Chapter 6: We Distribute, Yet Things Multiply - Complete with SVG Diagrams

Chapter 6: We Distribute, Yet Things Multiply

Class 8 Mathematics – NCERT Ganita Prakash

Complete Interactive Question Bank with SVG Diagrams

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

3.

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

Fig: Geometric proof of (a+b)² - Square divided into a², b², and two ab rectangles

Geometric Proof:
1. Draw a square of side length (a+b)
2. Divide horizontally at distance 'a' from left
3. Divide vertically at distance 'a' from top
4. This creates four regions:
- Top-left: Square of side a (area = a²)
- Top-right: Rectangle a×b (area = ab)
- Bottom-left: Rectangle b×a (area = ab)
- Bottom-right: Square of side b (area = b²)
5. Total area = a² + ab + ab + b² = a² + 2ab + b² = (a+b)²

The diagram visually proves the algebraic identity using area decomposition

5.

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

Fig: (m+n)² square with four corner rectangles (mn each) removed, leaving center square (n-m)²

Geometric Proof:
1. Start with square of side (m+n), area = (m+n)²
2. Remove four corner rectangles, each of size m×n
3. Total area removed = 4mn
4. Remaining center square has side = (n-m)
5. Area of center square = (n-m)²
6. Therefore: (m+n)² - 4mn = (n-m)²

The diagram shows removing corner rectangles from large square leaves smaller center square

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

1.

Derive all three identities geometrically using diagrams.

Three Algebraic Identities - Geometric Proofs

Identity 1A: (a+b)²

Identity 1B: (a-b)²

Identity 1C: (a+b)(a-b) = a²-b²

Geometric Derivations:

1. (a+b)² = a² + 2ab + b²
• Draw square of side (a+b)
• Divide into a², b², and two ab rectangles
• Total area = sum of all parts

2. (a-b)² = a² - 2ab + b²
• Start with a² square
• Remove two ab rectangles (overlaps b² area)
• Add back b² (was subtracted twice)
• Remaining = a² - 2ab + b²

3. (a+b)(a-b) = a² - b²
• Start with a² square
• Remove b×a rectangle from side
• Also remove a×b rectangle from bottom
• b² area was removed twice, so add back once
• Result = a² - ab - ab + b² = a² - b²

All three identities can be proven geometrically using area decomposition and rearrangement

6.

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

Fig: 2×2 Calendar square pattern - Diagonal products always differ by 7

Algebraic Proof:

In a 2×2 calendar square:
Top-left number = a
Top-right number = a+1
Bottom-left number = a+7 (next week same day)
Bottom-right number = a+8

Diagonal Products:
D₁ = a × (a+8) = a² + 8a
D₂ = (a+1) × (a+7) = a² + 8a + 7

Difference:
D₂ - D₁ = (a² + 8a + 7) - (a² + 8a)
= a² + 8a + 7 - a² - 8a
= 7

Therefore, the diagonal products always differ by 7, regardless of the starting number 'a'.

This works because calendar rows are 7 days apart, creating the constant difference

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

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.

Fig: Interactive diagram - Hover over parts to see details

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

\((m+n)^2 = m^2+2mn+n^2\) (expand using identity)

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 \( m \times n \)

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 for the same area

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: Pattern Diagrams

The number of circles in a pattern grows as shown:

Fig: Triangular number pattern - Each step adds one more row than previous

1.

How many circles in Step 4?

a) 10

b) 15

c) 20

d) 25

b) 15

Step 4 = 1+2+3+4+5 = 15 circles (triangular number T₅)