Answer:
1) [tex]30\ in\ by\ 12\ in[/tex] --------> [tex]120\ in^{2}[/tex]
2) [tex]30\ in\ by\ 18\ in[/tex] --------> [tex]180\ in^{2}[/tex]
3) [tex]12\ in\ by\ 15\ in[/tex] --------> [tex]60\ in^{2}[/tex]
4) [tex]18\ in\ by\ 15\ in[/tex] --------> [tex]90\ in^{2}[/tex]
Step-by-step explanation:
Verify each case
case 1) [tex]30\ in\ by\ 12\ in[/tex]
Find the area with the original dimensions
[tex]A=30(12)=360\ in^{2}[/tex]
Reduce the area of the poster by one-third
[tex]360/3=120\ in^{2}[/tex]
therefore
[tex]30\ in\ by\ 12\ in[/tex] --------> [tex]120\ in^{2}[/tex]
case 2) [tex]30\ in\ by\ 18\ in[/tex]
Find the area with the original dimensions
[tex]A=30(18)=540\ in^{2}[/tex]
Reduce the area of the poster by one-third
[tex]540/3=180\ in^{2}[/tex]
therefore
[tex]30\ in\ by\ 18\ in[/tex] --------> [tex]180\ in^{2}[/tex]
case 3) [tex]12\ in\ by\ 15\ in[/tex]
Find the area with the original dimensions
[tex]A=12(15)=180\ in^{2}[/tex]
Reduce the area of the poster by one-third
[tex]180/3=60\ in^{2}[/tex]
therefore
[tex]12\ in\ by\ 15\ in[/tex] --------> [tex]60\ in^{2}[/tex]
case 4) [tex]18\ in\ by\ 15\ in[/tex]
Find the area with the original dimensions
[tex]A=18(15)=270\ in^{2}[/tex]
Reduce the area of the poster by one-third
[tex]270/3=90\ in^{2}[/tex]
therefore
[tex]18\ in\ by\ 15\ in[/tex] --------> [tex]90\ in^{2}[/tex]
