Question

Figure ABCD is graphed on a coordinate plane.

Answer 1

20.9 Is your answer

Answer 2

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. $CD^2=HD^2+CH^2$

$CD^2=2^2+4^2=4+16=20$

$CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47$

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)