Question 59609
The sum of the perimeters of two squares is 40 inches.  the sum of their areas is 58 square inches.  What are the dimensions of the squares?
:
Let the sides of one square be: x
Let the sides of the other square be:y
:
The perimeter of a square is: {{{highlight(P=4s)}}}, p=perimeter, s=side
E1) Therefore the sum of the perimeters is 40 is: {{{4x+4y=40}}}
:
The area of a square is: {{{highlight(A=s^2)}}}, A=area, s=side
E2) Therefore the sunm of their area is 58 is: {{{x^2+y^2=58}}}
:
Therefore you have two equations and two unknowns.  Solve E1 for either variable, I'm going with y.
4x+4y=40
-4x+4x+4y=-4x+40
4y=-4x+40
{{{4y/4=-4x/4+40/4}}}
y=-x+10
Substitute that into E2 for y and solve for x:
{{{x^2+(-x+10)^2=58}}}
{{{x^2+(-x+10)(-x+10)=58}}}
{{{x^2+x^2-10x-10x+100=58}}}
{{{2x^2-20x+100=58}}}
{{{2x^2-20x+100-58=58-58}}}
{{{2x^2-20x+42=0}}}
{{{2x^2/2-20x/2+42/2=0/2}}}
{{{x^2-10x+21=0}}} Factor
(x-3)(x-7)=0  set each parentheses = to 0
x-3=0 and x-7=0
x-3+3=0+3 and x-7+7=0+7
x=3 and x=7
If you were to substitute either x into E1 you'd get the other x
4(3)+4y=40
12+4y=40
4y=28
4y/4=28/4
y=7 You can check 7 for yourself, you'll get 3.
One square is 3 in x 3 in
the other is 7 in x 7 in
Happy Calculating!!!