you can solve this by setting up 2 equations and solving quadratically.
perimeter of a rectangle is found by the formula 2*base + 2*height = perimeter.
area is found by base*height=area.
2b+2h = 24 and b*h=27 solving one equation for base or height and substituting in the other equation will give us an equation we can use for this particular problem. solving b*h=27 for b will give us b= 27/h. substituting this into the perimeter equation we get 2(27/h) + 2h = 24. If we use algebra to manipulate the equation we get 24-2h= 2(27/h) so 24-2h = 54/h multiplying both sides by h give (24-2h)h = 54 moving the 24h-2h^2 to the other side gives 0 = 2h^2-24h+54 solving the quadratic for h gives (2h+6)(h+9)=0 so solving for h we get h=6/2 = 3 or h=9 is we use h =9 and plug it into the original equation for 2b+2h=24 then we get 2b+2(9)= 24 so then 2b = 24-18 so b = 6/2 =3. so we get 2(3)+2(9)=24 so 6+18=24 for perimeter and for b*h=27 we get 3*9=27 so the dimensions are 3x9 3 is the shorter side and 9 is the longer side = 3x9
