A wire of length 20 cm is bent in different ways to form rectangles. For example, we can have a rectangle with length 7 cm and width 3 cm. We can also have one of length 5.5 cm and width 4.5 cm. (Think of a few more ways of forming such rectangles.) Can you write an expression for the area of such rectangles?
Step-by-step solution
To find: An expression for the area of every such rectangle
Idea: The whole wire becomes the perimeter, so 2(length + width) = 20, which means length + width = 10. Once we choose the length, the width is fixed.
- The wire is the perimeter: 2(length + width) = 20, so length + width = 10 cm. Check with the rectangles in the question: 7 + 3 = 10 and 5.5 + 4.5 = 10.
- More such rectangles (any two lengths adding to 10): 6 cm × 4 cm, 8 cm × 2 cm, 9 cm × 1 cm, 6.5 cm × 3.5 cm, and even the square 5 cm × 5 cm.1 mark
- Let the length be x cm. Then the width = 10 − x cm.1 mark
- Area = length × width = x(10 − x) = 10x − x2 cm2.1 mark
Check: For length 7: 10(7) − 72 = 70 − 49 = 21 = 7 × 3 ✓. For length 5.5: 55 − 30.25 = 24.75 = 5.5 × 4.5 ✓.
Answer to write in the exam
Perimeter = 20 cm ⇒ length + width = 10 cm
Other rectangles: 6 cm × 4 cm, 8 cm × 2 cm, 9 cm × 1 cm, 5 cm × 5 cm
Let length = x cm; width = (10 − x) cm
∴ Area = x(10 − x) = 10x − x2 cm2
Common mistakes that cost marks
- Taking the width as 20 − x. The wire goes round all four sides, so length + width is half of 20, which is 10.
- Writing the area as 10 − x2. Multiply x by both terms in the bracket: x × 10 − x × x = 10x − x2.
- Writing the area in cm instead of cm2.
How this can come in the exam
A 36 cm long wire is bent into a rectangle of length y cm. Its area in cm2 is
- y(36 − y)
- y(18 − y)
- 18y
- 36 − y2
Show answer
(B) y(18 − y)
Length + width = 36 ÷ 2 = 18, so width = 18 − y and area = y(18 − y).
A 28 cm wire is bent into a rectangle of length x cm. Write its area as an expression and find the area when x = 9.
Show answer
Width = 14 − x, so area = x(14 − x) = 14x − x2 cm2 (1 mark). At x = 9: 126 − 81 = 45 cm2 (= 9 × 5) (1 mark).Try one yourself
A rope of length 50 m is laid out as a rectangle with length t m. Write the area as an expression, and find it when t = 15.
Show answer
Width = 25 − t. Area = t(25 − t) = 25t − t2 m2. At t = 15: 375 − 225 = 150 m2.
More questions like this
- The expression for the area of these rectangles is x(10 − x) or 10x − x2.
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. Can you point out any similarity or difference between the algebraic expressions obtained for Raju’s pens and pencils (4x + 5y + 3) and for these rectangles? - Find the degrees of the following polynomials:
- Write polynomials of degrees 1, 2 and 3.
- What are the coefficients of x2 and x3 in the polynomial x4 − 3x3 + 6x2 − 2x + 7?
- What is the coefficient of z in the polynomial 4z3 + 5z2 − 11?