# A rectangular playground is 4 meters longer than it is wide. If the length is increased by 5 m,...

## Question:

A rectangular playground is 4 meters longer than it is wide. If the length is increased by 5 m, and the width decreased by 1 m, the area is increased by 15 {eq}m^2 {/eq}. Find the original dimensions of the playground.

The area of a rectangle is computed using the formula:

• {eq}A = l\times w {/eq}, where {eq}l {/eq} is the length and {eq}w {/eq} is the width of the rectangle.

Let the width of the original rectangle be {eq}w = x\; \rm m {/eq}. If the length is 4 meters longer than the width, then the expression for the length is:

• {eq}l = x + 4 {/eq}

Thus, the area of the original rectangle is:

• {eq}A = x(x + 4) = x^2 + 4x {/eq}

If the length is increased by 5 meters and the width reduced by 1 meter, the new length and width are:

• {eq}l_1 = x + 4 + 5 = x + 9 {/eq}
• {eq}w_1 = x - 1 {/eq}

The area of the new rectangle is:

• {eq}A_1 = (x - 1)(x + 9) = x^2 + 8x - 9 {/eq}

If the area of the new rectangle is 15 square meters more than the area of the original rectangle, then:

• {eq}A_1 - A = 15\; \rm m^2 {/eq}
• {eq}x^2 + 8x - 9 - x^2 - 4x = 15 {/eq}
• {eq}4x = 15 + 9 {/eq}
• {eq}4x = 24 {/eq}

Solving for x:

• {eq}x = \dfrac{24}{4} {/eq}
• {eq}x = 6\; \rm m {/eq}

Therefore, the dimensions of the original rectangular playground are:

• {eq}\boxed{w = x = 6\; \rm m} {/eq}
• {eq}\boxed{l = x + 4 = 6 + 4 = 10\; \rm m} {/eq} 