Chapter 6: We Distribute, Yet Things Multiply - Question Bank

Chapter 6: We Distribute, Yet Things Multiply

Class 8 Mathematics – NCERT Ganita Prakash

Complete Interactive Question Bank

Section A: Multiple Choice Questions (20 Questions)

1.

What is the expanded form of \((x + 4)(x + 3)\)?

a) \(x^2 + 7x + 12\)

b) \(x^2 + 12x + 7\)

c) \(x^2 + 7x + 7\)

d) \(x^2 + 12x + 12\)

a) \(x^2 + 7x + 12\)

Using distributive property: \((x+4)(x+3) = x(x+3) + 4(x+3) = x^2 + 3x + 4x + 12 = x^2 + 7x + 12\)

2.

The expression \((a - 7)^2\) equals:

a) \(a^2 - 49\)

b) \(a^2 - 14a + 49\)

c) \(a^2 + 14a + 49\)

d) \(a^2 - 7a + 49\)

b) \(a^2 - 14a + 49\)

Using identity: \((a-b)^2 = a^2 - 2ab + b^2\). Here \(b=7\), so \(a^2 - 2×a×7 + 49 = a^2 - 14a + 49\)

3.

Which identity is used to find \(98 \times 102\) quickly?

a) \((a + b)^2\)

b) \((a - b)^2\)

c) \((a + b)(a - b)\)

d) \(a(b + c)\)

c) \((a + b)(a - b)\)

\(98 \times 102 = (100-2)(100+2) = 100^2 - 2^2\). This uses difference of squares identity.

4.

The product \(45 \times 55\) can be written as:

a) \((50 - 5)(50 + 5)\)

b) \((40 + 5)(50 + 5)\)

c) \((50 + 5)^2\)

d) \((50 - 5)^2\)

a) \((50 - 5)(50 + 5)\)

\(45 = 50-5\) and \(55 = 50+5\), so \(45×55 = (50-5)(50+5)\)

5.

If \(a\) and \(b\) are integers, \((a - b)^2\) is always equal to:

a) \((b - a)^2\)

b) \(-(b - a)^2\)

c) \(a^2 - b^2\)

d) \(b^2 - a^2\)

a) \((b - a)^2\)

Squaring removes the sign: \((a-b)^2 = [-(b-a)]^2 = (b-a)^2\)

6.

The value of \(101^2\) using the identity is:

a) 10201

b) 10001

c) 10101

d) 11001

a) 10201

\(101^2 = (100+1)^2 = 100^2 + 2×100×1 + 1^2 = 10000 + 200 + 1 = 10201\)

7.

The expression \(3p(2q - 5)\) expands to:

a) \(6pq - 15p\)

b) \(6pq - 5\)

c) \(3p - 15q\)

d) \(6pq + 15p\)

a) \(6pq - 15p\)

\(3p(2q-5) = 3p×2q - 3p×5 = 6pq - 15p\)

8.

If the product of two numbers is \(ab\), and both are increased by 1, the new product is:

a) \(ab + 1\)

b) \(ab + a + b + 1\)

c) \(ab + a + b\)

d) \(ab + 2\)

b) \(ab + a + b + 1\)

\((a+1)(b+1) = ab + a + b + 1\)

9.

The product \((x + 2)(x - 2)\) simplifies to:

a) \(x^2 - 4\)

b) \(x^2 + 4\)

c) \(x^2 - 2\)

d) \(x^2 + 2x - 4\)

a) \(x^2 - 4\)

Using difference of squares: \((x+2)(x-2) = x^2 - 2^2 = x^2 - 4\)

10.

Which of these is NOT an identity?

a) \(a(b + c) = ab + ac\)

b) \(a^2 - b^2 = (a + b)(a - b)\)

c) \(a + b = b + a\)

d) \(a^2 + b^2 = (a + b)^2\)

d) \(a^2 + b^2 = (a + b)^2\)

\((a+b)^2 = a^2 + 2ab + b^2\), not \(a^2 + b^2\)

11.

The increase in \(23 \times 27\) if 27 is increased by 1 is:

a) 23

b) 27

c) 1

d) 50

a) 23

If second number increases by 1, product increases by first number

12.

The expression \((2x + 5)^2\) equals:

a) \(4x^2 + 25\)

b) \(4x^2 + 10x + 25\)

c) \(4x^2 + 20x + 25\)

d) \(4x^2 + 20x + 10\)

c) \(4x^2 + 20x + 25\)

\((2x+5)^2 = (2x)^2 + 2×2x×5 + 5^2 = 4x^2 + 20x + 25\)

13.

Which pattern follows \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\)?

a) Pattern 1 from the chapter

b) Pattern 2 from the chapter

c) Pattern for 11, 101, 1001

d) Coin triangle pattern

a) Pattern 1 from the chapter

Pattern 1 shows \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\)

14.

The product \(3874 \times 11\) using distributive property is:

a) 42614

b) 42514

c) 42624

d) 42714

a) 42614

\(3874×11 = 3874×(10+1) = 38740 + 3874 = 42614\)

15.

If \(m + n = 10\) and \(mn = 21\), then \(m^2 + n^2 = ?\)

a) 58

b) 100

c) 42

d) 79

a) 58

\(m^2+n^2 = (m+n)^2 - 2mn = 100 - 42 = 58\)

16.

Which is equivalent to \(k^2 + 2k\)?

a) \((k+1)^2 - 1\)

b) \(k(k+1) + k\)

c) \(k(k+2)\)

d) All of these

d) All of these

All simplify to \(k^2 + 2k\)

17.

The product \((a + 3)(a - 4)\) expands to:

a) \(a^2 - a - 12\)

b) \(a^2 + a - 12\)

c) \(a^2 - 12\)

d) \(a^2 + 7a - 12\)

a) \(a^2 - a - 12\)

\((a+3)(a-4) = a^2 - 4a + 3a - 12 = a^2 - a - 12\)

18.

To multiply by 101 quickly, we write the number as:

a) \(\times (100 + 1)\)

b) \(\times (10 + 1)\)

c) \(\times 100 + 1\)

d) \(\times 101 + 0\)

a) \(\times (100 + 1)\)

\(101 = 100 + 1\), so use distributive property

19.

The expression \((x - y)^2 + (x + y)^2\) simplifies to:

a) \(2x^2 + 2y^2\)

b) \(x^2 + y^2\)

c) \(4xy\)

d) \(2xy\)

a) \(2x^2 + 2y^2\)

\((x-y)^2 + (x+y)^2 = (x^2-2xy+y^2) + (x^2+2xy+y^2) = 2x^2 + 2y^2\)

20.

Which method is NOT used in "Mind the Mistake"?

a) Correcting sign errors

b) Expanding brackets

c) Using Pythagorean theorem

d) Combining like terms

c) Using Pythagorean theorem

"Mind the Mistake" deals with algebraic errors, not geometry theorems

Section B: Assertion & Reasoning Questions (20 Questions)

1.

Assertion (A): \((a + b)^2 = a^2 + 2ab + b^2\) for all integers \(a, b\).
Reason (R): The distributive property holds for integers.

a) Both A and R are true and R explains A.

b) Both A and R are true but R does not explain A.

c) A is true but R is false.

d) A is false but R is true.

a) Both A and R are true and R explains A.

The distributive property is used to derive \((a+b)^2 = a^2 + 2ab + b^2\)

2.

A: \((a - b)^2\) and \((b - a)^2\) are equal.
R: Squaring a negative gives a positive.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

\((a-b)^2 = [-(b-a)]^2 = (b-a)^2\) because square of negative is positive

3.

A: \(99 \times 101 = 9999\).
R: \(99 \times 101 = (100-1)(100+1) = 100^2 - 1^2\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows the method using difference of squares identity

4.

A: The product \(23 \times 27\) increases by 23 if 27 is increased by 1.
R: \(a(b+1) = ab + a\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R gives the general rule that explains A

5.

A: \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\) is an identity.
R: It follows from adding the two square identities.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

Adding \((a+b)^2\) and \((a-b)^2\) gives \(2(a^2+b^2)\)

6.

A: \(3874 \times 101 = 391374\).
R: \(3874 \times 101 = 387400 + 3874\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows the calculation method using distributive property

7.

A: \((x+2)(x+5) = x^2 + 7x + 10\).
R: Using distributive property: \(x(x+5) + 2(x+5)\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R shows how to expand using distributive property to get A

8.

A: \((a+b)^2\) is always greater than \(a^2 + b^2\).
R: Because \(2ab\) is always positive.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

d) A false, R true.

A is false because if a or b is negative, \(2ab\) can be negative. R is true about squares of negatives.

9.

A: The pattern \(k^2 + 2k\) can be written as \(k(k+2)\).
R: Both represent the same algebraic expression.

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

a) Both true, R explains A.

R explains why they are equivalent expressions

10.

A: The product of two numbers remains same if one is increased by 2 and other decreased by 2.
R: \((a+2)(b-2) = ab + 2b - 2a - 4\).

a) Both true, R explains A.

b) Both true, R does not explain A.

c) A true, R false.

d) A false, R true.

d) A false, R true.

A is false - product changes unless special case. R is true as expansion.

11.

A: \((a+b)(a-b) = a^2 - b^2\) for all real numbers.
R: This is the difference of squares identity.

a) Both true, R explains A.

R names the identity that A represents

12.

A: \(104^2 = 10816\).
R: \(104^2 = (100+4)^2 = 100^2 + 2×100×4 + 4^2\).

a) Both true, R explains A.

R shows the calculation using identity to get A

13.

A: The product decreases when one number increases by 1 and other decreases by 1 if \(b < a + 1\).
R: \((a+1)(b-1) = ab + b - a - 1\).

a) Both true, R explains A.

From R: increase is \(b-a-1\), so product decreases if \(b-a-1 < 0\) i.e., \(b < a+1\)

14.

A: \(97^2 = 9409\).
R: \(97^2 = (100-3)^2 = 100^2 - 2×100×3 + 3^2\).

a) Both true, R explains A.

R shows calculation using \((a-b)^2\) identity

15.

A: \(2(5^2 + 6^2) = (11)^2 + (1)^2\).
R: Using identity \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\) with \(a=6,b=5\).

a) Both true, R explains A.

R applies the pattern identity to verify A

16.

A: \((a+b)^2 - (a-b)^2 = 4ab\).
R: Subtracting the two square identities gives this result.

a) Both true, R explains A.

R explains how to derive A from known identities

17.

A: \((x+1)^3 = x^3 + 3x^2 + 3x + 1\).
R: \((x+1)^3 = (x+1)(x+1)^2 = (x+1)(x^2+2x+1)\).

a) Both true, R explains A.

R shows the expansion steps to get A

18.

A: The distributive property can be used for subtraction: \(a(b-c) = ab - ac\).
R: Subtraction is adding the negative: \(a(b-c) = a[b + (-c)]\).

a) Both true, R explains A.

R explains why distributive property works for subtraction

19.

A: \(73^2 = 5329\).
R: \(73^2 = (70+3)^2 = 70^2 + 2×70×3 + 3^2\).

a) Both true, R explains A.

R shows the decomposition method to calculate square

20.

A: The product \(46 \times 54\) is 2484.
R: \(46 \times 54 = (50-4)(50+4) = 50^2 - 4^2\).

a) Both true, R explains A.

R uses difference of squares to calculate product efficiently

Section C: True/False Questions (10 Questions)

1.

\((p + q)^2 = p^2 + q^2\)

False

Correct identity: \((p+q)^2 = p^2 + 2pq + q^2\)

2.

\((a - b)^2 = (b - a)^2\)

True

Squaring removes the sign difference

3.

\(a(b + c) = ab + ac\) for all real numbers

True

This is the distributive property

4.

\(11 \times 12 = 132\) can be found using \(10 \times 12 + 1 \times 12\)

True

\(11 \times 12 = (10+1) \times 12 = 10 \times 12 + 1 \times 12 = 120 + 12 = 132\)

5.

\((x + 3)(x - 3) = x^2 - 9\)

True

Difference of squares: \((x+3)(x-3) = x^2 - 3^2 = x^2 - 9\)

6.

\(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\)

True

Adding the two square identities gives this result

7.

The product \(23 \times 27\) increases by 50 if both numbers increase by 1

False

Increase is \(a + b + 1 = 23 + 27 + 1 = 51\), not 50

8.

\(101^2 = 10201\)

True

\((100+1)^2 = 10000 + 200 + 1 = 10201\)

9.

\(k^2 + 2k = (k+1)^2 - 1\)

True

\((k+1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k\)

10.

\((a+b)(a-b) = a^2 + b^2\)

False

Should be \(a^2 - b^2\)

Section D: Short Answer I (2 Marks × 15 Questions)

1.

Expand: \((3x + 2)(x + 5)\)

\(3x^2 + 17x + 10\)

\((3x+2)(x+5) = 3x(x+5) + 2(x+5) = 3x^2 + 15x + 2x + 10 = 3x^2 + 17x + 10\)

2.

Find \(96^2\) using \((a - b)^2\) identity.

9216

\(96^2 = (100-4)^2 = 100^2 - 2×100×4 + 4^2 = 10000 - 800 + 16 = 9216\)

3.

Simplify: \(4a(3b - 2) + 5a\)

\(12ab - 3a\)

\(4a(3b-2) + 5a = 12ab - 8a + 5a = 12ab - 3a\)

4.

How much does \(23 \times 27\) increase if 23 is increased by 1?

27

When first number increases by 1, product increases by second number

5.

Verify \(2(a^2 + b^2) = (a+b)^2 + (a-b)^2\) for \(a=5, b=3\).

LHS: \(2(25+9)=68\), RHS: \((8)^2 + (2)^2 = 64+4=68\) ✓

Both sides equal 68, identity verified

6.

Multiply \(48 \times 52\) using a suitable identity.

2496

\(48×52 = (50-2)(50+2) = 50^2 - 2^2 = 2500 - 4 = 2496\)

7.

Expand \((2m + 3n)^2\).

\(4m^2 + 12mn + 9n^2\)

\((2m+3n)^2 = (2m)^2 + 2×2m×3n + (3n)^2 = 4m^2 + 12mn + 9n^2\)

8.

Find \(999^2\) using \((1000 - 1)^2\).

998001

\(999^2 = (1000-1)^2 = 1000000 - 2000 + 1 = 998001\)

9.

Simplify: \((x+4)^2 - (x-4)^2\)

\(16x\)

\((x^2+8x+16) - (x^2-8x+16) = 16x\)

10.

Expand: \((a + b - c)(a + b + c)\)

\(a^2 + 2ab + b^2 - c^2\)

Consider \((a+b)\) as single term: \((a+b)^2 - c^2 = a^2+2ab+b^2-c^2\)

11.

Find the product: \(234 \times 11\) using distributive method.

2574

\(234×11 = 234×(10+1) = 2340 + 234 = 2574\)

12.

If \(a = 7, b = 3\), find \((a+b)^2 - (a-b)^2\).

84

\((10)^2 - (4)^2 = 100 - 16 = 84\) or using identity: \(4ab = 4×7×3 = 84\)

13.

Write \(k^2 + 2k\) in two other equivalent forms.

\((k+1)^2 - 1\) and \(k(k+2)\)

Both simplify to \(k^2 + 2k\)

14.

Expand: \((p - 8)(p + 8)\)

\(p^2 - 64\)

Difference of squares: \(p^2 - 8^2 = p^2 - 64\)

15.

Find \(73^2\) using \((70 + 3)^2\).

5329

\(73^2 = (70+3)^2 = 70^2 + 2×70×3 + 3^2 = 4900 + 420 + 9 = 5329\)

Section E: Short Answer II (3 Marks × 10 Questions)

1.

Expand \((a + b)(a^2 + 2ab + b^2)\) and simplify.

\(a^3 + 3a^2b + 3ab^2 + b^3\)

\((a+b)(a^2+2ab+b^2) = a(a^2+2ab+b^2) + b(a^2+2ab+b^2) = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 = a^3 + 3a^2b + 3ab^2 + b^3\)

2.

Multiply \(3874 \times 101\) using distributive property in one line.

391374

\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)

3.

Show geometrically that \((a+b)^2 = a^2 + 2ab + b^2\) (describe with diagram).

Draw square of side (a+b). It consists of: one square of side a (area a²), one square of side b (area b²), and two rectangles of sides a and b (each area ab). Total area = a² + b² + 2ab = (a+b)².

Geometric proof using area decomposition

4.

If \(x = 8, y = 3\), find the area of the shaded region from page 95.

25

Area = \((n-m)^2\) where n=8, m=3 gives \((8-3)^2 = 5^2 = 25\)

5.

Prove: \((m+n)^2 - 4mn = (n-m)^2\).

LHS: \(m^2+2mn+n^2 - 4mn = m^2 - 2mn + n^2 = (n-m)^2\)

Expanding and simplifying gives the identity

6.

Find three examples where product decreases when one number is increased by 1 and the other decreased by 1.

1. (2,8)→(3,7): 16→21 (actually increases)
2. (5,5)→(6,4): 25→24 (decreases)
3. (10,1)→(11,0): 10→0 (decreases)

Product decreases when \(b < a+1\)

7.

Verify Pattern 1: \(2(5^2 + 6^2) = (11)^2 + (1)^2\).

LHS: \(2(25+36)=122\), RHS: \(121+1=122\) ✓

Using identity \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\) with a=6,b=5

8.

Expand \((a - b)(a^3 + a^2b + ab^2 + b^3)\) and find the pattern.

\(a^4 - b^4\)

Pattern: \((a-b)(a^n + a^{n-1}b + ... + b^n) = a^{n+1} - b^{n+1}\)

9.

Correct the mistake: \((5m + 6n)^2 = 25m^2 + 36n^2\).

Correct: \(25m^2 + 60mn + 36n^2\)

Missing middle term \(2×5m×6n = 60mn\)

10.

Use Identity 1C to find \(45 \times 55\).

2475

\(45×55 = (50-5)(50+5) = 50^2 - 5^2 = 2500 - 25 = 2475\)

Section F: Long Answer Questions (5 Marks × 10 Questions)

1.

Derive all three identities: \((a+b)^2, (a-b)^2, (a+b)(a-b)\) using distributive property.

1. \((a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\)
2. \((a-b)^2 = (a-b)(a-b) = a(a-b) - b(a-b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2\)
3. \((a+b)(a-b) = a(a-b) + b(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2\)

All derived using distributive property \(a(b+c) = ab + ac\)

2.

A park has two square green plots each of side \(g\) and a walking path of width \(w\) around them. Write an expression for the tiled area.

Tiled area = \((2g+2w)^2 - 2g^2 = 4g^2 + 8gw + 4w^2 - 2g^2 = 2g^2 + 8gw + 4w^2\)

Total area minus green area gives tiled area

3.

For the pattern in "This Way or That Way" (p. 96), show that all four expressions simplify to \(k^2 + 2k\).

1. \((k+1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k\)
2. \(k^2 + 2k\) (given)
3. \(k(k+1) + k = k^2 + k + k = k^2 + 2k\)
4. \(k(k+2) = k^2 + 2k\)

All four methods give same algebraic expression

4.

Explain the method for fast multiplication by 11 and 101 with two examples each.

By 11: Multiply by (10+1): \(123×11 = 1230+123 = 1353\), \(456×11 = 4560+456 = 5016\)
By 101: Multiply by (100+1): \(123×101 = 12300+123 = 12423\), \(456×101 = 45600+456 = 46056\)

Use distributive property: \(a×11 = a×(10+1)\), \(a×101 = a×(100+1)\)

5.

From "Mind the Mistake," identify and correct any 5 errors with explanations.

1. \(-3p(-5p+2q) = 15p^2 - 6pq\) (not \(-3p+5p-2q\))
2. \(2(x-1)+3(x+4) = 2x-2+3x+12 = 5x+10\) (not \(5x+3\))
3. \(y+2(y+2) = y+2y+4 = 3y+4\) (not \((y+2)^2\))
4. \((5m+6n)^2 = 25m^2 + 60mn + 36n^2\) (not \(25m^2+36n^2\))
5. \(5w^2 + 6w\) cannot combine (not \(11w^2\))

Common errors: forgetting to multiply all terms, incorrect expansion of squares, combining unlike terms

6.

Prove algebraically that the diagonal products in a 2×2 calendar square differ by 7.

Numbers: \(a, a+1, a+7, a+8\)
Diagonal products: \(a(a+8) = a^2+8a\) and \((a+1)(a+7) = a^2+8a+7\)
Difference: \((a^2+8a+7) - (a^2+8a) = 7\)

The difference is always 7 regardless of \(a\)

7.

If a number leaves remainder 3 when divided by 7 and another leaves remainder 5, find remainders for their sum, difference, and product when divided by 7.

Let \(A = 7x+3\), \(B = 7y+5\)
Sum: \(A+B = 7(x+y) + 8 = 7(x+y+1) + 1\) → remainder 1
Difference: \(A-B = 7(x-y) - 2 = 7(x-y-1) + 5\) → remainder 5
Product: \(AB = (7x+3)(7y+5) = 49xy + 35x + 21y + 15 = 7(7xy+5x+3y+2) + 1\) → remainder 1

Using algebraic representation of numbers with remainders

8.

Show that \((6n+2)^2 - (4n+3)^2\) is 5 less than a perfect square.

\((6n+2)^2 - (4n+3)^2 = (10n+5)(2n-1) = 20n^2 - 5\)
Add 5: \(20n^2\) which is a perfect square when \(n\) is an integer

Using difference of squares and simplification

9.

Find \(406^2, 72^2, 145^2\) using suitable identities.

\(406^2 = (400+6)^2 = 160000 + 4800 + 36 = 164836\)
\(72^2 = (70+2)^2 = 4900 + 280 + 4 = 5184\)
\(145^2 = (150-5)^2 = 22500 - 1500 + 25 = 21025\)

Decompose numbers into easy-to-square parts

10.

Explore the coin triangle flipping problem: Find minimum moves for 15 coins and generalize for triangular number \(T_n\).

For triangle with \(T_n = n(n+1)/2\) coins:
- 3 coins (n=2): 1 move
- 6 coins (n=3): 2 moves
- 10 coins (n=4): 3 moves
- 15 coins (n=5): 5 moves
Pattern: Minimum moves = ⌊n/2⌋ for n>1

Triangular number pattern with practical problem-solving

Section G: Case-Based Questions (5 Cases × 4 Sub-Questions Each)

Case 1: Fast Multiplication Tricks

Rahul learns that the distributive property can be used to multiply numbers quickly. He sees the example: \( 3874 \times 11 = 3874 \times (10 + 1) = 38740 + 3874 = 42614 \). He also learns that for a 4-digit number \( dcba \): \( dcba \times 101 = dcba \times (100 + 1) = dcba00 + dcba \).

1.

Which property is Rahul using here?

a) Commutative property

b) Distributive property

c) Associative property

d) Identity property

b) Distributive property

Using \(a(b+c) = ab + ac\)

2.

What is \(495 \times 11\) using this method?

a) 5445

b) 5440

c) 4945

d) 4950

a) 5445

\(495 \times 11 = 495 \times (10+1) = 4950 + 495 = 5445\)

3.

Using the rule for multiplying by 101, what is \(3874 \times 101\)?

a) 391374

b) 387400

c) 391474

d) 387474

a) 391374

\(3874 \times 101 = 3874 \times (100+1) = 387400 + 3874 = 391374\)

4.

Which is NOT true about multiplying by 11?

a) You add the number to itself shifted left by one digit

b) It works for any number of digits

c) It only works for 2-digit numbers

d) It uses \( \times (10 + 1) \)

c) It only works for 2-digit numbers

The method works for numbers with any number of digits

Case 2: Calendar Number Patterns

In a calendar, Priya marks a 2×2 square of numbers. She labels the top-left number as \( a \), so the square becomes:

| \( a \) | \( a+1 \) |
| \( a+7 \) | \( a+8 \) |

1.

What is the difference between the diagonal products?

a) 1

b) 7

c) 8

d) 0

b) 7

\(a(a+8) - (a+1)(a+7) = a^2+8a - (a^2+8a+7) = -7\), absolute difference = 7

2.

For \( a = 9 \), what are the diagonal products?

a) 153 and 160

b) 144 and 153

c) 135 and 144

d) 126 and 135

a) 153 and 160

\(9×17=153\), \(10×16=160\)

3.

The algebraic expression for the difference of diagonal products is:

a) \((a+8)(a) - (a+1)(a+7)\)

b) \((a+1)(a+7) - a(a+8)\)

c) \(a(a+8) - (a+1)(a+7)\)

d) \((a+7)(a+8) - a(a+1)\)

c) \(a(a+8) - (a+1)(a+7)\)

This represents product of top-left & bottom-right minus product of top-right & bottom-left

4.

When simplified, this difference always equals:

a) 5

b) 7

c) 10

d) 14

b) 7

\(a(a+8) - (a+1)(a+7) = a^2+8a - (a^2+8a+7) = -7\), so absolute difference = 7

Case 3: Geometric Proof of Identities

A square of side length \( (m+n) \) is drawn. Four rectangles of dimensions \( m \times n \) are removed from the corners, leaving a smaller shaded square in the center.

1.

What is the area of the large square?

a) \( m^2 + n^2 \)

b) \( (m+n)^2 \)

c) \( m^2 + 2mn + n^2 \)

d) Both b and c

d) Both b and c

\((m+n)^2 = m^2+2mn+n^2\)

2.

What is the total area of the four removed rectangles?

a) \( mn \)

b) \( 2mn \)

c) \( 4mn \)

d) \( m^2n^2 \)

c) \( 4mn \)

Four rectangles each of area \( mn \)

3.

The area of the shaded square is:

a) \( (m-n)^2 \)

b) \( (n-m)^2 \)

c) \( m^2 + n^2 - 2mn \)

d) All of these

d) All of these

All are equivalent expressions

4.

If \( m = 3 \) and \( n = 7 \), what is the area of the shaded square?

a) 16

b) 25

c) 36

d) 49

a) 16

\((7-3)^2 = 4^2 = 16\)

Case 4: Algebraic Pattern Recognition

Anika observes this pattern in her notebook:
\( 2(2^2 + 1^2) = 3^2 + 1^2 \)
\( 2(3^2 + 1^2) = 4^2 + 2^2 \)
\( 2(5^2 + 3^2) = 8^2 + 2^2 \)
She realizes this follows the identity: \( 2(a^2 + b^2) = (a+b)^2 + (a-b)^2 \)

1.

For \( a = 6, b = 4 \), what is \( 2(a^2 + b^2) \)?

a) 100

b) 104

c) 124

d) 144

b) 104

\(2(36+16) = 2×52 = 104\)

2.

Using the identity, \( (a+b)^2 + (a-b)^2 \) for \( a=6, b=4 \) is:

a) 100 + 4

b) 104 + 4

c) 100 + 4

d) 104

d) 104

\((10)^2 + (2)^2 = 100 + 4 = 104\)

3.

Does this identity work for negative numbers? For \( a = -3, b = 5 \):

a) Yes, both sides equal 68

b) Yes, both sides equal 34

c) No, it only works for positive numbers

d) No, squares of negatives are positive

a) Yes, both sides equal 68

LHS: \(2(9+25)=68\), RHS: \((2)^2+(-8)^2=4+64=68\)

4.

Another identity in the same chapter is:

a) \( (a+b)^3 = a^3 + b^3 \)

b) \( a^2 - b^2 = (a+b)(a-b) \)

c) \( a(b+c) = ab + c \)

d) \( (a-b)^2 = a^2 + b^2 \)

b) \( a^2 - b^2 = (a+b)(a-b) \)

This is Identity 1C (difference of squares)

Case 5: Error Analysis in Algebra

Mr. Sharma gives his class these expansions to check:
1. \( (5m + 6n)^2 = 25m^2 + 36n^2 \)
2. \( ab^2 + a^2b + a^2b^2 = ab(a + b + ab) \)
3. \( -3p(-5p + 2q) = -3p + 5p - 2q \)

1.

What is the correct expansion of \( (5m + 6n)^2 \)?

a) \( 25m^2 + 36n^2 \)

b) \( 25m^2 + 60mn + 36n^2 \)

c) \( 5m^2 + 30mn + 6n^2 \)

d) \( 25m^2 + 30mn + 36n^2 \)

b) \( 25m^2 + 60mn + 36n^2 \)

Missing middle term \(2×5m×6n = 60mn\)

2.

The correct simplification of \( ab^2 + a^2b + a^2b^2 \) is:

a) \( ab(a + b + ab) \)

b) \( ab(b + a + ab) \)

c) \( ab(ab + a + b) \)

d) Cannot be factored further

b) \( ab(b + a + ab) \)

Factor \(ab\) from each term: \(ab(b + a + ab)\)

3.

The error in \( -3p(-5p + 2q) \) is:

a) Sign error

b) Forgot to multiply

c) Added instead of multiplied

d) All of these

d) All of these

Multiple errors: sign, multiplication, addition

4.

Which mistake is most common when expanding \( (a+b)^2 \)?

a) Writing \( a^2 + b^2 \)

b) Writing \( a^2 + 2ab - b^2 \)

c) Writing \( a^2 + ab + b^2 \)

d) Writing \( 2a + 2b \)

a) Writing \( a^2 + b^2 \)

Most common error is forgetting the middle term \(2ab\)

Class 8 Maths (Ganita Prakash) - Chapter 6 We DISTRIBUTE YET THINGS MULTIPLY Figure it out Q & Answers

 

Chapter 6
We DISTRIBUTE YET THINGS MULTIPLY
Figure it out Q & Answers

Figure it Out 142 

1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid

2. Expand the following products.
(i) (3 + u) (v – 3) 
(ii) 2/3 (15 + 6a) = 10 +4a
(iii) (10a + b) (10c + d) 
(iv) (3 – x) (x – 6) 
(v) (–5a + b) (c + d) 
(vi) (5 + z) (y + 9)

ANSWER: 

(i) (3 + u) (v – 3) = 3v – 3u + uv -9

(ii) 2/3 (15 + 6a) = 10 +4a

(iii) (10a + b) (10c + d)  = 100ac+10ad+10bc+bd

(iv) (3 – x) (x – 6) = -x² + 9x – 18

(v) (–5a + b) (c + d) = -5ac -5ad + bc + bd

(vi) (5 + z) (y + 9) = 5y+45+yz+9z

3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.

Solution: 

Let the numbers be a and b. 

Then, ab = (a + 2)(b - 4) 

⇒ ab = ab - 4a +2b - 8 

=> ab - ab + 4a + 8 = 2b

4a + 8 = 2b  [Divide throughout by 2]

 2a + 4 = b 

For a = 1 , b = 2 * 1 + 4 = 6 

ab = 1 * 6 = 6 

and (a+2) (b-4)=3×2=6 

 Hence, ab = (a + 2)(b - 4) Let a = 2 then b = 2 * 2 + 4 = 8 

Let a = 3 then b = 2 * 3 + 4 = 10 

Three such pairs are 1 and 6, 2 and 8, 3 and 10.

4. Expand (i) (a + ab – 3b²) (4 + b), and (ii) (4y + 7) (y + 11z – 3)

Solution:

 (i) (a + ab-3b²) (4+b)

= (4 + b)(a + ab - 3b²)

= 4(a + ab - 3b²) + b(a + ab - 3b²) 

= 4a + 4ab - 12b² + ab + ab² - 3b³ 

= 4a + 5ab - 12b² + ab² - 3b³

(ii) (4y + 7)(y + 11z - 3) 

= 4y(y + 11z - 3) + 7(y + 11z - 3) 

= 4y²  + 44yz - 12y + 7y + 77z - 21

= 4y² + 44yz - 5y + 77z - 21

5. Expand (i) (a – b) (a + b), (ii) (a – b) (a² + ab + b²) and (iii) (a – b)(a³ + a²b + ab² + b³), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?

Solution: (i) (a - b)(a + b) 

= a(a + b) - b(a + b) 

= a² + ab - ab – b²

= a² - b² 

(ii) (a - b)(a² + ab + b²) 

= a(a² + ab + b²) - b(a² + ab + b²) 

= a³ + a²b + ab² - a²b - a * b² - b³

= a³ - b³

(iii) (a - b)(a³ + a²b + ab² + b³) 

= a(a³ + a² * b + ab² + b³) -b(a³+a²b+ ab² + b³ ) 

= 𝑎^4  + a³b + a²b² + ab³ - a³b - a²b² -ab³ - 𝑏^4 

= 𝑎^4  - 𝑏^4 

The pattern follows

 (a-b)(𝑎^𝑛 + 𝑎^(𝑛−1) b+...+ 𝑏^𝑛 )= 𝑎^(𝑛+1)  - 𝑏^(𝑛+1) 

Next identity in the pattern

 (a - b)(𝑎^4 + a³b + a²b² + ab³ + 𝑏^4) = 𝑎^5 - 𝑏^5

Figure it Out 149

1.Which is greater: (a – b)² or (b – a)²? Justify your answer. 

Solution:

 (a - b)² = a² + b² - 2ab ... (1)

 and 

(b - a)² = b² + a² - 2ba

b² + a²= a² + b² 

and

 ba = ab 

:. (b - a)² = a² + b² - 2ab ... (2)

 Comparing (1) and (2), we get (a - b)² = (b - a)²

2.Express 100 as the difference of two squares. 

Solution:

 a² - b² = 100 

(a + b)(a - b) = 100 

[100=1×100, 2 × 50, 4 x 25, 5 x 20, 10 × 10] 

take any one 

Let us take

 50 * 2 = 100 

Hence, (a + b)(a - b) = 50 * 2

a + b = 50 ... (1)

a - b = 2 ... (2)

 Adding (1) and (2) 

2a = 52

Substituting a = 26in(1)  

⇒ 26 + b = 50 

b = 50 - 26 = 24

 Let us check 

26² - 24²

676 – 576 = 100 

Hence 

26² - 24² = 100 

a = 26

3. Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far. 

Solution:

(1) 406² = (400 + 6)² = 400² + 2 * 400 * 6 + 6² = 160000 + 4800 + 36 = 164836

(ii) 72² = (50 + 22)² = 50² + 2 * 50 * 22 + 22² =2500 + 2200 + 484 = 5184

(iii) 145² = (150 - 5)² = 150² - 2 * 150 * 5 + 5² =22500-1500+25 = 21025

 (iv) 1097² = (1100-3)² =1100² - 2×1100 × 3 + 3² = 1210000 – 6600 + 9 = 1203409

(ν) 124² = (100 + 24)² = 10000 + 4800 + 576 = 15376

4. Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer

Pattern 12 (a² + b² ) = (a+b)² + (a – b)²

Case I Let a = 4, b= 2

LHS = 2(4²+2²) = 2x (16+4) = 2 * 20 = 40

RHS = (4 + 2)² + (4 - 2)² = 36 + 4 = 40 

LHS = RHS, Pattern 1 holds for counting numbers

Case II Let a = -4, b = -2

LHS = 2((-4)²+(-2)²) = 2x (16+4) = 2 * 20 = 40

RHS = (- 4 + (- 2))² + (- 4 - (- 2))² = (- 4 - 2)² + (- 4 + 2)² 

= (- 6)² + (- 2)² =36+4=40 

LHS = RHS 

Pattern 1 holds for negative integers also. 

Case-III Let a =1/2, b = 1/3 

LHS 

= 2 [ (1/2)²  + (1/3)² ] 

= 2 [ 1/4   + 1/9 ] 

= 2 * 13/36   

= 13/18   

RHS 

(1/2  + 1/3)² + (1/2  - 1/3)² 

= (5/6)² + (1/6)² 

= 25/36 + 1/36

= 26/36 = 13/18

Pattern 1 holds for fractions also.Pattern 2  (a² - b² ) = ( a + b )(a – b)

Case I Let a = 5, b= 3

 LHS   5² - 3² = 25 – 9 = 16 

RHS  (5+3) (5-3)=8×2=16 . 

LHS = RHS  Pattern 2 holds for counting numbers.

Case II Let a = - 5, b = - 3

LHS = (-5)² - (-3)² = 25 – 9 = 16

RHS = [(- 5) + (- 3)][(-5)-(-3)] =(-8)(-2) = 16

LHS = RHS 

Pattern 2 holds for negative integers also. 

Case-III Let a =1/2, b = 1/3 

LHS 

=  [ (1/2)²  - (1/3)² ] 

=  [ 1/4   - 1/9 ] 

=  5/36   

RHS 

(1/2  + 1/3)(1/2  - 1/3)

= (5/6)(1/6) = 5/36   

Pattern 2 holds for fractions also.

Figure it Out 154-156

1. Compute these products using the suggested identity. 
(i) (46²) using Identity 1A: ((a + b)² = a² + 2ab + b²)

Let (a = 40), (b = 6):[ 46² = (40 + 6)² = 40² + 2*40*6 + 6² = 1600 + 480 + 36 = 2116]

(ii) 397 × 403 using Identity 1C for (a + b) (a – b) 

Let (a = 400), (b = 3):[ 397 X 403 = (400 - 3)(400 + 3) = 400² - 3² = 160000 - 9 = 159991]

(iii) 91² using Identity 1B for (a – b)²

Let (a = 100), (b = 9):[ 91² = (100 - 9)² = 100² - 2*100*9 + 9² = 10000 - 1800 + 81 = 8281 ]

 (iv) 43 × 45 using Identity 1C for (a + b) (a – b)

Let (a = 44), (b = 1):[ (43)(45) = (44 - 1)(44 + 1) = 44² - 1² = 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) → Identity 1C  = p² + 11p - p - 11 = p² + 10p - 11

 

(ii) (3a – 9b) (3a + 9b) → Identity 1C  = (3a)² - (9b)² = 9a² - 81b²

(iii) –(2y + 5) (3y + 4) → Distributive  = - [6y² + 8y + 15y + 20] = - [6y² + 23y + 20] = [-6y² - 23y - 20 ]

(iv) (6x + 5y)² = → Identity 1A  = 36x² + 2 * 6x * 5y + 25y² = 36x² + 60xy + 25y² 

(v) (2x – 1/2 )² → Identity 1B  = (2x)² - 2*2x *1/2 + (1/2)² = 4x² - 2x + 1/4 

(vi) (7p) × (3r) × (p + 2) → Distributive  = 21pr(p + 2) = 21p²r + 42pr = 21pr² + 42pr

3. For each statement identify the appropriate algebraic expression(s).

(i) Two more than a square number. 

2 + s (s + 2)² s² + 2 ANSWER S² + 4 2s²

 (ii) The sum of the squares of two consecutive numbers 

m² + n² (m + n)² m² + 1 m² + (m + 1)² ANSWER m² + (m – 1)² (m + (m + 1))² (2m)² + (2m + 1)²

Let the numbers be (m) and (m+1).Then the sum is: [ m² + (m+1)² = m² + m² + 2m + 1 = 2m² + 2m + 1 ]

✅ Correct expression: (m² + (m+1)²)Also valid: (m² + (m-1)²) — if the consecutive pair is (m) and (m-1)

4. Consider any 2 by 2 square of numbers in a calendar, as shown in the figure
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

Observation: In any 2×2 calendar square:

Diagonal 1: (a *(a + 8))

Diagonal 2: ((a + 1) *(a + 7))

Let’s compute the difference:

[ a(a + 8) - (a + 1)(a + 7) = a² + 8a - [a² + 7a + a + 7] = a² + 8a - (a² + 8a + 7) = -7 ]

✅ The diagonal products always differ by 7 — the number of days in a week! 

Why this happens: The calendar is structured in weeks (7 days), so the diagonals always span across a fixed offset. The algebra confirms that the difference between the two diagonal products is always 7, regardless of the starting number (a). Case 1 6 x 14 =84, 13 x 7 = 91, Difference = 91 – 84 = 7 

Case 2 9 x 17 = 153, 16 x 10 = 160, Difference = 160 – 153 = 7 

5. Verify which of the following statements are true.

(i) (k + 1) (k + 2) – (k + 3) is always 2. 

(k+1)(k+2)−(k+3) 

Expand: (k + 1)(k + 2) = k² + 3k + 2, k² + 3k + 2 - k - 3 = k² + 2k - 1 

❌ Not always 2 — it depends on k. So statement is sometimes true

Example: Let k = 5, then ( 5+ 1) (5+2)-(5+3) = 6 x 7 – 8 =42 – 8 = 34 is a multiple of 2, 4 x 5 -6 = 20 – 6 = 14 is a multiple of 2.

K = 4, then 5 x 6 – 7 = 30 -7 = 23

(ii) (2q + 1) (2q – 3) is a multiple of 4. 

Expand: [ = 4q² - 6q + 2q - 3 = 4q² - 4q - 3 ]

This expression is not always divisible by 4 due to the (-3).Verdict: False

Example: Let q = 3, then (6 +1) (6 - 3) = 7 x 3 =21 is not a multiple of 4. so statement is false

(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8. 

Example: 2² = 4 = 4 x 1, 4² = 16 = 4 x 4, 6² = 36 = 4 x 9 , always true.

Example: 3² = 9 = 8 + 1, 5² = 25 = 8 x 3 + 1, 7² = 49 = 8 x 6 + 1 , always true.

Let (n = 2k):[ (2k)² = 4k² multiple of 4 ]

Squares of odd numbers:Let (n = 2k + 1):[ (2k + 1)² = 4k² + 4k + 1 = multiple of 4 + 1 ]

✅ Always 1 more than a multiple of 8Verdict: True

(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.

(iv) ((6n + 2)² − (4n + 3)²)

Use identity:[ a² - b² = (a + b)(a - b) ]

Let (a = 6n + 2,; b = 4n + 3):[ = (10n + 5)(2n - 1) ]

This is a product — not necessarily 5 less than a square. Try (n = 1):[ (6 * 1 + 2)² = 64,; (4 * 1 + 3)² = 49,; 64 - 49 = 1515 = 20 - 5 = {5 less than } 20 ]

But 20 isn’t a square.Verdict: False

Example: Let n = 2, ( 6 x 2 +2 )² - (4 x 2 + 3)² = 14² - 11² = 196 – 121 = 75 = 80 – 5 but 80 is not a square number. 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? 

6. Remainders modulo 7

Let (a ≡ 3 (mod 7),; b ≡ 5 (mod 7)

Sum: (a + b ≡ 3 + 5 = 8 ≡ 1 (mod 7)

Difference: (a - b ≡ 3 - 5 = -2 ≡ 5 (mod 7)

Product: (a * b ≡ 3 *5 = 15 ≡ 1 (mod 7)

✅ Remainders:

Sum → 1

Difference → 5

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.

7. Three consecutive numbers

Let them be (n - 1,; n,; n + 1)

Square the middle: (n²)

Product of others: ((n - 1)(n + 1) = n² - 1)

Subtract: n² - (n² - 1) = 1

✅ Always 1 — a beautiful identity!

Statement and pattern

Choose: ((n-1),, n,, (n+1)).

Operation: Square the middle and subtract the product of the other two.

[ n²- (n-1)(n+1) = n²- (n²- 1) = 1 ]

Conclusion: The result is always (1), an identity:

[ n²- (n-1)(n+1) \equiv 1 ]

Classroom hook: Link this to the difference of squares ((n+1)(n-1) = n²- 1) and let students test several triples to “feel” the invariance.

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.

Solution: Let the two numbers be a and b.

Step 1: a + b

Step 2: (a + b) * 1/2 * (a + b) 

= (a + b) * 1/2 * (a + b)

 = 1/2 * (a + b)²

Product with half of the sum expression

Description and proof

Let: Two numbers be (a) and (b).

Instruction: Add them, then multiply by half their sum.

[ (a+b) * 1/2  (a+b) = 1/2 (a+b)² = 1/2 (a² + 2ab + b² ]

Conclusion: The result is half the square of the sum.

9. Which is larger? Find out without fully computing the product. (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74

Solution: (1) Let p = 14 * 26p’ = 16 * 24 =(14 + 2)(26 - 2) 

p' = 14 * 26 + 2 * 26 - 14 * 2 - 2 * 2 

= 14 * 26 + 2(26 - 14 - 2) 

= 14 * 26 + 2 * 10

 p' = p + 2 * 10 

:. p' > p or 16 * 24 > 14 * 26

Solution:

(ii) Let p = 25 * 75

p' = 26 * 74 = (25 + 1)(75 - 1) 

= 25 * 75 + 75 * 1 - 25 * 1 - 1 * 1 

= p + (75 - 25 - 1) = p + 49

 :. p'> p or 26 × 74 > 25 × 75

Comparing products without full computation

Principle

Key idea: For a fixed sum, the product is larger when the pair is closer together (by AM-GM or completing a square).

(i) (14 * 26) vs (16 * 24)

[ 14 *26 = (20-6)(20+6) = 20²- 6²= 400 - 36 = 364 ] 

[ 16 *24 = (20-4)(20+4) = 20²- 4²= 400 - 16 = 384 ]

Verdict: (16 *24) is larger.

(ii) (25 *75) vs (26 *74)

[ 25 *75 = (50-25)(50+25) = 50²- 25²= 2500 - 625 = 1875 ]

[ 26 *74 = (50-24)(50+24) = 50²- 24²= 2500 - 576 = 1924 ]

Verdict: (26 *74) is larger.

Quick heuristic: Same sum ⇒ use (S²- d² with mean (S/2) and deviation (d); smaller (d) means larger product.

10. A tiny park is coming up in Dhauli. w ft. 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.

Solution: Interpretation : the two green square plots each have side length g. The walkway of width www surrounds and separates them so that the overall outer rectangle has horizontal length 2g+3w and vertical length g+2w. (This matches the usual arrangement with a path of width w on both outer sides and between the two plots.)

Total outer rectangle area =(2g+3w)(g+2w).
Area of the two green squares =2g².
So area to be tiled (walking path) = total area −2g²
(2g+3w)(g+2w)−2g².
You can expand if you want:
(2g+3w)(g+2w)−2g² = (2g²+4gw+3gw+6w²)−2g² = 7gw+6w²=w(7g+6w) 
 Tiny park problem (two square plots of area g² each and walkway width w)

Park tiling area with two g² plots and path width w

The tiling area equals “total outer area” minus “green cover.” The exact total area depends on whether there’s a path between the two squares in addition to the outer border. Here are the two natural interpretations:

Case A: Path only around the outside (no path between squares)

Layout: Two (g*g) squares placed side-by-side with a border path of width (w) around the combined rectangle.

Outer dimensions: Width (= 2g + 2w), Height (= g + 2w).

Tiled area:

[ A tiled = (2g + 2w)(g + 2w) - 2g²]

Case B: Path around the outside and between the squares

Layout: Side-by-side (g*g) squares, a path of width (w) between them, and a border path of width (w) around.

Outer dimensions: Width (= 2g + 3w), Height (= g + 2w).

Tiled area:

[ A tiled = (2g + 3w)(g + 2w) - 2g² ]

If your figure shows a strip between the squares, use Case B; if it’s only the outer border, use Case A. Share the exact placement and I’ll give a clean diagram or simplify the expression further.

11. For each pattern shown below,

(i) Draw the next figure in the sequence. 
(ii) How many basic units are there in Step 10? 
(iii) Write an expression to describe the number of basic units in Step y.

Step 1: 2 vertical strips of 3 units each + 1 vertical strip of 3 units 

= 3 strips of 3 units each = 9 units squares = (1 + 2)²| unit squares

Step 2: 4 strips of 4 units each = 16 units squares = (2 + 2)²unit squares

Step 3: 5 strips of 5 units each =25 units squares =(3+2)² unit squares

Step 4: (1) 6 strips of 6 units each = 2 are vertical and 4 are horizontal

(ii) Number of unit squares in step 10 = (10 + 2)²= 144

(iii) Number of unit squares in step y = (y + 2)²11. For each pattern shown below,

(b) Number of unit squares in step 1 = 5 = 2²+ 1

Number of unit squares in step 2 = = 11 = 3²+ 2

Number of unit squares in step 3 = 19 = 4²+ 3

(ii) Step 1 has (1 + 1)²+ 1 or 5 squares

Step 2 has (2 + 1)²+ 2 or 11 squares

Step 3 has (3 + 1)²+ 3 or 19 squares

step 10 has

(10 + 1)²+ 10 or 131 squares

(iii) Step y has [(y + 1)²+ y] squares

Yellow Pattern Step 1: Analyze the pattern

The number of basic units in each step is: Step 1: 5 units Step 2: 9 units  Step 3: 13 units

The pattern is an arithmetic sequence where each subsequent step adds 4 units. The common difference is 4. 

Step 2: Draw the next figure (i)  

10. Tiny park problem (two square plots of area g² each and walkway width w)

The next figure (Step 4) will have  13+4=17 units.

 It is formed by adding 4 units to the top-most vertical part of Step 3, extending it upwards. Step 3: Calculate units in Step 10 

(ii) The number of units in Step 10 can be found using the formula for the n-th term of an arithmetic sequence: There are 41 basic units in Step 10. 

Step 4: Write an expression for Step y (iii) Using the same formula with n = y The expression to describe the number of basic units in Step y is 4y + 1 

Blue Pattern

Step 1: Analyze the pattern

The number of basic units in each step is:

Step 1: 4 units

Step 2: 9 units

Step 3: 16 units

The pattern is a sequence of perfect squares 2², 3², 4². The number of units in Step n is (n+1)²

Step 2: Draw the next figure (i) The next figure (Step 4) will have (4+1)² = 25+4=29 units = 5 x 5 square+4units

Step 3: Calculate units in Step 10 (ii)

The number of units in Step 10 is (10 + 1)² = 121 units. There are 121+10 = 131 basic units in Step 10.

Step 4: Write an expression for Step y (iii)

he expression to describe the number of basic units in Step y is ( y + 1)²+y