# A rectangular grass area of a park measures 50 yards \times 100 yards . The city wishes to put...

## Question:

A rectangular grass area of a park measures {eq}50 yards \times 100 yards {/eq}. The city wishes to put a uniform sidewalk around the grass area which would increase the area by {eq}459 yd ^2 {/eq}. What is the width of the sidewalk?

## Dimensions and Area of a Rectangle

A planner closed geometry with four sides and four right angles such that opposite sides are parallel and equal to each other are called the rectangle.

All four angles of a rectangle are of {eq}90^{\circ}. {/eq}

It has two dimensions length and width such that the opposite dimensions are equal in length so a square is also a rectangle.

A rectangle has two diagonals both are of equal length.

#### The Pythagorean theorem in a Rectangle

{eq}\displaystyle d^{2} = l^{2}+w^{2} {/eq}

here **d** is the length of diagonal

**l** and **w** are the length and width of the diagonal.

#### Area

{eq}\displaystyle A = l \times w {/eq}

where **A** is the area of the rectangle

## Answer and Explanation:

Given dimensions of the grass is -

{eq}\displaystyle l = 50 ~yds {/eq}

{eq}\displaystyle w = 100 ~yds {/eq}

So the area of the rectangular grass-

{eq}\displaystyle A = 50 \times 100 = 5000 yd^{2} {/eq}

Given that the city wishes to put a uniform sidewalk around the grass area which would increase the area {eq}\displaystyle 459 ~yd^{2}. {/eq}

So the area of the grass with the sidewalk-

{eq}\displaystyle A' = 5000+459 {/eq}

{eq}\displaystyle A' = 5459 ~yd^{2} -------(1) {/eq}

Let the width of the sidewalk is {eq}\displaystyle x ~yd {/eq}. then from equation(1)-

{eq}\displaystyle (50+2x)(100+2x) = 5459 {/eq}

{eq}\displaystyle 5000+100x+200x+4x^{2}-5459 = 0 {/eq}

{eq}\displaystyle 4x^{2}+300x-459 = 0 {/eq}

So by using the quadratic formula-

{eq}\displaystyle x = \frac{-300 \pm \sqrt{300^{2}-4(4)(-459)}}{2(4)} {/eq}

{eq}\displaystyle x = \frac{-300 \pm \sqrt{90000+7344}}{8} {/eq}

{eq}\displaystyle x = \frac{-300 \pm \sqrt{97344}}{8} {/eq}

{eq}\displaystyle x = \frac{-300 \pm 312}{8} {/eq}

so

{eq}\displaystyle x = \frac{-300 + 312}{8} = 1.5 {/eq}

or

{eq}\displaystyle x = \frac{-300 - 312}{8} = -76.5 {/eq}

because the width of the sidewalk can not be negative

{eq}\displaystyle \Rightarrow x = 1.5 ~yds {/eq}

So the width of the sidewalk is {eq}1.5 ~yds. {/eq}

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#### Learn more about this topic:

from Geometry: High School

Chapter 8 / Lesson 7