Question 169672
Let x=length of field and y=width of field



Remember, the perimeter of any triangle is simply the sum of its sides. In this case, two of the sides are the length and width (x and y) and the third side is the diagonal of the field



So the perimeter is {{{P=x+y+15}}}


Now plug in {{{P=36}}} to get {{{36=x+y+15}}}. After solving for y, we get {{{y=21-x}}}



Now since the diagonal splits the field into two triangles, where the length and width form the legs and the diagonal forms the hypotenuse, this means that we can use Pythagorean's Theorem


{{{a^2+b^2=c^2}}} Start with Pythagorean's Theorem



{{{x^2+y^2=15^2}}} Plug in {{{a=x}}}, {{{b=y}}} (the length and width) and {{{c=15}}} (the diagonal)



{{{x^2+y^2=225}}} Square 15 to get 225



{{{x^2+(21-x)^2=225}}} Plug in {{{y=21-x}}} (the previous isolated equation)



{{{x^2+441-42x+x^2=225}}} FOIL



{{{x^2+441-42x+x^2-225=0}}} Subtract 225 from both sides



{{{2x^2-42x+216=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-42}}}, and {{{c=216}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-42) +- sqrt( (-42)^2-4(2)(216) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-42}}}, and {{{c=216}}}



{{{x = (42 +- sqrt( (-42)^2-4(2)(216) ))/(2(2))}}} Negate {{{-42}}} to get {{{42}}}. 



{{{x = (42 +- sqrt( 1764-4(2)(216) ))/(2(2))}}} Square {{{-42}}} to get {{{1764}}}. 



{{{x = (42 +- sqrt( 1764-1728 ))/(2(2))}}} Multiply {{{4(2)(216)}}} to get {{{1728}}}



{{{x = (42 +- sqrt( 36 ))/(2(2))}}} Subtract {{{1728}}} from {{{1764}}} to get {{{36}}}



{{{x = (42 +- sqrt( 36 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (42 +- 6)/(4)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{x = (42 + 6)/(4)}}} or {{{x = (42 - 6)/(4)}}} Break up the expression. 



{{{x = (48)/(4)}}} or {{{x =  (36)/(4)}}} Combine like terms. 



{{{x = 12}}} or {{{x = 9}}} Simplify. 



So the answers are {{{x = 12}}} or {{{x = 9}}}




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Answer:



So the dimensions of the field are 12 m by 9 m