Answer:
The width of the pathway is [tex]4[/tex] meters.
Step-by-step explanation:
It is important that we are able to understand the situation in the problem. An organized solution to this problem helps us to present well the situation. It is also helpful to create a visualization to better understand the problem. We may employ mathematical symbols and statements to solve this problem.
1. Analyzation
2. Create visualization
3. Use mathematical symbols and equations
4. Write what are the known items.
5. Evaluate your solutions.
Since the park is 350 m long and 200 m wide, then the shape of the park must be rectangular. We also know that the area of a rectangle is given by [tex]A= l\times{w}[/tex]. The park is surrounded by a pathway of uniform width, so we can let [tex]x[/tex] be the width of the pathway. The opposite sides of the park would have an extension of
The park and the pathway would be [tex]350+2x[/tex] m long and [tex]200+2x[/tex] m wide. Since the total area of the park and pathway is 74464 sq. m then, we have the equation
[tex](350+2x)(200+2x)=74464[/tex].
Let's solve for [tex]x[/tex] in the equation above.
Solution:
[tex]\begin{aligned}(350+2x)(200+2x)&=74,464\\70,000+700x+400x+4x^2&=74,464\\4x^2+1100x+70,000&=74,464\\4x^2+1100x+70,000-74,464&=74,464-74,464\\4x^2+1100x-4464&=0\\4(x^2+275x-1116)&=0\\4(x^2+279x-4x-1116)&=0\\4[x(x+279)-4(x+279)]&=0\\4(x-4)(x+279)&=0\\4&\ne0\:\:\text{not a solution}\\x-4&=0\\x&=4\\&\text{or}\\x+279&=0\\x&=-279\end{aligned}[/tex]
Take note that we are dealing with distances and distances can't be a negative number, so the possible value of [tex]x[/tex] is 4.
Thus, the width of the pathway is [tex]4[/tex] meters.
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