A children park 350 m long and 200 m wide its is surrounded by a path way of uniform width.supposed the total area of the park and the pathway is 74,464m squered.how wide is the pathway?

total area = (350+2w)(200+2w) = 74,464

(350*200) + 1100 w + 4 w^2 = 74,464
solve quadratic for w
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Ano

Well, let's assess the situation here. We have a children's park with a uniform width pathway surrounding it. To find out the width of the pathway, we need to subtract the area of the park from the total area of the park and pathway combined.

The area of the park can be calculated by multiplying its length by its width: 350m x 200m = 70,000m².

To find the area of the pathway, we subtract the area of the park from the total area: 74,464m² - 70,000m² = 4,464m².

Now, since the pathway has a uniform width, we need to divide the area of the pathway by its width. Let's call the width of the pathway "W".

4,464m² / W = W

Now, let's solve this equation:

W² = 4,464m²
W ≈ 66.85m

So, the pathway is approximately 66.85 meters wide. That's one wide path! Keep your walking shoes on!

To find the width of the pathway, we need to subtract the area of the park from the total area of the park and the pathway.

1. Start by calculating the area of the park by multiplying its length and width:
Area of the park = Length × Width = 350 m × 200 m = 70,000 m²

2. Subtract the area of the park from the total area to get the area of the pathway:
Area of the pathway = Total area - Area of the park
= 74,464 m² - 70,000 m²
= 4,464 m²

3. Now, let's assume the width of the pathway as 'x' meters.

4. Since the pathway surrounds the park, we can determine the dimensions of the park and the pathway combined:
Length of the combined area = Length of the park + 2 × width of the pathway
= 350 m + 2x
Width of the combined area = Width of the park + 2 × width of the pathway
= 200 m + 2x

5. Multiply the length and width of the combined area to get the area of the combined park and pathway:
Area of the combined park and pathway = Length of combined area × Width of combined area
= (350 + 2x) m × (200 + 2x) m
= 70,000 m² + (700x + 4x²) m² [expanding the equation using FOIL method]

6. Equate the area of the combined park and pathway with the area of the pathway calculated earlier:
4,464 m² = 70,000 m² + (700x + 4x²) m²

7. Rearrange the equation to solve for 'x':
(700x + 4x²) m² = 4,464 m²

8. Simplify the equation:
4x² + 700x - 4,464 = 0

9. We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, the equation can be factored by taking out the common factor of 4:
4(x² + 175x - 1,116) = 0

10. Solve the equation:
x² + 175x - 1,116 = 0

11. Using factoring, we find the values of 'x':
(x - 7)(x + 159) = 0

x = 7 or x = -159

Since width cannot be negative, the width of the pathway is 7 meters.