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

\frac{15}{3}

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\frac{3}{15}

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\frac{1}{15}

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Solution

The correct option is A $\frac{15}{3}$ yd
$Area of the tree house = \frac{12}{5} s q y d .$

$Length = \frac{4}{3} y d$

$We know, Area = Length \times Width$

$∴ Width = \frac{A r e a}{L e n g t h}$

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

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

$= \frac{9}{5} y d$

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