Question

The area of a rectangle is 108m^2 and its diagonal is 15m. Find the perimeter of the rectangle. Please without quadratics.

Answer 1

$\text{l-length}\\\text{w-width}\\lw-area\\108\ m^2-area\\d-diagonal\\\\\text{The equation}\ \#1:\\\\lw=108\to l=\dfrac{108}{w}\\\\\text{The equation}\ \#2:\\\\\text{Use the Pythagorean theorem:}\\\\l^2+w^2=d^2\to l^2+w^2=15^2\to l^2+w^2=225\\\\\text{Subtitute from}\ \#1\ \text{to}\ \#2:$

$\left(\dfrac{108}{w}\right)^2+w^2=225\\\\\dfrac{11,664}{w^2}+w^2=225\qquad\text{multiply both sides by }\ w^2\neq0\\\\11,664+w^4=225w^2\qquad\text{subtract}\ 225w^2\ \text{from both sides}\\\\w^4-225w^2+11,664=0\qquad\text{substitute}\ t=w^2 > 0\\\\t^2-225t+11,664=0\\\\\text{use the quadratic formula or solve by factoring}.\\\\t^2-144t-81t+11,664=0\\\\t(t-144)-81(t-144)=0\\\\(t-144)(t-81)=0\iff t-144=0\ or\ t-81=0\\\\\boxed{t=144}\ or\ \boxed{t=81}\to w^2=144\ or\ w^2=81\\\\w=\sqrt{144}\ or\ w=\sqrt{81}\\\\\boxed{w=12\ m}\ or\ \boxed{w=9\ m}$

$\text{Put the values of}\ w\ \text{to the equation}\ \#1:\\\\l=\dfrac{108}{12}\to\boxed{l=9\ m}\ or\ l=\dfrac{108}{9}\to\boxed{l=12}\\\\\text{The formula of a perimeter of a rectangle:}\\\\P=2l+2w\\\\P=2(12)+2(9)=24+18=42\\\\Answer:\ \text{The perimeter of the rectangle is equal 42 m}.$