Question

You want the area of the platform in the tree house to be 12/5 sq yd. If the length of the platform is 4/3 yd, then the width of the platform is

• A. $\frac{15}{3}$ yd
• B. $\frac{3}{15}$ yd
• C. $\frac{1}{15}$ yd

Answer 1

The correct option is A $\frac{15}{3}$ yd

$\text{Area of the tree house =}\frac{12}{5}sqyd.$

$\text{Length =}\frac{4}{3}yd$

$\text{We know, Area = Length}\times \text{Width}$

$\text{∴ Width =}\frac{Area}{Length}$

$=\frac{12}{5}÷\frac{4}{3}$

$=\frac{12}{5}\times \frac{3}{4}$

$=\frac{9}{5}yd$