Question

10.5.PS-18 Question content area top Part 1 The diagram shows a track composed of a rectangle with a semicircle on each end. The area of the rectangle is square meters 11200. What is the perimeter of the​ track? Use 3.14 for pi.

Answer 1

First, we need to find the length and width of the rectangle:

Let's suppose the width of the rectangle is "w" and the length is "l".

The area of the rectangle is given as 11200 square meters:

lw = 11200

Now, we need to find the radius of each semicircle. Since the semicircle completes the circle with the rectangle width "w", the radius would be half of the width:

r = w/2

The perimeter of the track can be found by adding the perimeter of the rectangle with the circumference of both semicircles:

Perimeter = 2(l + w) + 2πr
Perimeter = 2(l + w) + 2(3.14)(w/2)
Perimeter = 2(l + w) + 3.14w

We know the area of the rectangle, which is lw = 11200, but we need to find the values of l and w. We can do this by trying different values of l and w that multiply to give 11200.

Here are a few possibilities:
l = 280, w = 40 -> Perimeter = 2(280 + 40) + 3.14(40) = 681.6
l = 560, w = 20 -> Perimeter = 2(560 + 20) + 3.14(20) = 1166.8
l = 1120, w = 10 -> Perimeter = 2(1120 + 10) + 3.14(10) = 2253.6

Therefore, the perimeter of the track could be 681.6 meters, 1166.8 meters, or 2253.6 meters, depending on the values of length and width. Without additional information, we cannot determine the exact perimeter of the track