You want to have the area of the platform in the treehouse as 12/5 sq m. If the length of the platform is 4/3 m, then the width of the platform is .

\frac{15}{3}

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\frac{3}{4}

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\frac{5}{12}

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\frac{9}{5}

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Solution

The correct option is D $\frac{9}{5}$ m
Area of the treehouse = $\frac{12}{5}$ sq m

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

Area = Length $\times$ Width

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

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

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

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

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You want to have the area of the platform in the treehouse as 12/5 sq m. If the length of the platform is 4/3 m, then the width of the platform is .

The correct option is D 95 mArea of the treehouse = 125 sq m Length = 43 Area = Length × Width Width = AreaLength = 125 ÷ 43 = 125 × 34 = 95 m

The correct option is D $\frac{9}{5}$ m
Area of the treehouse = $\frac{12}{5}$ sq m

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

Area = Length $\times$ Width

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

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

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

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