CBSE Class 9 Maths Ganita Manjari Part 1 Ch 04 Exploring Algebraic Identities Long Answer and Case Study Questions with Answers

5 Mark Questions

Question 1. A rectangular school stage is \( (x + 6) \) m long and \( (x + 2) \) m wide.
(a) Obtain its area in expanded form.
(b) A second stage is a square of side \( (x + 4) \) m. Compare the two areas and simplify the difference.
(c) For x = 10, find both areas and verify the simplified difference.
Answer:
(a) \( (x + 6)(x + 2) = x^2 + 8x + 12 \) m²
(b) Square area = \( (x + 4)^2 = x^2 + 8x + 16 \) m²
Difference = \( (x^2 + 8x + 16) - (x^2 + 8x + 12) = 4 \) m². The square stage is always 4 m² larger.
(c) At x = 10: rectangle = 16 × 12 = 192 m²; square = 14² = 196 m²
196 − 192 = 4 m² ✓
In simple words: Both stages have the same x² and x terms, so the square is always exactly 4 m² bigger.

Exam Tip: When comparing, write both expanded forms one under the other and subtract term by term.

Question 2. A square park has side \( (2x + 5) \) m. A smaller square lawn of side \( (2x - 5) \) m is removed from one corner.
(a) Find the remaining area using an identity.
(b) For x = 8, calculate the remaining area.
(c) Explain why the difference-of-squares identity is appropriate.
Answer:
(a) Remaining area = \( (2x + 5)^2 - (2x - 5)^2 \)
\( = [(2x + 5) - (2x - 5)][(2x + 5) + (2x - 5)] = 10 \times 4x = 40x \) m²
(b) At x = 8: \( 40 \times 8 = 320 \) m². Check: \( 21^2 - 11^2 = 441 - 121 = 320 \) ✓
(c) The remaining area is one square minus another square, which is exactly the form \( a^2 - b^2 \). The identity \( a^2 - b^2 = (a - b)(a + b) \) avoids expanding both squares separately.
In simple words: Big square minus small square is a difference of squares, which simplifies straight to 40x.

Exam Tip: Look for the pattern 'square − square' in word problems; it signals \( a^2 - b^2 \).

Question 3. During a number game, a student is given \( x + \frac{1}{x} = 6 \), \( x \neq 0 \).
(a) Find \( x^2 + \frac{1}{x^2} \).
(b) Find \( x^3 + \frac{1}{x^3} \).
(c) State the identities used and show the key steps.
Answer:
(a) \( \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \Rightarrow 36 = x^2 + 2 + \frac{1}{x^2} \Rightarrow x^2 + \frac{1}{x^2} = 34 \)
(b) \( \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) \)
\( 216 = x^3 + \frac{1}{x^3} + 3(6) \Rightarrow x^3 + \frac{1}{x^3} = 216 - 18 = 198 \)
(c) Identities used: \( (a + b)^2 = a^2 + 2ab + b^2 \) and \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), with \( a = x \), \( b = \frac{1}{x} \), so \( ab = 1 \).
In simple words: Square the given sum for part (a) and cube it for part (b); the product x × 1/x is always 1.

Exam Tip: Remember \( x \times \frac{1}{x} = 1 \); it is what makes these problems simple.

Question 4. A rectangular sheet has length \( (y + 9) \) cm and breadth \( (y - 9) \) cm. Its area is 319 cm².
(a) Find y.
(b) If the length is increased by 2 cm while the breadth is unchanged, find the new area.
(c) Find the increase in area and verify it algebraically.
Answer:
(a) \( (y + 9)(y - 9) = y^2 - 81 = 319 \Rightarrow y^2 = 400 \Rightarrow y = 20 \) (y must be more than 9 for the breadth to be positive)
Length = 29 cm, breadth = 11 cm (29 × 11 = 319 ✓)
(b) New length = 31 cm. New area = 31 × 11 = 341 cm²
(c) Increase = 341 − 319 = 22 cm²
Algebraically: \( (y + 11)(y - 9) - (y + 9)(y - 9) = (y - 9)[(y + 11) - (y + 9)] = 2(y - 9) \). At y = 20: \( 2 \times 11 = 22 \) ✓
In simple words: The sheet is 29 cm by 11 cm; adding 2 cm to the length adds a strip of 2 × 11 = 22 cm².

Exam Tip: An increase in only the length always adds (increase × breadth) to the area.

Case Study Based Questions

Question 5. A school is laying square tiles in a corridor. The side of each tile is \( (x + 1) \) cm.
(i) Write the area of one tile in expanded form.
(ii) For x = 9, find the area.
(iii) The same tile is compared with a square of side \( (x - 1) \) cm. Find the difference between their areas.
Answer:
(i) \( (x + 1)^2 = x^2 + 2x + 1 \) cm²
(ii) At x = 9: \( 10^2 = 100 \) cm²
(iii) \( (x + 1)^2 - (x - 1)^2 = 4x \) cm² (using \( (a + b)^2 - (a - b)^2 = 4ab \) with a = x, b = 1). At x = 9 this is 36 cm² (100 − 64 ✓).
In simple words: Each tile is x² + 2x + 1, and it is 4x bigger than the (x − 1) square.

Exam Tip: For (ii), substituting into the side first (9 + 1 = 10) is quicker than using the expanded form.

Question 6. A science exhibition model uses a rectangular panel of dimensions \( (p + 6) \) cm by \( (p - 6) \) cm.
(i) Write the area using a suitable identity.
(ii) Find the area for p = 10.
(iii) If another panel has area \( p^2 + 36 \) cm², state whether it is always equal to the first panel's area and justify.
Answer:
(i) \( (p + 6)(p - 6) = p^2 - 36 \) cm²
(ii) At p = 10: \( 100 - 36 = 64 \) cm² (16 × 4 ✓)
(iii) No. \( (p^2 + 36) - (p^2 - 36) = 72 \), so the second panel is always 72 cm² larger. They are never equal.
In simple words: Sum × difference gives p² − 36, which is always 72 less than p² + 36.

Exam Tip: A common mistake is to write \( (p + 6)(p - 6) = p^2 + 36 \); the sign must be minus.

Question 7. A shopkeeper uses mental calculation to find the cost of \( 99 \times 101 \) items at ₹1 each. He rewrites the product as \( (100 - 1)(100 + 1) \).
(i) Find the value quickly using an identity.
(ii) Explain the identity used.
(iii) If the price per item is ₹5, find the total cost.
Answer:
(i) \( (100 - 1)(100 + 1) = 100^2 - 1^2 = 10000 - 1 = 9999 \)
(ii) \( (a - b)(a + b) = a^2 - b^2 \). The product of the sum and difference of two numbers equals the difference of their squares.
(iii) Total cost = 9999 × ₹5 = ₹49,995
In simple words: 99 × 101 is just 100² − 1, which is 9999.

Exam Tip: For (iii), 9999 × 5 = 50000 − 5 is a quick mental method.