Question

The length of Rectangle A is 5 inches more than its width. The perimeter of Rectangle B is 17. The difference between the perimeters of the rectangles is less than 9. What are the possible integer values for the width of Rectangle A? 1in. 2in. 3in. 4in.5in. 6in. 7in. 8in.

Answer 1

Explanation:

Let L and W be the length and width of a triangle.

Rectangle perimeter (Px) = 2L + 2W, where x is triangle a or b.

Rectangle A: La = 5Wa [The length of Rectangle A is 5 inches more than its width.]

Perimeter Rectangle A: Pa = 2La + 2Wa

Substitute L = 5Wa:

Pa = 2La + 2Wa

Pa = 5Wa + 2Wa

Pa = 7Wa

Perimeter Rectangle B: Pb = 17 [The perimeter of Rectangle B is 17]

Pb = 17

Difference between Pa and Pb: Is it < 9?

See the attached table. The perimeters of rectangle A are computed for each of the answer options that define the width. Length (rectangle A) is 5 inches more than the width. The perimeter is 2*Length + 2*Width, and shown as Pa.

The perimeter of rectangle b is 17, shown as Pb on the table. The last column of the table is the absolute value of the difference of Pa and Pb. If it is less than 9, the cell is green (it is a possible integer value for the width of rectangle A), Red mean it does not meet the "less than 9" requirement, One cell has the difference as 9. The problem states "less than 9," so it does not meet the requirement.