Perimeter of a Trapezoid: Finding Area based off Perimeter and Vice Versa

Since ABCD is a trapezoid, one can determine that:

$AD=BC=7$

Thus the formula to find the area will be

$$$S_{ABCD}=\frac{(AB+DC)\times h}{2}$

Since we are given the perimeter of the trapezoid, we can find$AB+DC$

$P_{ABCD}=7+AB+7+DC$

$34=14+AB+DC$

$34-14=AB+DC$

$20=AB+DC$

Now we will place the data we obtained into the formula in order to calculate the area of the trapezoid:

$S=\frac{20\times5}{2}=\frac{100}{2}=50$

We can find the height BE by calculating the trapezoidal area formula:

$S=\frac{(AB+CD)}{2}\times h$

We replace the known data: $9=\frac{(3+6)}{2}\times BE$

We multiply by 2 to get rid of the fraction:

$9\times2=9\times BE$

$18=9BE$

We divide the two sections by 9:

$\frac{18}{9}=\frac{9BE}{9}$

$2=BE$

If we draw the height from A to CD we get a rectangle and two congruent triangles. That is:

$AF=BE=2$

$AB=FE=3$

$ED=CF=1.5$

Now we can find one of the legs through the Pythagorean theorem.

We focus on triangle BED:

$BE^2+ED^2=BD^2$

We replace the known data:

$2^2+1.5^2=BD^2$

$4+2.25=DB^2$

$6.25=DB^2$

We extract the root:

$\sqrt{6.25}=DB$

$2.5=DB$

Now that we have found DB, it can be argued that:

$AC=BD=2.5$

We calculate the perimeter of the trapezoid:$6+3+2.5+2.5=$

$9+5=14$