Question

A rectangular park measuring 32 yards by 24 yards is surrounded by a trail of uniform width. If the area of the park and the trail combine is 1748 square yards, what is the width of the park

Answer 1

Answer:

The width = 38 yard

Step-by-step explanation:

Given

Dimension of Park = 32 by 24 yard

Area = 1748 yd²

Required

Find the width of the park

Given that the park is surrounded by a trail;

Let the distance between the park and the trail be represented with y;

Such that, the dimension of the park becomes (32 + y + y) by (24 + y + y) because it is surrounded on all sides

Area of rectangle is calculated as thus;

Area = Length * Width

Substitute 1748 for area; 32 + 2y and 24 + 2y for length and width

The formula becomes

$1748 = (32 + 2y) * (24 +2y)$

Open Bracket

$1748 = 32(24 + 2y) + 2y(24 + 2y)$

$1748 = 768 + 64y + 48y + 4y^2$

$1748 = 768 + 112y + 4y^2$

Subtract 1748 from both sides

$1748 -1748 = 768 -1748 + 112y + 4y^2$

$0 = 768 -1748 + 112y + 4y^2$

$0 = -980 + 112y + 4y^2$

Rearrange

$4y^2 + 112y -980 = 0$

Divide through by 4

$y^2 + 28y - 245 = 0$

Expand

$y^2 + 35y -7y - 245 = 0$

Factorize

$y(y+35) - 7(y + 35) = 0$

$(y-7)(y+35) = 0$

Split the above into two

$y - 7 = 0\ or\ y + 35 = 0$

$y = 7\ or\ y = -35$

But y can't be less than 0;

$So,\ y = 7$

Recall that the dimension of the park is 32 + 2y by 24 + 2y

So, the dimension becomes 32 + 2*7 by 24 + 2*7

Dimension = 32 + 14 yard by 24 + 14 yard

Dimension = 46 yard by 38 yard

Hence, the width = 38 yard