Question

The vertices of a backyard are W(10,30),X(10,100),Y(110,100),Z(50,30). The coordinates are measured in feet. The line segment XZ separates the backyard into a lawn and a garden. The area of the lawn is greater than the area of the garden. How many times larger is the lawn than the garden?

Answer 1

2.5

Explanation:

Answer 2

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.
`WX=√((10-10)^2+(30-100)^2)=√(70^2)=70\ ft;`

2.
`XY=√((110-10)^2+(100-100)^2)=√(100^2)=100\ ft;`

3.
`YZ=√((110-50)^2+(100-30)^2)=√(60^2+70^2)=10√(85)\ ft;`

4.
`ZW=√((50-10)^2+(30-30)^2)=√(40^2)=40\ ft;`

5.
`XZ=√((10-50)^2+(100-30)^2)=√(40^2+70^2)=10√(65)\ ft.`

Then:

1. the area

`A_(WXZ)=\sqrt{(70+40+10√(65))/(2)\\\cdot ((70+40+10√(65))/(2)-70)\cdot ((70+40+10√(65))/(2)-40)\cdot ((70+40+10√(65))/(2)-10√(65))}=\\ \\=1400\ ft^2.`

2. the area

`A_(XYZ)=\sqrt{(100+10√(85)+10√(65))/(2)\cdot ((100+10√(85)+10√(65))/(2)-100)}\cdot\\ \\\cdot\sqrt{((100+10√(85)+10√(65))/(2)-10√(85))\cdot ((100+10√(85)+10√(65))/(2)-10√(65))}=3500\ ft^2.`

Then

`(A_(XYZ))/(A_(WXZ))=(3500)/(1400)=2.5` times greater.