ANSWER
20m and 40m
EXPLANATION
A rectangular ballroom has two dimensions: length and width,
The perimeter is the sum of all the sides of the rectangle,
[tex]P=L+L+W+W=2L+2W[/tex]The area is the product of these two dimensions,
[tex]A=L\cdot W[/tex]We know that the perimeter is 120m and the area is 800m². Replacing in the equations above, we have two equations with two variables L and W,
[tex]\begin{gathered} 120=2L+2W \\ 800=LW \end{gathered}[/tex]Solve the first equation for W. Subtract 2L from both sides,
[tex]\begin{gathered} 120-2L=2L-2L+2W \\ 120-2L=2W \end{gathered}[/tex]And divide both sides by 2,
[tex]\begin{gathered} \frac{120-2L}{2}=\frac{2W}{2} \\ 60-L=W \end{gathered}[/tex]The next step is to replace W by this expression in the second equation,
[tex]800=L(60-L)[/tex]Now we have to solve this equation for L. Apply the distributive property on the right side of the equation,
[tex]800=60L-L^2[/tex]Subtract 800 from both sides,
[tex]\begin{gathered} 800-800=-L^2+60L-800 \\ 0=-L^2+60L-800 \end{gathered}[/tex]We have a quadratic equation to find the zeros of the polynomial. To solve it, we can use the quadratic formula,
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In this case x = L, a = -1, b = 60 and c = -800,
[tex]L=\frac{-60\pm\sqrt[]{60^2-4\cdot(-1)\cdot(-800)}}{2\cdot(-1)}[/tex][tex]L=\frac{-60\pm\sqrt[]{3600-3200}}{-2}=\frac{-60\pm\sqrt[]{400}}{-2}=\frac{-60\pm20}{-2}[/tex]We have two possible results for L,
[tex]L_1=\frac{-60+20}{-2}=20[/tex][tex]L_2=\frac{-60-20}{-2}=40[/tex]To find the width of the ballroom we have to replace L into the equation we found when we solved the first equation for W,
[tex]W=60-L[/tex]So we have two possible results for W as well,
[tex]W_1=60-L_1=60-20=40[/tex][tex]W_2=60-L_2=60-40=20[/tex]Note that the pairs are 20m and 40m or 40m and 20m. Hence, we can say that the dimensions of the ballroom are 20m and 40m
