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

Answer:

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) $18\ in\ by\ 15\ in$ --------> $90\ in^{2}$

Step-by-step 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

$18\ in\ by\ 15\ in$ --------> $90\ in^{2}$