Question

A rectangular garden is 5 feet longer than twice its width. It has a sidewalk 3 feet wide on two of its sides. The area of the sidewalk is 213 square feet. Find the dimensions of the garden.

Answer 1

The length and width of the garden are 47 ft and 21 ft respectively.

Explanation

Suppose, the width of the garden is $x$ feet.

As the garden is 5 feet longer than twice its width, so the length will be: $(2x+5) feet$

So, the area of the garden $= length*width=x(2x+5)ft^2$

Now, the garden has a sidewalk 3 feet wide on two of its sides. That means, the length of the garden including the sidewalk $=(2x+5+3)ft =(2x+8)ft$ and the width including the sidewalk $=(x+3)ft$

Given that, the area of the sidewalk is 213 ft². So the equation will be.....

$(2x+8)(x+3)-x(2x+5)=213\\ \\ 2x^2+14x+24-2x^2-5x=213\\ \\ 9x+24=213\\ \\ 9x=189\\ \\ x=\frac{189}{9}=21$

So, the width of the garden is 21 feet and the length is $(2*21+5)ft= 47 ft$