Solve the problem.
-The width of a rectangle is 6 kilometers less than twice its length. If its area is 56 square kilometers, find the dimensions of the rectangle.
A) length $= 4 \mathrm {~km}$, width $= 2 \mathrm {~km}$
B) length $= 3 \mathrm {~km}$, width $= \frac { 56 } { 3 } \mathrm {~km}$
C) length $= 7 \mathrm {~km}$, width $= 8 \mathrm {~km}$
D) width $= 7 \mathrm {~km}$, length $= 8 \mathrm {~km}$

Question

Solve the problem.
-The width of a rectangle is 6 kilometers less than twice its length. If its area is 56 square kilometers, find the dimensions of the rectangle. 
A) length $= 4 \mathrm {~km}$, width $= 2 \mathrm {~km}$
B) length $= 3 \mathrm {~km}$, width $= \frac { 56 } { 3 } \mathrm {~km}$
C) length $= 7 \mathrm {~km}$, width $= 8 \mathrm {~km}$
D) width $= 7 \mathrm {~km}$, length $= 8 \mathrm {~km}$

Quizplus · Accepted Answer

Let the length be $ x $ km. Then the width is $ 2x - 6 $ km. The area is given by $ x(2x - 6) = 56 $. Solving $ 2x^2 - 6x - 56 = 0 $ gives $ x = 7 $ km (length) and width $ = 2(7) - 6 = 8 $ km.

Solve the Problem =4 km= 4 \mathrm {~km}=4 km , Width =2 km= 2 \mathrm {~km}=2 km

Solve the Problem $= 4 \mathrm {~km}$ , Width $= 2 \mathrm {~km}$