In trapezoid \(ABCD\), sides \(AB\) and \(CD\) are parallel, and the lengths of sides \(AD\), \(AB\), and \(BC\) are equal.
If the perpendicular distance between \(AB\) and \(CD\) is \(8\) units, and the length of \(CD\) is equal to \(4\) more than half the sum of the other three side lengths, determine the area and perimeter of trapezoid \(ABCD\).
Let \(x\) represent the length of \(AD\). Then \(AD=AB=BC=x\). Since \(CD\) is equal to \(4\) more than half the sum of the other three side lengths, \(CD=\frac{3x}{2}+4\). Note that since \(x > 0\), \(\frac{3}{2}x + 4 > x\) and so \(CD\) is longer than \(AB\).
Draw altitudes from \(A\) and \(B\) meeting \(CD\) at \(E\) and \(F\), respectively. Then \(AE=BF=8\).
Let \(y\) represent the length of \(DE\). We can show that \(DE=CF\) using the Pythagorean Theorem as follows: \(DE^2=AD^2-AE^2=x^2-8^2=x^2-64\) and \(CF^2=BC^2-BF^2=x^2-8^2=x^2-64\). Then \(CF^2=x^2-64=DE^2\), so \(CF=DE=y\) since \(CF>0\).
Since \(\angle AEF = \angle BFE=90\degree\) and \(AB\) is parallel to \(CD\), it follows that \(\angle BAE=\angle ABF=90\degree\) and \(ABFE\) is a rectangle so \(AB=EF=x\).
We can now determine the relationship between \(x\) and \(y\). \[\begin{aligned} CD &= DE+EF+CF\\ \frac{3x}{2}+4 &= y+x+y\\ \frac{x}{2}+4 &= 2y\\ \frac{x}{4}+2 &= y \end{aligned}\] So \(DE=CF=\frac{x}{4}+2\). We then use the Pythagorean Theorem in \(\triangle AED\). \[\begin{aligned} AD^2 &= AE^2+DE^2\\ x^2 &= 8^2+\left(\frac{x}{4}+2\right)^2\\ x^2 &= 64 + \frac{x^2}{16}+x+4\\ 0 &= \frac{15x^2}{16}-x-68 \end{aligned}\] Using the quadratic formula, we can determine the value of \(x\). \[\begin{aligned} x &= \frac{1 \pm \sqrt{1-4\left(\tfrac{15}{16}\right)(-68)}}{2\left(\tfrac{15}{16}\right)}\\ &= \frac{1 \pm \sqrt{256}}{\tfrac{15}{8}}\\ &= \frac{8 \pm 128}{15} \end{aligned}\] Thus, \(x=-8\) or \(x=\frac{136}{15}\). Since \(x>0\), we can conclude that \(x=\frac{136}{15}\), so \(AD=AB=BC=\frac{136}{15}\). Then \(CD=\left(\frac{3}{2}\right)\left(\frac{136}{15}\right)+4=\frac{88}{5}\). Then we can calculate the area and perimeter of trapezoid \(ABCD\). \[\begin{aligned} \text{Area of }ABCD &= \frac{1}{2}\times AE \times (AB + CD)\\ &= \frac{1}{2}\times 8 \times \left( \frac{88}{5}+\frac{136}{15}\right)\\ &=\frac{320}{3} \end{aligned}\] \[\begin{aligned} \text{Perimeter of }ABCD &= DA + AB + BC + CD\\ &= 3\left(\frac{136}{15}\right) + \frac{88}{5}\\ &= \frac{224}{5} \end{aligned}\]
Therefore the area of trapezoid \(ABCD\) is \(\frac{320}{3}\text{ units}^2\) and the perimeter is \(\frac{224}{5}\text{ units}\).