Question 182199
Let x = the length of the rectangle and y = the width of the rectangle.
The area is x*y = 48 sq.m.
The diagonal is 10 m, so {{{x^2+y^2 = 10^2}}} from the Pythagorean theorem.
So we have two equations:
1) {{{x*y = 48}}} Rewrite this equation as: {{{x = 48/y}}} and substitute into equation 2).
2) {{{x^2+y^2 = 100}}}
2a) {{{(48/y)^2+y^2 = 100}}} Simplifying this, we get:
2a) {{{2304/y^2 + y^2 = 100}}}
2a) {{{(2304 + y^4)/y^2 = 100}}} Multiply both sides by {{{y^2}}}
2a) {{{2304 + y^4 = 100y^2}}} Rearrange this into standard "quadratic" form.
{{{y^4-100y^2+2034 = 0}}} Solve by factoring, noting that {{{y^4 = (y^2)^2}}}
{{{(y^2)^2 - 100(y^2)+2304 = 0}}}
{{{((y^2)-64)((y^2)-36) = 0}}} Applying the zero product rule, we get:
{{{y^2-64 = 0}}} or {{{y^2-36 = 0}}} Factoring the left sides, we get:
{{{(y+8)(y-8) = 0}}} or {{{(y+6)(y-6) = 0}}} From which we get:
{{{y = -8}}}
{{{y = 8}}}
{{{y = -6}}}
{{{y = 6}}}
Discard the negative solutions as the lengths of the rectangle sides can only be positive.
So we have:
y = 6 or 8 and x = 8 or 6
The dimensions of the rectangle would be:
Length = 8m and width = 6m
Please note that while we solved only for y, the x-value is obtained by substituting y = 8 into eqation 1):
{{{x*y = 48}}} Substitute {{{y = 8}}}
{{{x*(8) = 48}}} Dividing both sides by 8.
{{{x = 6}}}