Answer:
Let's represent the width of the region by "w". According to the problem, the length of the region is 10 miles longer than the width, so we can represent the length as "w + 10".
The formula for the area of a rectangle is:
Area = Length x Width
So we can write an equation for the area of this region:
299 = (w + 10) x w
Expanding the right side, we get:
299 = w^2 + 10w
Now we can rearrange this equation into standard quadratic form:
w^2 + 10w - 299 = 0
We can solve for "w" by using the quadratic formula:
w = (-b ± sqrt(b^2 - 4ac)) / 2a
Where a = 1, b = 10, and c = -299. Plugging in these values, we get:
w = (-10 ± sqrt(10^2 - 4(1)(-299))) / 2(1)
w = (-10 ± sqrt(1180)) / 2
w = (-10 ± 34.351) / 2
We can ignore the negative solution, since the width of the region cannot be negative. So the width is:
w = (-10 + 34.351) / 2
w = 12.176
We can round the width to the nearest mile, since we can't have a fractional width. So the width is approximately 12 miles.
Now we can use the equation we derived earlier to find the length:
299 = (w + 10) x w
299 = (12 + 10) x 12
299 = 22 x 12
So the length is 22 miles.
Therefore, the width of the region is approximately 12 miles and the length is 22 miles.