Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
- (i) 25a2 − 30ab + 9b2
- (ii) 36s2 − 49t2
Step-by-step solution
Idea: Area of a rectangle = length × breadth. So if we factorise the area expression into two factors, those two factors are possible expressions for the length and the breadth. We use identities to factorise.
(i) 25a2 − 30ab + 9b2
- 25a2 = (5a)2 and 9b2 = (3b)2, and the middle term is negative, so try x2 − 2xy + y2 = (x − y)2 with x = 5a, y = 3b.
- Check the middle term: 2 × 5a × 3b = 30ab ✓. So 25a2 − 30ab + 9b2 = (5a − 3b)2 = (5a − 3b) × (5a − 3b).½ mark
- Area = length × breadth, so a possible answer is length = (5a − 3b) units and breadth = (5a − 3b) units. Length and breadth are equal, so this rectangle is in fact a square.½ mark
- A length must be positive, so this works when 5a > 3b.
(ii) 36s2 − 49t2
- Two perfect squares with a minus between them: 36s2 = (6s)2, 49t2 = (7t)2. Use a2 − b2 = (a + b)(a − b).
- 36s2 − 49t2 = (6s + 7t)(6s − 7t)½ mark
- The bigger factor is usually taken as the length: length = (6s + 7t) units, breadth = (6s − 7t) units (this needs 6s > 7t).½ mark
Check: (i) a = 2, b = 1: area 100 − 60 + 9 = 49 and (10 − 3) × (10 − 3) = 49 ✓. (ii) s = 2, t = 1: area 144 − 49 = 95 and (12 + 7) × (12 − 7) = 19 × 5 = 95 ✓.
Answer to write in the exam
(i)
Area = 25a2 − 30ab + 9b2
= (5a)2 − 2(5a)(3b) + (3b)2 [a2 − 2ab + b2 = (a − b)2]
= (5a − 3b)2 = (5a − 3b)(5a − 3b)
∴ Length = (5a − 3b) units, breadth = (5a − 3b) units
(ii)
Area = 36s2 − 49t2 = (6s)2 − (7t)2
= (6s + 7t)(6s − 7t) [a2 − b2 = (a + b)(a − b)]
∴ Length = (6s + 7t) units, breadth = (6s − 7t) units
Common mistakes that cost marks
- Writing 25a2 − 30ab + 9b2 = (5a + 3b)2. The middle term is negative, so the bracket has a minus sign: (5a − 3b)2.
- Thinking a rectangle cannot have length equal to breadth. A square is a special rectangle, so length = breadth = 5a − 3b is a correct answer.
- Factorising 36s2 − 49t2 as (6s − 7t)2. That gives 36s2 − 84st + 49t2; a difference of squares needs one plus bracket and one minus bracket.
How this can come in the exam
The area of a rectangle is (x2 − 16) square units. Possible length and breadth are
- (x − 4) and (x − 4)
- (x + 4) and (x − 4)
- (x + 8) and (x − 8)
- (x − 16) and x
Show answer
(B) (x + 4) and (x − 4)
x2 − 16 = x2 − 42 = (x + 4)(x − 4).
The area of a square is (4x2 + 12x + 9) square units. Find its side and its perimeter.
Show answer
4x2 + 12x + 9 = (2x)2 + 2(2x)(3) + 32 = (2x + 3)2 (1 mark). So side = (2x + 3) units and perimeter = 4(2x + 3) = (8x + 12) units (1 mark).Try one yourself
Find possible expressions for the length and breadth of a rectangle whose area is (49p2 − 25q2) square units.
Show answer
49p2 − 25q2 = (7p)2 − (5q)2 = (7p + 5q)(7p − 5q). Length = (7p + 5q) units, breadth = (7p − 5q) units.
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