Question 914554
My diagram shoes two squares.The sum of their area is 58m2.The sum of the lengths of their side is10m..Find values for x and y(y >x)..no idea how to go about this..Thanks
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One side of 1st square: x
Area of 1st square: {{{x^2}}}
One side of 2nd square: y
Area of 2nd square: {{{y^2}}}
Sum of their areas is {{{58m^2}}}, so: {{{x^2 + y^2 = 58}}} ------ eq (i)
Sum of the lengths of a side of each square is 10, so: x + y = 10____y = 10 - x  ------ eq (ii)
{{{x^2 + (10 - x)^2 = 58}}} ------- Substituting 10 - x for y in eq (i)
{{{x^2 + 100 - 20x + x^2 = 58}}}
{{{x^2 + x^2 - 20x + 100 - 58 = 0}}}
{{{2x^2 - 20x + 42 = 0}}}
{{{2(x^2 - 10x + 21) = 2(0)}}}
{{{x^2 - 10x + 21 = 0}}}
(x - 7)(x - 3) = 0
{{{highlight_green(x = 7)}}}       OR       {{{highlight_green(x = 3)}}}
If x = 7, then y = 3
If x = 3, then y = 7
However, since y > x, then the latter is the correct choice.