Question

A rectangle has each side length of x+2 inches and each side width of 7 inches. For what values of x is the area of the rectangle greater than the perimeter?

Answer 1

To find when the area of a rectangle with dimensions of (x+2) by 7 inches exceeds its perimeter, set up the inequality A > P and solve for x to find that x must be greater than 0.8 inches.

Step-by-step explanation:

To determine for what values of x the area of the rectangle is greater than the perimeter, we need to first express both the area and the perimeter with respect to x. The area (A) of the rectangle is given by length times width, which in this case is (x+2) × 7 square inches. The perimeter (P) is the sum of all the sides, or 2((x+2) + 7) inches.

Now, we set up the inequality A > P, which translates to:

• (x+2) × 7 > 2(x+2) + 2 × 7

Simplifying this inequality to find the values of x, we have:

• 7x + 14 > 2x + 4 + 14
• 7x + 14 > 2x + 18
• 5x > 4
• x > 4/5 or 0.8

Therefore, the value of x must be greater than 0.8 inches for the area of the rectangle to be greater than its perimeter.