CBSE Class 9 Maths Ganita Manjari Part 1 Ch 04 Exploring Algebraic Identities Short Answer Questions with Answers

2 Mark Questions

Question 1. A square photo frame has side \( (x + 2) \) cm. Its outer area is 100 cm². Find x, and hence the side length.
Answer:
\( (x + 2)^2 = 100 \Rightarrow x + 2 = 10 \) (a side length must be positive, so we take the positive root)
\( x = 8 \)
\( \therefore x = 8 \) and the side length is 10 cm.
In simple words: The square's area is 100, so its side is 10 cm, and x is 2 less than that.

Exam Tip: Reject the negative square root when the quantity is a length.

Question 2. Without direct long multiplication, using an algebraic identity evaluate \( 49^2 \).
Answer:
\( 49^2 = (50 - 1)^2 \)
Using \( (a - b)^2 = a^2 - 2ab + b^2 \): \( = 2500 - 2(50)(1) + 1 = 2500 - 100 + 1 \)
\( = 2401 \)
In simple words: Write 49 as 50 − 1 so the working uses easy numbers.

Exam Tip: Choose a nearby round number (like 50 or 100) to split the given number.

Question 3. The expression \( 9a^2 - 24ab + 16b^2 \) appears in a worksheet. Recognise it as a square and write its factorised form. Briefly justify.
Answer:
\( 9a^2 = (3a)^2 \), \( 16b^2 = (4b)^2 \) and \( 24ab = 2(3a)(4b) \).
So \( 9a^2 - 24ab + 16b^2 = (3a)^2 - 2(3a)(4b) + (4b)^2 \), which matches \( a^2 - 2ab + b^2 \).
\( \therefore 9a^2 - 24ab + 16b^2 = (3a - 4b)^2 \)
In simple words: The first and last terms are squares, and the middle term is twice their product, so it is a perfect square.

Exam Tip: Always check the middle term equals 2 × (first root) × (second root) before writing a perfect square.

Question 4. A student claims: \( (2x + 3)(2x - 3) = 4x^2 + 9 \). Test the claim by choosing x = 2 and state the correct expression.
Answer:
LHS at x = 2: \( (4 + 3)(4 - 3) = 7 \times 1 = 7 \)
Student's RHS at x = 2: \( 4(4) + 9 = 25 \)
Since 7 ≠ 25, the claim is false.
Correct expression: \( (2x + 3)(2x - 3) = (2x)^2 - 3^2 = 4x^2 - 9 \) (check: 16 − 9 = 7 ✓)
In simple words: Trying x = 2 gives 7 on one side and 25 on the other, so the right answer must have −9, not +9.

Exam Tip: Sum × difference always gives a difference of squares (minus sign), never a plus sign.

Question 5. If the sum of two sides of a rectangular park is 12 m and its area is 20 m², find the sum of the squares of its two sides.
Answer:
Let the sides be p and q. Then \( p + q = 12 \) and \( pq = 20 \).
\( p^2 + q^2 = (p + q)^2 - 2pq = 144 - 40 = 104 \)
\( \therefore p^2 + q^2 = 104 \) m². (The sides are 10 m and 2 m: 100 + 4 = 104 ✓)
In simple words: Square the sum and remove twice the product to get the sum of the squares.

Exam Tip: \( a^2 + b^2 = (a + b)^2 - 2ab \) is one of the most useful rearranged identities.

3 Mark Questions

Question 6. A school wants a square-shaped display board whose side is \( (3x - 2) \) cm.
(i) Express its area in expanded form.
(ii) Find the area when x = 4.
Answer:
(i) Area = \( (3x - 2)^2 = 9x^2 - 12x + 4 \) cm²
(ii) At x = 4: \( 9(16) - 12(4) + 4 = 144 - 48 + 4 = 100 \) cm²
Check: side = 3(4) − 2 = 10 cm, and \( 10^2 = 100 \) ✓
In simple words: Expand the square with the (a − b)² identity, then substitute x = 4.

Exam Tip: Checking with the actual side length is a quick way to catch mistakes.

Question 7. A rectangular garden has dimensions \( (2x + 5) \) m and \( (2x - 5) \) m.
(i) Express its area using an identity.
(ii) Find the area when x = 6.
Answer:
(i) Area = \( (2x + 5)(2x - 5) = (2x)^2 - 5^2 = 4x^2 - 25 \) m²
(ii) At x = 6: \( 4(36) - 25 = 144 - 25 = 119 \) m²
Check: sides 17 m and 7 m, and 17 × 7 = 119 ✓
In simple words: Sum times difference gives a difference of squares, which is quick to evaluate.

Exam Tip: Name the identity used, \( (a + b)(a - b) = a^2 - b^2 \), for full marks.

Question 8. A mental-maths activity asks students to calculate \( 1003^2 \). Use an algebraic identity and explain why your method is efficient.
Answer:
\( 1003^2 = (1000 + 3)^2 = 1000^2 + 2(1000)(3) + 3^2 \)
\( = 1000000 + 6000 + 9 = \) 1006009
This is efficient because each part (squaring 1000, doubling 3000, squaring 3) is easy to do mentally, instead of a long four-digit multiplication.
In simple words: Split 1003 into 1000 + 3, and every piece becomes easy to work out in your head.

Exam Tip: Line up the place values carefully when adding 1000000, 6000 and 9.

Question 9. A student claims that \( 123 \times 73 \) is difficult without a calculator. Use an identity to find the value.
Answer:
The two numbers are equally spaced around their average: \( \frac{123 + 73}{2} = 98 \), and each is 25 away from 98.
\( 123 \times 73 = (98 + 25)(98 - 25) = 98^2 - 25^2 \)
\( 98^2 = (100 - 2)^2 = 10000 - 400 + 4 = 9604 \); \( 25^2 = 625 \)
\( 123 \times 73 = 9604 - 625 = 8979 \)
In simple words: Write both numbers as 'middle ± gap', and the product becomes a difference of two squares.

Exam Tip: This trick works whenever you can find the middle (average) of the two numbers.

Question 10. A designer compares two square panels of sides \( (n + 3) \) cm and \( (n - 3) \) cm. Show that the difference in their areas is \( 12n \) cm².
Answer:
Difference = \( (n + 3)^2 - (n - 3)^2 \)
Using \( a^2 - b^2 = (a - b)(a + b) \): \( = [(n + 3) - (n - 3)][(n + 3) + (n - 3)] \)
\( = 6 \times 2n = 12n \) cm²
Hence proved.
In simple words: Treat the two areas as two squares and use the difference-of-squares identity.

Exam Tip: You can also expand both squares and subtract; the answer must be the same, 12n.

Question 11. Simplify and factorise \( 4a^2 - 12ab + 9b^2 - 25 \). Interpret the result as a difference of two squares.
Answer:
\( 4a^2 - 12ab + 9b^2 = (2a)^2 - 2(2a)(3b) + (3b)^2 = (2a - 3b)^2 \)
So the expression = \( (2a - 3b)^2 - 5^2 \), a difference of two squares.
\( = (2a - 3b - 5)(2a - 3b + 5) \)
In simple words: The first three terms form one perfect square, and 25 is another, so we subtract two squares.

Exam Tip: Group the first three terms first; the −25 is the second square.