Answer:
Line segment AD
[tex]AD = \sqrt{(c-0)^{2}+(d-0)^{2}}[/tex]
[tex]AD = \sqrt{c^{2}+d^{2}}[/tex]
Line segment BC
[tex]BC = \sqrt{[(b+c)-b]^{2}+(d-0)^{2}}[/tex]
[tex]BC = \sqrt{c^{2}+d^{2}}[/tex]
Line segment AB
[tex]AB = \sqrt{(b-0)^{2}+(0-0)^{2}}[/tex]
[tex]AB = \sqrt{b^{2}+0^{2}}[/tex]
[tex]AB = b[/tex]
Line segment CD
[tex]CD = \sqrt{[c-(b+c)]^{2}+(d-d)^{2}}[/tex]
[tex]CD = \sqrt{b^{2}+0^{2}}[/tex]
[tex]CD = b[/tex]
Step-by-step explanation:
We defined the length of each side by the Equation of the Line Segment, which is a particular case of the Pythagorean Theorem. Let [tex]A(x,y) = (0,0)[/tex], [tex]B(x,y) = (b,0)[/tex], [tex]C(x,y) = (b+c, d)[/tex] and [tex]D(x,y) = (c,d)[/tex], we construct the equations below:
Line segment AD
[tex]AD = \sqrt{(c-0)^{2}+(d-0)^{2}}[/tex]
[tex]AD = \sqrt{c^{2}+d^{2}}[/tex]
Line segment BC
[tex]BC = \sqrt{[(b+c)-b]^{2}+(d-0)^{2}}[/tex]
[tex]BC = \sqrt{c^{2}+d^{2}}[/tex]
Line segment AB
[tex]AB = \sqrt{(b-0)^{2}+(0-0)^{2}}[/tex]
[tex]AB = \sqrt{b^{2}+0^{2}}[/tex]
[tex]AB = b[/tex]
Line segment CD
[tex]CD = \sqrt{[c-(b+c)]^{2}+(d-d)^{2}}[/tex]
[tex]CD = \sqrt{b^{2}+0^{2}}[/tex]
[tex]CD = b[/tex]
