Question

Joan wants to reduce the area of her posters by one-third. Draw lines to match the original dimensions in the left column with the correct new area in the right column. Not all dimensions will have a match.​

Answer 1

1)
`30\ in\ by\ 12\ in` -------->
`120\ in^(2)`

2)
`30\ in\ by\ 18\ in` -------->
`180\ in^(2)`

3)
`12\ in\ by\ 15\ in` -------->
`60\ in^(2)`

4)
`18\ in\ by\ 15\ in` -------->
`90\ in^(2)`

Explanation:

Verify each case

case 1)
`30\ in\ by\ 12\ in`

Find the area with the original dimensions

`A=30(12)=360\ in^(2)`

Reduce the area of the poster by one-third

`360/3=120\ in^(2)`

therefore

`30\ in\ by\ 12\ in` -------->
`120\ in^(2)`

case 2)
`30\ in\ by\ 18\ in`

Find the area with the original dimensions

`A=30(18)=540\ in^(2)`

Reduce the area of the poster by one-third

`540/3=180\ in^(2)`

therefore

`30\ in\ by\ 18\ in` -------->
`180\ in^(2)`

case 3)
`12\ in\ by\ 15\ in`

Find the area with the original dimensions

`A=12(15)=180\ in^(2)`

Reduce the area of the poster by one-third

`180/3=60\ in^(2)`

therefore

`12\ in\ by\ 15\ in` -------->
`60\ in^(2)`

case 4)
`18\ in\ by\ 15\ in`

Find the area with the original dimensions

`A=18(15)=270\ in^(2)`

Reduce the area of the poster by one-third

`270/3=90\ in^(2)`

therefore

`18\ in\ by\ 15\ in` -------->
`90\ in^(2)`