Question

a coach gives her players the option of running around their field twice or around their entire stadium once.
the following diagram shows the field and stadium dimensions on a coordinate grid.
coordinate values are in meters.
which run is longer, and how much longer?

Explanation:

Step1: Analyze the field (a square)

The field is a square (since it has right angles and equal - looking sides). To find the side length, we can use the distance formula or observe the grid. Let's take two adjacent vertices, say from \((-60,0)\) to \((0,60)\) (approximate, but actually, looking at the right - angled sides, the length of each side of the square: using the distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For a side with endpoints \((-60,0)\) and \((0,60)\), \(d=\sqrt{(0 + 60)^2+(60 - 0)^2}=\sqrt{3600 + 3600}=\sqrt{7200}=60\sqrt{2}\) meters? Wait, no, actually, looking at the grid, the horizontal and vertical changes: from \((-60,0)\) to \((0,60)\), the horizontal change is \(60\) (from \(x=-60\) to \(x = 0\)) and vertical change is \(60\) (from \(y = 0\) to \(y=60\)). But actually, the field is a square with side length \(s\). Wait, another way: the distance between \((-60,0)\) and \((0,60)\) is \(s=\sqrt{(0+60)^2+(60 - 0)^2}=\sqrt{7200}=60\sqrt{2}\)? No, wait, maybe the side length is calculated as follows: the square has vertices, and the distance between two adjacent vertices (right - angled) can be found by looking at the grid. Let's take two points: \((-60,0)\) and \((0,60)\). The horizontal difference is \(60\) (from \(x=-60\) to \(x = 0\)) and vertical difference is \(60\) (from \(y = 0\) to \(y = 60\)). So the length of the side of the square \(s=\sqrt{60^{2}+60^{2}}=\sqrt{7200}=60\sqrt{2}\) meters. The perimeter of a square is \(P_{field}=4s\). But wait, maybe the square has side length \(s = \sqrt{(60)^2+(60)^2}\)? No, wait, actually, looking at the diagram, the field is a square with side length equal to the distance between \((-60,0)\) and \((0,60)\). Wait, maybe a better approach: the stadium is a circle with radius \(r = 120\) meters (since it goes from \(x=-120\) to \(x = 120\) on the x - axis, so diameter \(d = 240\), radius \(r = 120\)). The circumference of the stadium is \(C_{stadium}=2\pi r=2\pi\times120 = 240\pi\) meters.

Now, the field: let's find the side length. Take two adjacent vertices of the square, say \((-60,0)\) and \((0,60)\). The distance between them is \(d=\sqrt{(0+60)^2+(60 - 0)^2}=\sqrt{3600 + 3600}=\sqrt{7200}=60\sqrt{2}\) meters. Wait, but the square has 4 sides, so the perimeter of the field \(P_{field}=4\times60\sqrt{2}=240\sqrt{2}\) meters. Running around the field twice: \(2\times P_{field}=480\sqrt{2}\approx480\times1.414 = 678.72\) meters.

The circumference of the stadium: \(C_{stadium}=2\pi r\), with \(r = 120\) (since the circle goes from \(x=-120\) to \(x = 120\), so radius is \(120\)). So \(C_{stadium}=2\pi\times120 = 240\pi\approx240\times3.14 = 753.6\) meters.

Wait, but this can't be right. Wait, maybe the field is a square with side length \(s\), and we can calculate the side length as follows: looking at the coordinates, the square has vertices at \((-60,0)\), \((0,60)\), \((60,0)\), \((0,-60)\)? Wait, no, the diagram shows a square with right angles. Wait, maybe the side length is \(s=\sqrt{(60)^2+(60)^2}\)? No, wait, let's check the distance between \((-60,0)\) and \((0,60)\): \(\Delta x=60\), \(\Delta y = 60\), so \(s=\sqrt{60^{2}+60^{2}}=60\sqrt{2}\approx84.85\) meters. Then the perimeter of the field is \(4\times84.85 = 339.4\) meters. Running twice: \(2\times339.4 = 678.8\) meters.

The stadium: radius \(r = 120\) meters (since the circle is centered at the origin and goes to \(x = 120\), \(y = 120\) etc.). Circumference \(C=2\pi r=2\pi\times120 = 240\pi\approx753.6\) meters.

Now, find the difference: \(240\pi-480\sqrt{2}\approx753.6 - 678.8 = 74.8\) meters? Wait, no, maybe I made a mistake in the field's shape. Wait, the field is a square? Wait, the diagram shows a square with right angles, but maybe the side length is calculated as the distance between \((-60,0)\) and \((60,0)\)? No, that would be a horizontal line. Wait, no, the field has vertices at \((-60,0)\), \((0,60)\), \((60,0)\), \((0,-60)\). So it's a square rotated by \(45\) degrees. The distance from the center (origin) to each vertex (the radius of the circumscribed circle around the square) is \(60\sqrt{2}\)? Wait, no, the distance from \((0,0)\) to \((60,0)\) is \(60\), but to \((0,60)\) is \(60\), and to \((60,60)\)? No, the vertices are \((-60,0)\), \((0,60)\), \((60,0)\), \((0,-60)\). So the length of each side: between \((-60,0)\) and \((0,60)\): \(\sqrt{(0 + 60)^2+(60 - 0)^2}=\sqrt{7200}=60\sqrt{2}\approx84.85\) meters. So perimeter of field: \(4\times60\sqrt{2}=240\sqrt{2}\approx339.4\) meters. Running twice: \(2\times339.4 = 678.8\) meters.

Stadium: circumference \(C = 2\pi r\), \(r = 120\) meters (since the circle is centered at the origin and has a radius of \(120\) meters, as it reaches \(x = 120\), \(y = 120\) etc.). So \(C=2\pi\times120 = 240\pi\approx753.6\) meters.

Now, calculate the difference: \(240\pi-480\sqrt{2}\approx753.6 - 678.8 = 74.8\) meters. Wait, but maybe the field is a square with side length \(s = 120\) meters? No, that doesn't match the coordinates. Wait, another approach: the field is a square, and the distance between two opposite vertices (the diagonal) is \(120\) meters (from \(x=-60\) to \(x = 60\) is \(120\) meters, and \(y = 0\) to \(y = 0\) for one diagonal, but no, the other diagonal is from \(y=-60\) to \(y = 60\), length \(120\) meters). For a square, the diagonal \(d=s\sqrt{2}\), so if \(d = 120\), then \(s=\frac{120}{\sqrt{2}}=60\sqrt{2}\) meters, which matches our previous calculation. So perimeter of field is \(4s=4\times60\sqrt{2}=240\sqrt{2}\) meters. Running twice: \(480\sqrt{2}\) meters.

Stadium circumference: \(2\pi r\), \(r = 120\) meters, so \(240\pi\) meters.

Now, \(240\pi\approx753.6\), \(480\sqrt{2}\approx678.8\). So the stadium run is longer. The difference is \(240\pi-480\sqrt{2}\approx753.6 - 678.8 = 74.8\) meters, approximately \(75\) meters, or more accurately, \(240\pi-480\sqrt{2}=240(\pi - 2\sqrt{2})\approx240(3.1416 - 2.8284)=240\times0.3132 = 75.168\) meters, approximately \(75\) meters (or using more precise values, \(\pi\approx3.14159\), \(2\sqrt{2}\approx2.82843\), so \(\pi-2\sqrt{2}\approx0.31316\), \(240\times0.31316\approx75.16\) meters).

Wait, but maybe I made a mistake in the field's perimeter. Let's re - check: the field is a square. Let's find the length of one side. Take two adjacent vertices: \((-60,0)\) and \((0,60)\). The horizontal change is \(60\) (from \(x=-60\) to \(x = 0\)), vertical change is \(60\) (from \(y = 0\) to \(y = 60\)). So the length of the side is \(\sqrt{60^{2}+60^{2}}=\sqrt{7200}=60\sqrt{2}\approx84.85\) meters. The perimeter of the square is \(4\times84.85 = 339.4\) meters. Running around the field twice: \(2\times339.4 = 678.8\) meters.

The stadium: it's a circle with radius \(120\) meters (since the distance from the origin to the edge of the stadium on the x - axis is \(120\) meters). The circumference of a circle is \(C = 2\pi r\), so \(C=2\pi\times120 = 240\pi\approx753.6\) meters.

Now, subtract the two distances: \(240\pi-678.8\approx753.6 - 678.8 = 74.8\) meters, which is approximately \(75\) meters (or more precisely, \(240\pi-480\sqrt{2}\) meters).

Step2: Calculate the two distances

- Distance for running around the field twice:
The field is a square. Let the side length of the square be \(s\). Using the distance formula between two adjacent vertices \((x_1,y_1)=(-60,0)\) and \((x_2,y_2)=(0,60)\), \(s=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}=\sqrt{(0 + 60)^2+(60 - 0)^2}=\sqrt{7200}=60\sqrt{2}\) meters.
The perimeter of the square \(P = 4s=4\times60\sqrt{2}=240\sqrt{2}\) meters.
Running around the field twice: \(D_{field}=2\times P = 480\sqrt{2}\) meters.

- Distance for running around the stadium once:
The stadium is a circle with radius \(r = 120\) meters (from the coordinate grid, the circle extends to \(x=\pm120\) and \(y=\pm120\)).
The circumference of a circle is \(C = 2\pi r\), so \(D_{stadium}=2\pi\times120 = 240\pi\) meters.

Step3: Find the difference and determine which is longer

Calculate the numerical values:
\(480\sqrt{2}\approx480\times1.4142 = 678.816\) meters.
\(240\pi\approx240\times3.1416 = 753.984\) meters.

Since \(753.984>678.816\), the stadium run is longer.

The difference is \(240\pi - 480\sqrt{2}\approx753.984 - 678.816 = 75.168\) meters, approximately \(75\) meters (or more precisely, \(240(\pi - 2\sqrt{2})\) meters).

Answer:

The run around the stadium is longer. It is approximately \(75\) meters (or \(240\pi - 480\sqrt{2}\approx75\) meters) longer. (If we need an exact form, it's \(240(\pi - 2\sqrt{2})\) meters, and approximately \(75\) meters; if we use more precise decimal values, approximately \(75.2\) meters)