Exercise 11.5 — Geometric Representations
Geometrical representations of algebraic identities.
What is Exercise 11.5 About?
Exercise 11.5 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) is one of the most visually rich and logically satisfying lessons in the entire chapter. Instead of simply stating algebraic identities as rules to memorise, this exercise asks students to verify three fundamental identities geometrically — by drawing squares and rectangles and showing that their areas match both sides of each identity.
This approach builds deep conceptual understanding: you see why the identities are true, not just that they are true. The three identities covered are the cornerstones of all algebra from Class 8 through Class 10 and beyond.
The Three Algebraic Identities at a Glance
Before diving into the exercise problems, here is a quick overview of the three identities you will work with. Understanding what each identity means geometrically is the key to answering every question in this exercise.
| Identity | Name | Formula | Geometric Shape Used |
|---|---|---|---|
| 1 | Square of a Sum | (a+b)² = a² + 2ab + b² | Square of side (a+b) split into 4 regions |
| 2 | Square of a Difference | (a−b)² = a² − 2ab + b² | Square of side a with inner paths of width b |
| 3 | Difference of Two Squares | a² − b² = (a+b)(a−b) | Square of side a minus corner square of side b |
Geometrical Proof of (a + b)² ≡ a² + 2ab + b²
Imagine drawing a square whose side length is (a + b). Now divide this square into four smaller regions by drawing one horizontal and one vertical line — at distance a from the top-left corner in each direction. This creates four pieces whose areas together must equal the total area (a + b)².
The big square of side (a+b) is divided into 4 pieces — their areas sum to a² + ab + ab + b² = a² + 2ab + b²
Step-by-Step Proof
Question 1 — Verify (a + b)² ≡ a² + 2ab + b² Geometrically
This question asks you to verify the first identity by substituting actual numbers, drawing a square, and confirming that LHS = RHS. Three sets of values are given.
Setup
Draw a square of side a + b = 2 + 4 = 6 units. Divide it into 4 regions at the 2-unit mark.
= 6²
= 4 + 16 + 16
= 4 + 8 + 8 + 16
Setup
Draw a square of side a + b = 3 + 1 = 4 units. Divide it at the 3-unit mark.
Setup
Draw a square of side a + b = 5 + 2 = 7 units. Divide it at the 5-unit mark.
| Part | a | b | Side (a+b) | LHS (a+b)² | RHS a²+2ab+b² | Verified? |
|---|---|---|---|---|---|---|
| (i) | 2 | 4 | 6 units | 36 sq.u. | 4 + 8 + 8 + 16 = 36 | ✅ Yes |
| (ii) | 3 | 1 | 4 units | 16 sq.u. | 9 + 3 + 3 + 1 = 16 | ✅ Yes |
| (iii) | 5 | 2 | 7 units | 49 sq.u. | 25 + 10 + 10 + 4 = 49 | ✅ Yes |
Geometrical Proof of (a − b)² ≡ a² − 2ab + b²
This proof uses a clever subtraction approach. Start with a large square of side a. Inside it, draw two rectangular paths (strips) of width b along two adjacent sides. This divides the square into four labelled regions: I, II, III, and IV.
Part I (dashed border) is the inner square of side (a−b). Its area = total area − Parts II − III − IV.
Step-by-Step Proof
Question 2 — Verify (a − b)² ≡ a² − 2ab + b² Geometrically
Setup
Draw a square of side a = 3 units. Draw inner paths of width b = 1 unit along two sides. The inner square (Part I) has side a − b = 3 − 1 = 2 units.
= 2²
= 9 − 3 − 3 + 1
Setup
Draw a square of side a = 5 units. Draw inner paths of width b = 2 units. Inner square (Part I) has side 5 − 2 = 3 units.
| Part | a | b | Side (a−b) | LHS (a−b)² | RHS a²−2ab+b² | Verified? |
|---|---|---|---|---|---|---|
| (i) | 3 | 1 | 2 units | 4 sq.u. | 9 − 3 − 3 + 1 = 4 | ✅ Yes |
| (ii) | 5 | 2 | 3 units | 9 sq.u. | 25 − 10 − 10 + 4 = 9 | ✅ Yes |
Geometrical Proof of a² − b² ≡ (a + b)(a − b)
This proof uses a two-step cutting-and-rearranging technique. Start with a large square of side a (area = a²). Remove a small square of side b from one corner (area removed = b²). The remaining L-shaped piece has area = a² − b². Now cut this L-shape into two rectangles and rearrange — the resulting single rectangle has dimensions (a + b) and (a − b).
The L-shaped region (blue) has area a² − b². Cut Part II and flip it below Part I → you get a rectangle of size (a+b) × (a−b).
Step-by-Step Proof
Question 3 — Verify a² − b² ≡ (a + b)(a − b) Geometrically
Setup
Remove a square of side b = 2 from a corner of a square of side a = 3. Remaining area = 3² − 2² = 9 − 4 = 5 sq. units. Now cut and rearrange into a rectangle with breadth (a − b) = 3 − 2 = 1.
= 9 − 4
= 5 × 1
= a(a−b) + b(a−b)
= 3×1 + 2×1
Setup
Remove a square of side b = 1 from a corner of a square of side a = 2. Remaining area = 4 − 1 = 3 sq. units. Rearrange into a rectangle with breadth (a − b) = 2 − 1 = 1.
= 4 − 1
= 2×1 + 1×1
= 2 + 1
| Part | a | b | RHS (a²−b²) | LHS (a+b)(a−b) | Calculation | Verified? |
|---|---|---|---|---|---|---|
| (i) | 3 | 2 | 5 sq.u. | 5 sq.u. | 3×1 + 2×1 = 5 | ✅ Yes |
| (ii) | 2 | 1 | 3 sq.u. | 3 sq.u. | 2×1 + 1×1 = 3 | ✅ Yes |
Exercise 11.5 — All Answers at a Glance
| Question | Values | Identity | LHS | RHS | Result |
|---|---|---|---|---|---|
| Q1 (i) | a=2, b=4 | (a+b)² = a²+2ab+b² | 36 | 4+8+8+16=36 | ✅ |
| Q1 (ii) | a=3, b=1 | (a+b)² = a²+2ab+b² | 16 | 9+3+3+1=16 | ✅ |
| Q1 (iii) | a=5, b=2 | (a+b)² = a²+2ab+b² | 49 | 25+10+10+4=49 | ✅ |
| Q2 (i) | a=3, b=1 | (a−b)² = a²−2ab+b² | 4 | 9−3−3+1=4 | ✅ |
| Q2 (ii) | a=5, b=2 | (a−b)² = a²−2ab+b² | 9 | 25−10−10+4=9 | ✅ |
| Q3 (i) | a=3, b=2 | a²−b² = (a+b)(a−b) | 5 | 3×1+2×1=5 | ✅ |
| Q3 (ii) | a=2, b=1 | a²−b² = (a+b)(a−b) | 3 | 2×1+1×1=3 | ✅ |
Common Mistakes to Avoid
- Forgetting the middle term 2ab: Students sometimes write (a+b)² = a² + b², missing the crucial 2ab. The geometric proof makes this impossible to forget — the two rectangles are the 2ab term, visually.
- Sign error in Identity 2: In (a−b)², the middle term is −2ab, not +2ab. Remember: you are subtracting the two rectangular strips, not adding them.
- Confusing Identity 1 and Identity 2: (a+b)² has +2ab while (a−b)² has −2ab. The only difference is the sign of b in the bracket — this flips the sign of the middle term only.
- Wrong formula for Identity 3: a² − b² ≠ (a−b)². These are completely different identities. Use Identity 3 only when you see a² − b² (difference of squares), never when you see (a−b)².
- Calculation errors in LHS vs RHS verification: Always compute LHS and RHS separately, then compare. Do not assume they are equal before finishing the calculation.
- Skipping intermediate steps in board exams: TS/AP and CBSE examiners expect you to show the geometric interpretation — write the area of each sub-region explicitly before adding or subtracting.
All Three Identities — Quick Comparison
| Identity | Formula | Middle Term | Geometric Shape | Used When |
|---|---|---|---|---|
| 1 | (a+b)² = a²+2ab+b² | +2ab | Square of side (a+b) split into 4 pieces | Expanding square of a sum |
| 2 | (a−b)² = a²−2ab+b² | −2ab | Square of side a with b-width inner paths | Expanding square of a difference |
| 3 | a²−b² = (a+b)(a−b) | None | L-shaped region rearranged into rectangle | Factorising difference of two squares |
What Exercise 11.5 Prepares You For
The three identities you verify geometrically in Exercise 11.5 are not just Class 8 content — they are used extensively throughout your secondary school mathematics journey. In Exercise 11.4 (algebraic identities in abstract form) you will apply these same formulas to expand and simplify expressions without drawing diagrams.
In Class 9 Polynomials, these identities become the foundation for factorising quadratic expressions. The difference-of-squares identity (a² − b² = (a+b)(a−b)) is used constantly in factorisation and simplification throughout Class 9 and 10. In Class 10 Quadratic Equations, the identity (a+b)² and (a−b)² appear when completing the square and working with the quadratic formula.
For students in CBSE, Telangana, and Andhra Pradesh boards, mastering these three identities in Class 8 — both the geometric justification and the algebraic form — gives you a decisive advantage in every algebra topic going forward. The geometric method taught here is the best way to truly understand why the identities hold, making them far easier to recall correctly in exams.