johnny is directly in front of dougie, who is playing goalie as shown. johnny is 2.8m from both goal posts. he is also three times as far from dougie as dougie is from either post. he is also three times as far from dougie as dougie is from either post. determine the width of the net.

Question

johnny is directly in front of dougie, who is playing goalie as shown. johnny is 2.8m from both goal posts. he is also three times as far from dougie as dougie is from either post. he is also three times as far from dougie as dougie is from either post. determine the width of the net.

Answer 1

Let's assume the distance from Dougie to each goal post is "x".

Since Johnny is three times as far from Dougie as Dougie is from either post, the distance between Johnny and Dougie can be calculated as 3 * x.

According to the given information, Johnny is also 2.8m from both goal posts. Therefore, we can set up an equation:

2.8m + 2.8m + 3 * x = 3 * x

Simplifying the equation, we have:

5.6m = 3 * x

Now, we can solve for x by dividing both sides of the equation by 3:

x = 5.6m / 3

x ≈ 1.87m

Since x represents the distance from Dougie to each goal post, the width of the net is twice this distance:

Width of the net = 2 * x
Width of the net ≈ 2 * 1.87m
Width of the net ≈ 3.74m

Therefore, the width of the net is approximately 3.74 meters.

Answer 2

To determine the width of the net, let's break down the problem and use variables for the unknown values.

Let's assume the distance between the two goal posts is 'x'.

1. According to the problem, Johnny is directly in front of Dougie, meaning Johnny is in line with the center of the goal. Since Johnny is 2.8 meters away from both goal posts, the distance between Johnny and each goal post is 2.8 meters.

2. It is also given that Johnny is three times as far from Dougie as Dougie is from either post. This means if Dougie is 'd' meters away from either post, Johnny is 3d meters away from Dougie.

3. We can now set up an equation using the distances described. Since Dougie is equidistant from both posts, each distance will be equal.

So, the equation becomes:
2.8 + d + 3d = x

Simplifying the equation:
2.8 + 4d = x

Now, we need another equation to solve for 'x'.

4. We know that the total width of the net is twice the distance from the center to one post. So, the total width is 2 times (2.8 + d).

Therefore, our second equation becomes:
2(2.8 + d) = x

Simplifying this equation:
5.6 + 2d = x

Now, we have a system of equations:
2.8 + 4d = x
5.6 + 2d = x

By solving this system of equations, we can find the value of 'x' and thus determine the width of the net.