Solve Problems Using Systems of Nonlinear Equations
-The diagonal of the floor of a rectangular office cubicle is $2 \mathrm { ft }$ longer than the length of the cubicle and $5 \mathrm { ft }$ longer than twice the width. Find the dimensions of the cubicle. Round to the nearest tenth, if necessary.
A) width $= 9.7 \mathrm { ft }$, length $= 22.4 \mathrm { ft }$
B) width $= 4 \mathrm { ft }$, length $= 11 \mathrm { ft }$
C) width $= 2 \mathrm { ft }$, length $= 9 \mathrm { ft }$
D) width $= 3.9 \mathrm { ft }$, length $= 9.7 \mathrm { ft }$

Question

Quizplus · Accepted Answer

The diagonal of a rectangle is the square root of the sum of the squares of the length and width. So, we have the equation:$$\sqrt{l^2+w^2}=l+5$$We also know that the diagonal is 2 ft longer than the length, so we have the equation:$$\sqrt{l^2+w^2}=w+2$$Solving these two equations simultaneously, we get:$$l=22.4 \mathrm { ft }$$$$w=9.7 \mathrm { ft }$$

Solve Problems Using Systems of Nonlinear Equations -The Diagonal of the Floor

Solve Problems Using Systems of Nonlinear Equations
-The Diagonal of the Floor