Question

team task: stations!
enrique is hiking on the campground map. he will start from the nature center at ( f(0,0) ) and then to the waterfall at ( e(10,30) ). from there, he will walk to the river overlook at ( d(-50,50) ) before returning to the nature center. determine how long the hike will be.

Explanation:

Step1: Find distance \( F E \)

Use distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). For \( F(0,0) \) and \( E(10,30) \):
\( FE = \sqrt{(10 - 0)^2 + (30 - 0)^2} = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10} \)

Step2: Find distance \( E D \)

For \( E(10,30) \) and \( D(-50,50) \):
\( ED = \sqrt{(-50 - 10)^2 + (50 - 30)^2} = \sqrt{(-60)^2 + 20^2} = \sqrt{3600 + 400} = \sqrt{4000} = 20\sqrt{10} \)

Step3: Find distance \( D F \)

For \( D(-50,50) \) and \( F(0,0) \):
\( DF = \sqrt{(0 - (-50))^2 + (0 - 50)^2} = \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \) Wait, no, recalculate: \( \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \)? Wait, no, \( 50^2 + 50^2 = 2500 + 2500 = 5000 \), \( \sqrt{5000} = 50\sqrt{2} \approx 70.71 \), but wait, earlier steps: Wait, no, \( FE = 10\sqrt{10} \approx 31.62 \), \( ED = 20\sqrt{10} \approx 63.25 \), \( DF \): Wait, \( D(-50,50) \) to \( F(0,0) \): \( x \) difference \( 50 \), \( y \) difference \( -50 \), so \( \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \). Wait, but total distance is \( FE + ED + DF \). Wait, no, wait, \( FE = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10} \approx 31.62 \), \( ED = \sqrt{(-60)^2 + 20^2} = \sqrt{3600 + 400} = \sqrt{4000} = 20\sqrt{10} \approx 63.25 \), \( DF = \sqrt{50^2 + 50^2} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \)? Wait, no, that can't be. Wait, no, \( D(-50,50) \) to \( F(0,0) \): \( x \) from -50 to 0: 50, \( y \) from 50 to 0: -50. So \( (50)^2 + (-50)^2 = 2500 + 2500 = 5000 \), so \( \sqrt{5000} = 50\sqrt{2} \approx 70.71 \). But wait, let's check again. Wait, maybe I made a mistake. Wait, \( FE + ED + DF \): \( 10\sqrt{10} + 20\sqrt{10} + 50\sqrt{2} = 30\sqrt{10} + 50\sqrt{2} \approx 30*3.162 + 50*1.414 = 94.86 + 70.7 = 165.56 \). But wait, maybe there's a better way. Wait, no, the three points are \( F(0,0) \), \( E(10,30) \), \( D(-50,50) \), and back to \( F \). So the hike is the perimeter of triangle \( FED \). So we need to calculate the lengths of all three sides and sum them.

Wait, let's recalculate \( DF \) correctly. \( D(-50,50) \), \( F(0,0) \): \( \Delta x = 0 - (-50) = 50 \), \( \Delta y = 0 - 50 = -50 \). So \( DF = \sqrt{(50)^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \). \( FE = \sqrt{10^2 + 30^2} = \sqrt{1000} = 10\sqrt{10} \approx 31.62 \). \( ED = \sqrt{(-60)^2 + 20^2} = \sqrt{4000} = 20\sqrt{10} \approx 63.25 \). So total distance is \( 10\sqrt{10} + 20\sqrt{10} + 50\sqrt{2} = 30\sqrt{10} + 50\sqrt{2} \approx 30*3.1623 + 50*1.4142 = 94.869 + 70.71 = 165.579 \approx 165.58 \). But maybe we can factor differently. Wait, \( 30\sqrt{10} = 30*3.162 = 94.86 \), \( 50\sqrt{2} = 70.71 \), sum is ~165.57. Alternatively, maybe the problem expects integer values? Wait, no, maybe I made a mistake in coordinates. Wait, \( E(10,30) \), \( D(-50,50) \): \( x \) difference: -50 - 10 = -60, \( y \) difference: 50 - 30 = 20. So \( ED = \sqrt{(-60)^2 + 20^2} = \sqrt{3600 + 400} = \sqrt{4000} = 20\sqrt{10} \approx 63.25 \). \( FE = \sqrt{10^2 + 30^2} = \sqrt{1000} = 10\sqrt{10} \approx 31.62 \). \( DF = \sqrt{(-50 - 0)^2 + (50 - 0)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \). So total distance is \( 10\sqrt{10} + 20\sqrt{10} + 50\sqrt{2} = 30\sqrt{10} + 50\sqrt{2} \approx 165.58 \). But maybe the problem has a typo, or I misread the coordinates. Wait, \( D(-50,50) \), \( F(0,0) \): distance is \( \sqrt{(-50)^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \). Alternatively, maybe the coordinates are \( D(-50, 50) \), \( E(10,30) \), \( F(0,0) \). So the three sides:

\( FE \): from (0,0) to (10,30): \( \sqrt{10^2 + 30^2} = \sqrt{1000} = 10\sqrt{10} \approx 31.62 \)

\( ED \): from (10,30) to (-50,50): \( \sqrt{(-60)^2 + 20^2} = \sqrt{4000} = 20\sqrt{10} \approx 63.25 \)

\( DF \): from (-50,50) to (0,0): \( \sqrt{50^2 + 50^2} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \)

Summing them: \( 10\sqrt{10} + 20\sqrt{10} + 50\sqrt{2} = 30\sqrt{10} + 50\sqrt{2} \approx 30*3.1623 + 50*1.4142 = 94.869 + 70.71 = 165.579 \approx 165.58 \). So the length of the hike is approximately \( 30\sqrt{10} + 50\sqrt{2} \) or approximately 165.58 units.

But wait, maybe I made a mistake in the distance formula. Let's re-express:

Distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

For \( F(0,0) \) and \( E(10,30) \):

\( x_2 - x_1 = 10 - 0 = 10 \)

\( y_2 - y_1 = 30 - 0 = 30 \)

So \( d = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10} \approx 31.62 \)

For \( E(10,30) \) and \( D(-50,50) \):

\( x_2 - x_1 = -50 - 10 = -60 \)

\( y_2 - y_1 = 50 - 30 = 20 \)

So \( d = \sqrt{(-60)^2 + 20^2} = \sqrt{3600 + 400} = \sqrt{4000} = 20\sqrt{10} \approx 63.25 \)

For \( D(-50,50) \) and \( F(0,0) \):

\( x_2 - x_1 = 0 - (-50) = 50 \)

\( y_2 - y_1 = 0 - 50 = -50 \)

So \( d = \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = 50\sqrt{2} \approx 70.71 \)

Now, sum these three distances:

\( 10\sqrt{10} + 20\sqrt{10} + 50\sqrt{2} = (10 + 20)\sqrt{10} + 50\sqrt{2} = 30\sqrt{10} + 50\sqrt{2} \)

Calculating the numerical value:

\( \sqrt{10} \approx 3.1623 \), so \( 30\sqrt{10} \approx 30 * 3.1623 = 94.869 \)

\( \sqrt{2} \approx 1.4142 \), so \( 50\sqrt{2} \approx 50 * 1.4142 = 70.71 \)

Adding them together: \( 94.869 + 70.71 = 165.579 \approx 165.58 \)

So the length of the hike is approximately 165.58 units. If we want an exact form, it's \( 30\sqrt{10} + 50\sqrt{2} \), or we can factor out 10: \( 10(3\sqrt{10} + 5\sqrt{2}) \).

Answer:

The length of the hike is approximately \( \boldsymbol{165.58} \) units (or \( 30\sqrt{10} + 50\sqrt{2} \) units in exact form).