Question

Let A and P be the area and perimeter of a rectangle respectively. The length and breadth of the rectangle are (x + 2) cm and (x - 2) cm. If \(A \leq 140 cm^2\) and \(P \geq 20 cm\), then the range of possible values of x is

• A. [5, 12]
• B. [2, 12]
• C. [5, 10]
• D. [2, 6]
• E. [6, 12]

Hint:

In geometry problems involving variables, always check if your final range satisfies the condition that all physical dimensions (length, breadth) must be strictly positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to express area \(A\) and perimeter \(P\) in terms of \(x\) using the given dimensions, and then solve the system of inequalities.
Note: Dimensions must be positive, so \(x+2 > 0\) and \(x-2 > 0 \implies x > 2\).

Step 2: Key Formula or Approach:
1. \(A = \text{length} \times \text{breadth} = (x + 2)(x - 2) = x^2 - 4\).
2. \(P = 2(\text{length} + \text{breadth}) = 2(x + 2 + x - 2) = 4x\).

Step 3: Detailed Explanation:
Condition 1: Perimeter \(P \geq 20\)
\[ 4x \geq 20 \implies x \geq 5 \]
Condition 2: Area \(A \leq 140\)
\[ x^2 - 4 \leq 140 \implies x^2 \leq 144 \]
Taking the square root for positive \(x\):
\[ x \leq 12 \]
Combining the conditions:
We have \(x \geq 5\) and \(x \leq 12\).
Also, the dimension \(x-2\) must be positive, which means \(x > 2\). Since \(5 \geq 2\), the physical constraint is automatically satisfied.
Thus, the range is \(x \in [5, 12]\).

Step 4: Final Answer:
The range of possible values for \(x\) is [5, 12].