Tina usually brings her children every Sunday afternoon at the children's park in their barangay. The park is 350m long and 200m wide complete with slides and swings for children to play. It is surrounded by a pathway of uniform width for children who loves biking and jogging. Suppose the total area of the park and the pathway is 74,464 m?. a. How wide is the pathway of the park? B. If you were Tina, would you like your children to go biking and jogging in that park?

a. Let a be the width of the uniform pathway. Since the uniform pathway plus the park form a rectangle, of which the dimension of the park are 350 m long and 200 m wide respectively, then the dimensions of the park plus path way would be (350 + a) m long and (250 + a) m wide.

So the total area of the park plus pathway is (350 + a)(250 + a) which equals 74,464 m². So,

(350 + a)(200 + a) = 74,464

expanding the brackets, we have

70000 + 350a + 200a + a² = 74,464

collecting like terms, we have

a² + 550a = 74,464 - 70000

a² + 550a = 4,464

a² + 550a - 4,464 = 0

Using the quadratic formula, we find a,

So,

[tex]a = \frac{-550 +/-\sqrt{550^{2} - 4 X 1 X -4,464} }{2 X 1} \\a = \frac{-550 +/-\sqrt{302,500 + 17856} }{2} \\a = \frac{-550 +/-\sqrt{320,356} }{2} \\a = \frac{-550 +/-566} }{2} \\a = \frac{-550 - 566} }{2}or \frac{-550 + 566} }{2} \\a = \frac{-1116} }{2}or\frac{16} }{2} \\a = -558 or 8[/tex]

Since the width of the uniform pathway cannot be negative, so a = 8 m.

So, the width of the uniform pathway is 8 m.

b. If I were Tina, I would like my children to go biking or jogging in the park because, the park has a wide area = 350 m × 200 m = 70000 m² and its perimeter, P = 2(350 m + 200 m) = 2(550m) = 1100 m allows enough distance for biking and jogging.

The width of the pathway also allows for biking and jogging without the children bumping into each other.