Question 863534
b, bottom base, the longer base
t, top base
y, each side
-
{{{b=-5+2t}}}
and
{{{y=-3+t}}}
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SUM of the lengths of the sides is 90 meters, according to "perimeter of 90 meters".  Notice how b and y are described in terms of t.  


{{{90=(-5+2t)+t+2(-3+t)}}}
{{{-6+2t+t+2t-5=90}}}
{{{5t-11=90}}}
{{{5t=101}}}
{{{highlight(t=101/5)}}}.


The base angle at bottom is needed, so that the height can be found.  This isosceles trapezoid is composed of two right triangles, one on each side, and a rectangle in the middle.  You can draw this figure.


Call the base angle, measure a.
Let h = height of the triangle and also of the trapezoid.
The bottom leg OF EACH RIGHT TRIANGLE is {{{(b-t)/2}}}; and since you have a formula for b and have now found t, each bottom leg of the right triangle is {{{((2t-5)-t)/2}}}
{{{(t-5)/2}}}
{{{(101/5-5)/2}}}
{{{(101-25)/(5*2)}}}
{{{76/10}}}
{{{highlight_green(38/5)}}}, leg length of a right triangle.
-
The hypotenuse of each right triangle is y.  
{{{cos(a)=(38/5)/y}}}, and a formula for y is already described in terms of t.
{{{(38/5)/(t-3)=cos(a)}}}
{{{(38/5)/(101/5-3)=cos(a)}}}
{{{(38/5)/(101/5-15/5)=cos(a)}}}
{{{(38/5)/(86/5)=cos(a)}}}
{{{cos(a)=38/86}}}
{{{highlight(a=63.8)}}} degrees
-
{{{highlight(h=y*sin(63.8))}}}
{{{h=(t-3)sin(63.8)}}}
{{{h=(101/5-3)sin(63.8)}}}
{{{highlight(h=(86/5)sin(63.8))}}}-----The HEIGHT of the trapezoid.


AREA:
{{{((b+t)/2)h}}}
{{{((2t-5+t)/2)(86/5)sin(63.8)}}}
{{{((3t-5)/2)(86/5)sin(63.8)}}}
{{{((3(101/5)-5)/2)(86/5)sin(63.8)}}}
.... you can finish this computation.