1. BC is 1+3=4 (units)
2. AD is 3+5= 8 (units)
3. Draw the altitude CH from point C, as shown in the figure.
4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.
5. [tex]CD^2=HD^2+CH^2[/tex]
[tex]CD^2=2^2+4^2=4+16=20[/tex]
[tex]CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47 [/tex]
6. AB=CD=4.47 because the trapezoid is isosceles.
7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)
