Case Study on Chapter 4: ‘Exploring Algebraic Identities’ for Class 9 Maths

This case study on Chapter 4: Exploring Algebraic Identities offers short, focused problem scenarios with clear answers and step-by-step solutions to build exam-ready skills. Each question emphasises methodical reasoning, common shortcuts, and error-avoidance strategies to strengthen conceptual understanding and improve speed and accuracy for board exams. A free PDF with answer keys and timed practice sets is included for offline revision, classroom worksheets, and homework ideal for quick revision and targeted skill building before tests.

Case Study on Chapter 4:Exploring Algebraic Identities for Class 9 With Answers

CBSE's case-based questions usually give you a short passage describing a real situation, sometimes with a small figure, followed by 4 - 5 questions of increasing difficulty, a couple of MCQs to check basic understanding, a short-answer question that needs a calculation, and a longer question that asks you to interpret or justify your answer.

Case Study 1: Resizing the Society Pool Deck

A housing society's square pool deck currently measures 50 m × 50 m. To create more lounging space, the committee decides to add 6 m to both the length and the width, making it a larger square. Later, to lay a tiled walking border around this enlarged deck, they take 4 m off the length and 4 m off the width.

(i)  Which identity is most useful to calculate the area of the enlarged deck (56 m × 56 m)?

(ii)Find the exact area of the enlarged deck using that identity.

(iii)After the border is cut in, the deck becomes (56 − 4) m on each side. Find its new area using a suitable identity.

(iv) Find the difference between the area of the enlarged deck and the final deck.

OR 

Verify the area found in part (iii) by direct multiplication of 52 × 52.

Solution: 

(i) Since 56 = 50 + 6, the area is (50 + 6)², so the identity (a + b)² = a² + 2ab + b² is the one to use.

(ii) (50 + 6)² = 50² + 2(50)(6) + 6² = 2500 + 600 + 36 = 3136 m².

(iii) 56 − 4 = 52, so the area is (56 − 4)² = 56² − 2(56)(4) + 4² = 3136 − 448 + 16 = 2704 m², using the identity (a − b)² = a² − 2ab + b².

(iv) Difference = 3136 − 2704 = 432 m².

OR: 52 × 52 = 2704, which matches the value found using the identity in part (iii).

Case Study 2: The Carpenter's Cube-Shaped Gift Box

A carpenter is building a cube-shaped wooden gift box. The volume of the box, in cubic centimetres, works out to 8y³ + 36y² + 54y + 27, where y is a design parameter he uses across different box sizes.

(i) Express this volume in the form (a + b)³ by identifying a and b.

(ii) Write down the side length of the cube in terms of y.

(iii) If y = 3 cm, find the actual side length and the volume of the box.

(iv) Verify your volume from part (iii) by substituting y = 3 directly into the original expression.

OR

 If the volume were instead 8y³ − 36y² + 54y − 27, what would the side length be, and why does the sign pattern in the expression change?

Solution: 

(i) 8y³ = (2y)³ and 27 = 3³. Checking the middle terms: 3(2y)²(3) = 36y² and 3(2y)(3)² = 54y, both of which match. So a = 2y and b = 3, and the expression is (2y + 3)³.

(ii) Side length = (2y + 3) cm.

(iii) At y = 3: side = 2(3) + 3 = 9 cm, and volume = 9³ = 729 cm³.

(iv) Direct substitution: 8(27) + 36(9) + 54(3) + 27 = 216 + 324 + 162 + 27 = 729 cm³, which matches.

OR: The side length would be (2y − 3) cm, since the expression would then match the identity (a − b)³ = a³ − 3a²b + 3ab² − b³. The sign pattern changes because in (a − b)³, the four terms alternate between positive and negative, unlike (a + b)³ where every term is positive.

Case Study 3: The Community Garden Plot

A rectangular community garden plot is designed so that its breadth is 5 metres less than its length. The total area of the plot is 84 square metres.

(i) Taking the length as x metres, write an expression for the breadth and form a quadratic equation for the area.

(ii) Factorise this quadratic equation by splitting the middle term.

(iii) Find the length and breadth of the garden plot.

(iv) Why is the negative root of the quadratic rejected in this situation?

OR 

Verify your answer by multiplying the length and breadth you found.

Solution:

(i) Breadth = (x − 5) m. Since area = length × breadth: x(x − 5) = 84, which gives x² − 5x − 84 = 0.

(ii) We need a + b = −5 and ab = −84, which gives a = −12 and b = 7. So x² − 12x + 7x − 84 = x(x − 12) + 7(x − 12) = (x − 12)(x + 7) = 0.

(iii) x = 12 or x = −7. Taking x = 12: length = 12 m and breadth = 12 − 5 = 7 m.

(iv) x represents a length, and a physical length can never be negative, so x = −7 is rejected even though it satisfies the equation algebraically.

OR: 12 × 7 = 84 m², which matches the given area.

 

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Class 9 Chapter 4: Exploring Algebraic Identities Case Study PDF