Question

A rectangle has an area of 60 cm² and a perimeter of 32 cm. What is the length if it is greater than the width?

• A. 10 cm
• B. 12 cm
• C. 14 cm
• D. 16 cm

Hint:

Use area and perimeter to form a quadratic equation and select the solution where length exceeds width.

Answer 1

The Correct Option is A

We need the length of the rectangle, given it is greater than the width.
- Step 1: Define variables. Let length = \( l \), width = \( w \).
- Step 2: Set up equations.
- Area: \( l \cdot w = 60 \).
- Perimeter: \( 2(l + w) = 32 \Rightarrow l + w = 16 \).
- Step 3: Solve for one variable. From perimeter: \( w = 16 - l \). Substitute into area:
\[ l (16 - l) = 60 \] \[ l^2 - 16l + 60 = 0 \] - Step 4: Solve quadratic equation. Use quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -16 \), \( c = 60 \):
\[ l = \frac{16 \pm \sqrt{256 - 240}}{2} = \frac{16 \pm \sqrt{16}}{2} = \frac{16 \pm 4}{2} \] \[ l = 10 \text{ or } l = 6 \] - Step 5: Determine length and width. Since \( l>w \):
- If \( l = 10 \), \( w = 16 - 10 = 6 \). Check: \( 10 \times 6 = 60 \), \( 2(10 + 6) = 32 \). Correct.
- If \( l = 6 \), \( w = 16 - 6 = 10 \). \( l<w \), so discar(d)
- Step 6: Check options.
- (a) 10: Correct.
- (b) 12: Incorrect.
- (c) 14: Incorrect.
- (d) 16: Incorrect.
Thus, the answer is a.