Two line segments, \(CE\) and \(CF\), are perpendicular to each other, each with length \(10\). Rectangle \(ABCD\) is drawn so that \(D\) is on \(CE\), \(B\) is on \(CF\) with \(BF=4\), and the diagonal of \(ABCD\) has length \(10\). Line segments \(EA\) and \(AF\) are then drawn. Determine the perimeter of quadrilateral \(AFCE\), rounded to one decimal place.
Note: You may find the following useful:
The Pythagorean Theorem states, "In a right-angled triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides."
For example, if \(c\) is the length of the hypotenuse, and \(a\) and \(b\) are the lengths of the other two sides, then \(c^2=a^2+b^2\).
First, \(BC=CF-BF=10-4=6\). Since \(ABCD\) is a rectangle, \(AD=BC=6\). We then use the Pythagorean Theorem in \(\triangle ABC\). \[\begin{aligned} AB^2 &= AC^2-BC^2\\ &= 10^2 - 6^2\\ &= 100 - 36\\ &= 64 \end{aligned}\] Therefore \(AB=8\), since \(AB>0\). Since \(ABCD\) is a rectangle, \(CD=AB=8\). Then \(DE=CE-CD=10-8=2\). These lengths are shown on the diagram.
We then use the Pythagorean Theorem in \(\triangle ADE\). \[\begin{aligned} AE^2 &= AD^2+DE^2\\ &= 6^2 + 2^2\\ &= 36+4\\ &=40 \end{aligned}\] Therefore \(AE=\sqrt{40}\), since \(AE>0.\)
We then use the Pythagorean Theorem in \(\triangle ABF.\) \[\begin{aligned} AF^2 &= AB^2+BF^2\\ &= 8^2 + 4^2\\ &= 64+16\\ &=80 \end{aligned}\] Therefore \(AF=\sqrt{80}\), since \(AF>0.\)
Thus, the perimeter of \(AFCE\) is equal to \(AF+CF+CE+AE=\sqrt{80}+10+10+\sqrt{40}\approx 35.3.\)