9514 1404 393
Answer:
- width: 40 units
- length: 51 units
- area: 2040 square units
Explanation:
Let w represent the width of the rectangle. Then w+11 is the length. The perimeter is given by ...
P = 2(L +W)
182 = 2((w+11) +w)
182 = 4w +22 . . . . . simplify
160 = 4w . . . . . . . . . subtract 22
40 = w . . . . . . . . . . . divide by 4
40 +11 = length = 51 . . . . determine length
The dimensions of the rectangle are 40 × 51 units.
__
The area of the rectangle is ...
A = LW
A = (51)(40) = 2040 . . . . square units
The area is 2040 square units.
_____
Additional comment
The perimeter of a rectangle is double the sum of length and width, so that sum is 182/2 = 91 units. Now, this is a "sum and difference" problem, where the sum of length and width is 91, and the difference is 11. The solutions to such a problem are half the sum and half the difference of these numbers. Above, we found the smaller dimension (width) first: (91-11)/2 = 40. We could have found the length as (91+11)/2 = 51. The knowledge of the general solution to "sum and difference" problems often lets you work them with mental arithmetic.