Question

What are the side lengths of a rectangle if the area = 40 in and the perimeter = 48 in

Answer 1

4, 10, 4, 10. 10*4 gives area of 40. 10+10+4+4 gives 48 as shown in question

Answer 2

$a,b-the\ side\ lengths\ of\ a\ rectangle\\ \\a\cdot b=40\ [in^2]\ \ \wedge\ \ \ 2\cdot(a+b)=48\ [in]\\ \\ a+b=24\ \ \ \Rightarrow\ \ \ a=24-b\ \ \ \Rightarrow\ \ \ ab=(24-b)b=24b-b^2\\ \\ab=40\ \ \ \Rightarrow\ \ \ 24b-b^2=40\ \ \ \Rightarrow\ \ \ -b^2+24b-40=0\ /\cdot(-1)\\ \\b^2-24b+40=0\ \ \Rightarrow\ \Delta=(-24)^2-4\cdot40=576-160=416=16\cdot 26$

$\sqrt{\Delta} =4 \sqrt{26} \ \ \ \Rightarrow\ \ \ b_1= \frac{24-4 \sqrt{26} }{2}=12-2 \sqrt{26}\ \Rightarrow\ a_1=12+2 \sqrt{26} \\ \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b_2= \frac{24+4 \sqrt{26} }{2}=12+2 \sqrt{26}\ \Rightarrow\ a_2=12-2 \sqrt{26}$