide fairly
🔴 understanding logic is more important than memorising rules
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TEXTBOOK QUESTIONS
🔒 ❓ 1. Compute these products using the suggested identity.
🔒 ❓ (i) 46² using Identity 1A for (a + b)²
📌 ✅ Answer:
🟢 Step 1: Express the number as a sum
⬥ We can write 46 as (40 + 6).
⬥ Using Identity 1A: (a + b)² = a² + 2ab + b²
⬥ Here, a = 40 and b = 6.
🔵 Step 2: Apply the identity
⬥ (40 + 6)² = (40)² + 2(40)(6) + (6)²
⬥ = 1600 + 480 + 36
🟡 Step 3: Add the values
⬥ 1600 + 480 = 2080
⬥ 2080 + 36 = 2116
⬥ 46² = 2116
🔒 ❓ (ii) 397 × 403 using Identity 1C for (a + b) (a – b)
📌 ✅ Answer:
🟢 Step 1: Express numbers around a common base
⬥ 397 = (400 – 3)
⬥ 403 = (400 + 3)
⬥ The expression becomes (400 – 3)(400 + 3).
🔵 Step 2: Apply Identity 1C
⬥ Identity: (a – b)(a + b) = a² – b²
⬥ Here, a = 400 and b = 3.
⬥ = (400)² – (3)²
🟡 Step 3: Simplify
⬥ = 160000 – 9
⬥ 159991
🔒 ❓ (iii) 91² using Identity 1B for (a – b)²
📌 ✅ Answer:
🟢 Step 1: Express the number as a difference
⬥ We can write 91 as (100 – 9).
⬥ Using Identity 1B: (a – b)² = a² – 2ab + b²
⬥ Here, a = 100 and b = 9.
🔵 Step 2: Apply the identity
⬥ (100 – 9)² = (100)² – 2(100)(9) + (9)²
⬥ = 10000 – 1800 + 81
🟡 Step 3: Calculate
⬥ 10000 – 1800 = 8200
⬥ 8200 + 81 = 8281
⬥ 91² = 8281
🔒 ❓ (iv) 43 × 45 using Identity 1C for (a + b) (a – b)
📌 ✅ Answer:
⬥ Note: The question asks to use Identity 1C (a+b)(a-b).
⬥ However, 43 and 45 are (44 – 1) and (44 + 1).
🟢 Step 1: Rewrite using the middle number
⬥ 43 = (44 – 1)
⬥ 45 = (44 + 1)
⬥ Product = (44 – 1)(44 + 1)
🔵 Step 2: Apply Identity 1C
⬥ (a – b)(a + b) = a² – b²
⬥ = (44)² – (1)²
⬥ = 1936 – 1
⬥ 1935
🔒 ❓ 2. Use either a suitable identity or the distributive property to find each of the following products.
🔒 ❓ (i) (p – 1) (p + 11)
📌 ✅ Answer:
🟢 Step 1: Use the identity (x + a)(x + b) = x² + (a + b)x + ab
⬥ Here, x = p, a = –1, b = 11.
⬥ = p² + (–1 + 11)p + (–1)(11)
🔵 Step 2: Simplify
⬥ = p² + (10)p – 11
⬥ p² + 10p – 11
🔒 ❓ (ii) (3a – 9b) (3a + 9b)
📌 ✅ Answer:
🟢 Step 1: Identify the form (x – y)(x + y)
⬥ This matches the identity: a² – b²
⬥ First term = 3a, Second term = 9b.
🔵 Step 2: Apply identity
⬥ = (3a)² – (9b)²
⬥ = 9a² – 81b²
⬥ 9(a² – 9b²)
🔒 ❓ (iii) –(2y + 5) (3y + 4)
📌 ✅ Answer:
🟢 Step 1: Expand the binomials first
⬥ (2y + 5)(3y + 4) = 2y(3y + 4) + 5(3y + 4)
⬥ = 6y² + 8y + 15y + 20
⬥ = 6y² + 23y + 20
🔵 Step 2: Apply the negative sign
⬥ – (6y² + 23y + 20)
⬥ –6y² – 23y – 20
🔒 ❓ (iv) (6x + 5y)²
📌 ✅ Answer:
🟢 Step 1: Use Identity (a + b)² = a² + 2ab + b²
⬥ Here, a = 6x and b = 5y.
⬥ = (6x)² + 2(6x)(5y) + (5y)²
🔵 Step 2: Simplify terms
⬥ = 36x² + 60xy + 25y²
⬥ 36x² + 60xy + 25y²
🔒 ❓ (v) (2x – 1/2)²
📌 ✅ Answer:
🟢 Step 1: Use Identity (a – b)² = a² – 2ab + b²
⬥ Here, a = 2x and b = 1/2.
⬥ = (2x)² – 2(2x)(1/2) + (1/2)²
🔵 Step 2: Simplify
⬥ = 4x² – 2x + 1/4
⬥ 4x² – 2x + 1/4
🔒 ❓ (vi) (7p) × (3r) × (p + 2)
📌 ✅ Answer:
🟢 Step 1: Multiply the monomials first
⬥ (7p) × (3r) = 21pr
🔵 Step 2: Distribute into the binomial
⬥ 21pr × (p + 2)
⬥ = (21pr × p) + (21pr × 2)
⬥ 21p²r + 42pr
🔒 ❓ 3. For each statement identify the appropriate algebraic expression(s).
🔒 ❓ (i) Two more than a square number.
📌 ✅ Answer:
⬥ We look for a square (s²) plus 2.
⬥ Correct Expression: s² + 2
🔒 ❓ (ii) The sum of the squares of two consecutive numbers
📌 ✅ Answer:
⬥ Consecutive numbers can be m and (m+1) OR (m-1) and m.
⬥ Sum of their squares: m² + (m + 1)² OR m² + (m – 1)²
⬥ Correct Expressions from list: m² + (m + 1)² and m² + (m – 1)²
🔒 ❓ 4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure.
(Image shows February calendar. Selected 2×2: 4, 5, 11, 12)
Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens.
Hint: Label the numbers in each 2 by 2 square as
a (a + 1)
a + 7 (a + 8)
📌 ✅ Answer:
🟢 Step 1: Analyze the pattern with variables
⬥ Let the first number be a.
⬥ In a calendar (7-day week):
⬥ Next number in row = a + 1
⬥ Number directly below a = a + 7
⬥ Number diagonal to a = a + 8
🔵 Step 2: Calculate Diagonal Products
⬥ Diagonal 1 product: a × (a + 8) = a² + 8a
⬥ Diagonal 2 product: (a + 1) × (a + 7) = a(a + 7) + 1(a + 7)
⬥ = a² + 7a + a + 7
⬥ = a² + 8a + 7
🟡 Step 3: Find the difference
⬥ Difference = (Diagonal 2 Product) – (Diagonal 1 Product)
⬥ = (a² + 8a + 7) – (a² + 8a)
⬥ = 7
🔴 Conclusion:
⬥ The product of the diagonal (top-right to bottom-left) is always 7 more than the product of the other diagonal (top-left to bottom-right). This is because the expansion results in a constant difference of 7.
🔒 ❓ 5. Verify which of the following statements are true.
🔒 ❓ (i) (k + 1) (k + 2) – (k + 3) is always 2.
📌 ✅ Answer:
🟢 Step 1: Simplify the expression
⬥ Expansion: (k + 1)(k + 2) = k² + 2k + k + 2 = k² + 3k + 2
⬥ Subtract (k + 3): (k² + 3k + 2) – k – 3
⬥ = k² + 2k – 1
🔵 Step 2: Check validity
⬥ The result k² + 2k – 1 is not “always 2”. It depends on the value of k.
⬥ Statement is False.
🔒 ❓ (ii) (2q + 1) (2q – 3) is a multiple of 4.
📌 ✅ Answer:
🟢 Step 1: Expand the expression
⬥ = 2q(2q – 3) + 1(2q – 3)
⬥ = 4q² – 6q + 2q – 3
⬥ = 4q² – 4q – 3
🔵 Step 2: Analyze for multiple of 4
⬥ 4q² – 4q is divisible by 4 (it equals 4(q² – q)).
⬥ However, –3 is not divisible by 4.
⬥ The total expression is NOT a multiple of 4.
⬥ Statement is False.
🔒 ❓ (iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
📌 ✅ Answer:
🟢 Step 1: Check Even Numbers
⬥ Let even number = 2n.
⬥ Square = (2n)² = 4n². This is clearly a multiple of 4. (True)
🔵 Step 2: Check Odd Numbers
⬥ Let odd number = 2n + 1.
⬥ Square = (2n + 1)² = 4n² + 4n + 1
⬥ = 4n(n + 1) + 1
⬥ Since n(n + 1) is the product of consecutive integers, it is always even (let’s say 2k).
⬥ = 4(2k) + 1 = 8k + 1.
⬥ This is 1 more than a multiple of 8. (True)
⬥ Statement is True.
🔒 ❓ (iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
📌 ✅ Answer:
🟢 Step 1: Expand both squares
⬥ (6n + 2)² = 36n² + 24n + 4
⬥ (4n + 3)² = 16n² + 24n + 9
🔵 Step 2: Subtract
⬥ (36n² + 24n + 4) – (16n² + 24n + 9)
⬥ = 20n² – 5
⬥ A “square number” implies a perfect square (k²).
⬥ 20n² – 5 is not generally 5 less than a perfect square for all n.
⬥ Statement is False.
🔒 ❓ 6. A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?
📌 ✅ Answer:
🟢 Step 1: Define numbers
⬥ First number A = 7m + 3
⬥ Second number B = 7n + 5
🔵 Step 2: Check Sum (A + B)
⬥ (7m + 3) + (7n + 5) = 7(m + n) + 8
⬥ 8 divided by 7 gives remainder 1.
⬥ Remainder of Sum = 1
🟡 Step 3: Check Difference (A – B)
⬥ (7m + 3) – (7n + 5) = 7(m – n) – 2
⬥ In modular arithmetic, remainder –2 is equivalent to –2 + 7 = 5.
⬥ Remainder of Difference = 5
🔴 Step 4: Check Product (A × B)
⬥ (7m + 3)(7n + 5) = 49mn + 35m + 21n + 15
⬥ All terms except 15 are divisible by 7.
⬥ 15 divided by 7 gives remainder 1.
⬥ Remainder of Product = 1
🔒 ❓ 7. Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
📌 ✅ Answer:
🟢 Step 1: Observation
⬥ Example set: 3, 4, 5. Middle squared: 4² = 16. Product of others: 3 × 5 = 15. Difference: 16 – 15 = 1.
⬥ Example set: 9, 10, 11. Middle squared: 100. Product: 99. Difference: 1.
⬥ Pattern: The result is always 1.
🔵 Step 2: Algebraic Equation
⬥ Let numbers be (x – 1), x, (x + 1).
⬥ Middle squared = x²
⬥ Product of others = (x – 1)(x + 1)
⬥ Equation: x² – (x – 1)(x + 1) = 1
🟡 Step 3: Verification
⬥ LHS = x² – (x² – 1) [Using Identity a² – b²]
⬥ LHS = x² – x² + 1
⬥ LHS = 1
⬥ LHS = RHS. Identity verified.
🔒 ❓ 8. What is the algebraic expression describing the following steps—add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
📌 ✅ Answer:
🟢 Step 1: Formulate Expression
⬥ Let the two numbers be x and y.
⬥ “Add any two numbers”: (x + y)
⬥ “Multiply by half of the sum”: (x + y) × ½(x + y)
🔵 Step 2: Simplify
⬥ = ½(x + y)(x + y)
⬥ = ½(x + y)²
🟡 Step 3: Proof
⬥ The result ½(x + y)² literally translates to “half of the square of the sum”.
⬥ Square of sum = (x + y)²
⬥ Half of that = ½(x + y)²
⬥ Proven.
🔒 ❓ 9. Which is larger? Find out without fully computing the product.
🔒 ❓ (i) 14 × 26 or 16 × 24
📌 ✅ Answer:
🟢 Step 1: Analyze using Difference of Squares
⬥ Concept: If two pairs of numbers have the same sum, the pair with numbers closer to each other has the larger product.
🔵 Step 2: Compare sums
⬥ Both sum to 40.
⬥ 14 × 26 = (20 – 6)(20 + 6) = 400 – 36
⬥ 16 × 24 = (20 – 4)(20 + 4) = 400 – 16
🟡 Step 3: Conclusion
⬥ Subtracting 16 is better than subtracting 36.
⬥ 16 × 24 is larger.
🔒 ❓ (ii) 25 × 75 or 26 × 74
📌 ✅ Answer:
🟢 Step 1: Analyze sums
⬥ Both sum to 100.
🔵 Step 2: Compare gaps
⬥ 26 and 74 are closer to each other (gap 48) than 25 and 75 (gap 50).
🟡 Step 3: Conclusion
⬥ The pair with numbers closer together is larger.
⬥ 26 × 74 is larger.
🔒 ❓ 10. A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.
📌 ✅ Answer:
🟢 Step 1: Analyze dimensions from the figure
⬥ We have a large rectangle containing two square plots (area g² means side = g) and a surrounding path of width w.
⬥ Inner Area: Area of 2 green squares = g² + g² = 2g².
🔵 Step 2: Determine Total Dimensions
⬥ The diagram shows the path separates the squares and surrounds them.
⬥ Total Length: w (left path) + g (plot 1) + w (middle path) + g (plot 2) + w (right path) = 2g + 3w.
⬥ Total Width (Height): w (top path) + g (plot height) + w (bottom path) = g + 2w.
🟡 Step 3: Formulate Area Expression
⬥ Total Area = (Length) × (Width) = (2g + 3w)(g + 2w)
⬥ Tiled Area = Total Area – Inner Green Area
⬥ Area = (2g + 3w)(g + 2w) – 2g²
🔴 Step 4: Expand and Simplify
⬥ Expansion: 2g² + 4gw + 3gw + 6w²
⬥ Subtract 2g²: (2g² + 7gw + 6w²) – 2g²
⬥ Result: 7gw + 6w² (or w(7g + 6w))
🔒 ❓ 11. For each pattern shown below,
🔒 ❓ (i) Draw the next figure in the sequence.
📌 ✅ Answer:
🟢 Pattern A (Yellow)
⬥ Step 1: 4 units. Step 2: 7 units. Step 3: 10 units.
⬥ Next figure (Step 4) will have 13 blocks.
🔵 Pattern B (Blue)
⬥ Step 1: 5 units. Step 2: 8 units. Step 3: 11 units.
⬥ Next figure (Step 4) will have 14 blocks.
🔒 ❓ (ii) How many basic units are there in Step 10?
📌 ✅ Answer:
🟢 Pattern A (Yellow)
⬥ Formula: 3y + 1
⬥ Step 10: 3(10) + 1 = 31 units.
🔵 Pattern B (Blue)
⬥ Formula: 3y + 2
⬥ Step 10: 3(10) + 2 = 32 units.
🔒 ❓ (iii) Write an expression to describe the number of basic units in Step y.
📌 ✅ Answer:
🟢 Pattern A (Yellow)
⬥ Expression: 3y + 1
🔵 Pattern B (Blue)
⬥ Expression: 3y + 2
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OTHER IMPORTANT QUESTIONS
🔹 Part A — MCQs (Q1–Q10)
🔒 ❓ Q1.
Which identity is most suitable to evaluate 49 × 51 mentally?
🟢1️⃣ (a + b)(a − b) = a² − b²
🔵2️⃣ (a + b)² = a² + 2ab + b²
🟡3️⃣ (a − b)² = a² − 2ab + b²
🟣4️⃣ a(b + c) = ab + ac
✔️ Answer: 🟢1️⃣
🔒 ❓ Q2.
Which expression is always equal to (x + 3)(x − 3) + 9 ?
🟢1️⃣ x²
🔵2️⃣ x² − 9
🟡3️⃣ x² + 9
🟣4️⃣ (x + 3)²
✔️ Answer: 🟢1️⃣
🔒 ❓ Q3.
Which is larger without direct multiplication?
🟢1️⃣ 38 × 42
🔵2️⃣ 40²
🟡3️⃣ Both are equal
🟣4️⃣ Cannot be compared
✔️ Answer: 🟢1️⃣
🔒 ❓ Q4.
Which expression always represents a perfect square?
🟢1️⃣ (a + b)²
🔵2️⃣ (a + b)(a − b)
🟡3️⃣ a² − b²
🟣4️⃣ 2ab
✔️ Answer: 🟢1️⃣
🔒 ❓ Q5.
The value of (x + 5)² − (x − 5)² is:
🟢1️⃣ 20x
🔵2️⃣ 10x
🟡3️⃣ 25
🟣4️⃣ x²
✔️ Answer: 🟢1️⃣
🔒 ❓ Q6.
Which identity helps compare 27 × 33 with 30²?
🟢1️⃣ (a + b)(a − b) = a² − b²
🔵2️⃣ (a + b)²
🟡3️⃣ (a − b)²
🟣4️⃣ None
✔️ Answer: 🟢1️⃣
🔒 ❓ Q7.
If (p + q)² = p² + q², which must be true?
🟢1️⃣ One of p or q is 0
🔵2️⃣ p = q
🟡3️⃣ p + q = 0
🟣4️⃣ This is always true
✔️ Answer: 🟢1️⃣
🔒 ❓ Q8.
Which expression represents the difference of two squares?
🟢1️⃣ (a + b)(a − b)
🔵2️⃣ (a + b)²
🟡3️⃣ (a − b)²
🟣4️⃣ a² + b²
✔️ Answer: 🟢1️⃣
🔒 ❓ Q9.
If two numbers differ by 4, their product can be written as:
🟢1️⃣ n² − 4
🔵2️⃣ n² + 4
🟡3️⃣ (n + 4)²
🟣4️⃣ (n − 4)²
✔️ Answer: 🟢1️⃣
🔒 ❓ Q10.
Why do diagonal products in any 2 × 2 calendar block differ by a constant?
🟢1️⃣ Due to a² − b² identity
🔵2️⃣ Due to a² + b²
🟡3️⃣ Due to 2ab
🟣4️⃣ Due to division
✔️ Answer: 🟢1️⃣
🔹 Part B — Short Answer Questions (Q11–Q20)
🔒 ❓ Q11.
Explain why (x + y)² − (x − y)² is divisible by 4.
📌 ✅ Answer:
🔹 (x + y)² = x² + 2xy + y²
🔹 (x − y)² = x² − 2xy + y²
🔹 Difference = (x² + 2xy + y²) − (x² − 2xy + y²)
🔹 Difference = 4xy
🔹 Hence divisible by 4
🔒 ❓ Q12.
Why is the square of any even number divisible by 4?
📌 ✅ Answer:
🔹 Let the even number be 2n
🔹 Square = (2n)²
🔹 (2n)² = 4n²
🔹 Hence divisible by 4
🔒 ❓ Q13.
Compare 99 × 101 and 100² without calculation.
📌 ✅ Answer:
🔹 99 × 101 = (100 − 1)(100 + 1)
🔹 Using identity: a² − b²
🔹 = 100² − 1
🔹 Hence smaller than 100²
🔒 ❓ Q14.
Explain why (a + b)² ≥ a² + b².
📌 ✅ Answer:
🔹 (a + b)² = a² + 2ab + b²
🔹 Term 2ab is non-negative when a, b ≥ 0
🔹 Hence (a + b)² ≥ a² + b²
🔒 ❓ Q15.
Verify whether (x + 4)(x − 4) + 16 = x².
📌 ✅ Answer:
🔹 (x + 4)(x − 4) = x² − 16
🔹 Adding 16 gives x²
🔹 Statement is true
🔒 ❓ Q16.
Why is (p + q)² − (p² + q²) not a perfect square in general?
📌 ✅ Answer:
🔹 (p + q)² = p² + 2pq + q²
🔹 Difference = 2pq
🔹 2pq is not always a square
🔹 Hence not a perfect square
🔒 ❓ Q17.
Explain why (x − 1)(x + 1) is one less than a square.
📌 ✅ Answer:
🔹 (x − 1)(x + 1) = x² − 1
🔹 x² − 1 is one less than x²
🔒 ❓ Q18.
Is (a − b)² equal to a² − b²? Explain.
📌 ✅ Answer:
🔹 (a − b)² = a² − 2ab + b²
🔹 a² − b² does not contain −2ab
🔹 Hence they are not equal
🔒 ❓ Q19.
Why are products of numbers equidistant from a number close to its square?
📌 ✅ Answer:
🔹 Let numbers be n − d and n + d
🔹 Product = (n − d)(n + d)
🔹 = n² − d²
🔹 Difference from n² depends only on d²
🔒 ❓ Q20.
State two reasons why algebraic identities save time.
📌 ✅ Answer:
🔹 They reduce long multiplication
🔹 They give results using known patterns
🔹 Part C — Detailed Answer Questions (Q21–Q30)
🔒 ❓ Q21.
Prove that (a + b)² + (a − b)² = 2(a² + b²).
📌 ✅ Answer:
🔹 (a + b)² = a² + 2ab + b²
🔹 (a − b)² = a² − 2ab + b²
🔹 Adding both expressions:
🔹 a² + 2ab + b² + a² − 2ab + b²
🔹 = 2a² + 2b²
🔹 = 2(a² + b²)
🔒 ❓ Q22.
Derive the identity (a + b)(a − b) = a² − b².
📌 ✅ Answer:
🔹 (a + b)(a − b) = a(a − b) + b(a − b)
🔹 = a² − ab + ab − b²
🔹 = a² − b²
🔒 ❓ Q23.
Show that the difference of squares of two consecutive odd numbers is divisible by 8.
📌 ✅ Answer:
🔹 Let odd numbers be (2n − 1) and (2n + 1)
🔹 Their squares are (2n − 1)² and (2n + 1)²
🔹 Difference = (2n + 1)² − (2n − 1)²
🔹 = 4n² + 4n + 1 − (4n² − 4n + 1)
🔹 = 8n
🔹 Hence divisible by 8
🔒 ❓ Q24.
Explain algebraically why diagonal products in a 2 × 2 calendar differ by a constant.
📌 ✅ Answer:
🔹 Let the top-left number be a
🔹 Other numbers are a + 1, a + 7, a + 8
🔹 First diagonal product = a(a + 8)
🔹 Second diagonal product = (a + 1)(a + 7)
🔹 Difference = a² + 8a − (a² + 8a + 7)
🔹 Difference = −7
🔹 Constant difference obtained
🔒 ❓ Q25.
Prove that (x + y)² + (x − y)² is always even.
📌 ✅ Answer:
🔹 (x + y)² + (x − y)² = 2x² + 2y²
🔹 = 2(x² + y²)
🔹 Hence always even
🔒 ❓ Q26.
Show that (n + 2)² − (n − 2)² = 8n.
📌 ✅ Answer:
🔹 (n + 2)² = n² + 4n + 4
🔹 (n − 2)² = n² − 4n + 4
🔹 Difference = (n² + 4n + 4) − (n² − 4n + 4)
🔹 = 8n
🔒 ❓ Q27.
Verify that (2a + b)² − (2a − b)² = 8ab.
📌 ✅ Answer:
🔹 (2a + b)² = 4a² + 4ab + b²
🔹 (2a − b)² = 4a² − 4ab + b²
🔹 Difference = 8ab
🔒 ❓ Q28.
Explain why (a + b)² − (a² + b²) is never negative for a ≥ 0, b ≥ 0.
📌 ✅ Answer:
🔹 (a + b)² − (a² + b²) = 2ab
🔹 Product of non-negative numbers
🔹 Hence never negative
🔒 ❓ Q29.
Show that the product of two numbers differing by 6 is always 9 less than a square.
📌 ✅ Answer:
🔹 Let numbers be n − 3 and n + 3
🔹 Product = (n − 3)(n + 3)
🔹 = n² − 9
🔒 ❓ Q30.
Explain why identities are preferred over direct multiplication for large numbers.
📌 ✅ Answer:
🔹 Fewer calculation steps
🔹 Reduced chances of error
🔹 Faster mental computation
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