Question 138214
How do you draw a figure with an area of 64 cm squared and a perimeter of 64 cm
Trial and error will be difficult because I don't think the sides will be integers.
:
Let x = length
Let y = width
:
x*y = 64; area equation
and
2x + 2y = 64; perimeter equation
Simplify, divide by 2
x + y = 32
y = (32-x)
:
Substitute (32-x) for y in the area equation
x(32-x) = 64
32x - x^2 - 64 = 0
Multiply equation by -1, arrange as a quadratic equation:
x^2 - 32x + 64 = 0
:
Use the quadratic formula; a=1; b=-32; c=64
:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
:
{{{x = (-(-32) +- sqrt(-32^2 - 4 * 1 * 64 ))/(2*1) }}}
:
{{{x = (32 +- sqrt(1024 - 256 ))/(2) }}}

If you do the math here you will get two solutions
:
x = 29.8564 or x = 2.1436 One value is the length and the other is the width