I need to write concise formulas of these two:$\space \space \space$ $(1)$ finding the dimensions of a rectangle given the Area and the diagonal $\space \space \space$$(2)$ finding the dimensions of a rectangle given the Perimeter and the diagonal
Searching on google I can only find examples using number (eg. “Let’s assume the diagonal is $5$ and the area is $25$”etc) but I’m having trouble writing down formulas that use variables (so that I can use them in my code).
Would appreciate if someone could show me how to formulate the formulas (no pun intended) for these two. Thank you!
Let the dimensions of rectangle be $a$ and $b$ and let the Area be $A$, perimeter be $P$ and the diagonal be $D$ $$A=ab$$ $$D=\sqrt{a^2+b^2}$$ $$D^2=a^2+b^2$$ $$D^2=(a+b)^2-2ab=(a+b)^2-2A$$ $$a+b=\sqrt{D^2+2A}$$ Now $b=\dfrac{A}{a}$ $$a+\dfrac{A}{a}=\sqrt{D^2+2A}$$ $$a^2-a\sqrt{D^2+2A}+A=0$$ Now this is a quadratic equation in the form of $a$. Hence solving it gives $$a=\dfrac{\sqrt{D^2+2A}\space\pm\sqrt{D^2+2A-4A}}{2}=\dfrac{\sqrt{D^2+2A}\space\pm\sqrt{D^2-2A}}{2}$$ $$b=\dfrac{2A}{\sqrt{D^2+2A}\space\pm\sqrt{D^2-2A}}$$ The perimeter method $$P=2(a+b)$$ $$b=\dfrac{P-2a}{2}$$ $$D^2=a^2+b^2=a^2+\Big(\dfrac{P-2a}{2}\Big)^2$$ $$8a^2-4aP-4D^2+P^2=0$$ Now this is a quadratic equation in the form of $a$. Hence solving it gives $$a=\dfrac{P\pm \sqrt{8D^2-P^2}}{4}$$ $$b=\dfrac{P\mp \sqrt{8D^2-P^2}}{4}$$