Given vertices W(10,30), X(10,100), Y(110,100), Z(50,30) of a backyard, find distances WX, XY, YZ, ZW and XZ:
1. [tex]WX=\sqrt{(10-10)^2+(30-100)^2}=\sqrt{70^2}=70\ ft;[/tex]
2. [tex]XY=\sqrt{(110-10)^2+(100-100)^2}=\sqrt{100^2}=100\ ft;[/tex]
3. [tex]YZ=\sqrt{(110-50)^2+(100-30)^2}=\sqrt{60^2+70^2}=10\sqrt{85}\ ft;[/tex]
4. [tex]ZW=\sqrt{(50-10)^2+(30-30)^2}=\sqrt{40^2}=40\ ft;[/tex]
5. [tex]XZ=\sqrt{(10-50)^2+(100-30)^2}=\sqrt{40^2+70^2}=10\sqrt{65}\ ft.[/tex]
Then:
1. the area
[tex]A_{WXZ}=\sqrt{\frac{70+40+10\sqrt{65}}{2}\\\cdot (\frac{70+40+10\sqrt{65}}{2}-70)\cdot (\frac{70+40+10\sqrt{65}}{2}-40)\cdot (\frac{70+40+10\sqrt{65}}{2}-10\sqrt{65})}=\\ \\=1400\ ft^2.[/tex]
2. the area
[tex]A_{XYZ}=\sqrt{\frac{100+10\sqrt{85}+10\sqrt{65}}{2}\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-100)}\cdot\\ \\\cdot\sqrt{(\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{85})\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{65})}=3500\ ft^2.[/tex]
Then
[tex]\dfrac{A_{XYZ}}{A_{WXZ}}=\dfrac{3500}{1400}=2.5[/tex] times greater.
