Question

A rectangle whose perimeter is 80 m has an area of 384 m^2. Find the dimensions of the rectangle.

Answer 1

$2a+2b=80 \ /:2\ \ \ \Rightarrow\ \ \ \ a+b=40\ \ \ \Rightarrow\ \ \ a=40-b\\ \\a\cdot b=384\ \ \ \ \Rightarrow\ \ \ (40-b)\cdot b=384\ \ \ \ \Rightarrow\ \ \ -b^2+40b-384=0\\ \\ \Delta=40^2-4\cdot (-1)\cdot(-384)=1600-1536=64\ \ \ \Rightarrow\ \ \ \sqrt{\Delta} =8\\ \\b_1= \frac{-40-8}{2\cdot(-1)}= \frac{-48}{-2} =24\ \ \ \ \Rightarrow\ \ \ a_1=40-b_1=40-24=16 \\ \\b_2= \frac{-40+8}{2\cdot(-1)}= \frac{-32}{-2} =16\ \ \ \ \Rightarrow\ \ \ a_2=40-b_2=40-16=24$

$Ans.\ The\ dimensions\ of\ the\ rectangle:\ \ \ 16\ m\ \ \ and\ \ \ 24\ m$

Answer 2

Since area of a rectangle =  length X width I came up with an answer of 16 X 24 =384
Also, 16+16+24+24 = 80