Question

Puzzle FourIt's Friday and you are leaving school. You are going to the pizza palace andthen the movies with your friends. After the movies you go home andSaturday you wake up to spend the day at the park and then go to theschool to play on the basketball court. After a very active day you go hometo sleep. Determine the total distance you traveled on Friday and Saturdayif you always took the most direct route (HINT: Find the hypotenuse). Eachcell on the graph represents one mile. The total distance traveled willunlock puzzle fourl Round all your answers to the nearest hundredthMallSchoolPizza PalaceHomeParkMovisTheatre

Answer 1

This map is drawn like a cartesian plane. To answer this question, first we're going to write the coordinate of every place as an ordered pair. The first coordinate will be the horizontal coordinate(x) and the second coordinate will be the vertical coordinate(y).

We have:

School = (4, 3)

Pizza Palace = (-8, 2)

Movies = (-4, -6)

Home = (0, 0)

Park = (7, -5)

Now that we have the coordinates, we can use the formula to determinate the distance between two points in a cartesian plane.

`d((x_1,y_1)(x_2,y_2))=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2}`

Then, we start to calculate. First, we need the distance between the School and the Pizza Palace.

`d((4,3)(-8_{},2))=\sqrt[]{(4_{}-(-8)_{})^2+(3-2)^2}=\sqrt[]{12^2+1^2}=\sqrt[]{145}`

Now, from the pizza palace to the movies.

`d((-8_{},2)(-4,-6))=\sqrt[]{(-8-(-4)_{})^2+(2-(-6))^2}=\sqrt[]{16+64}=\sqrt[]{80}`

From the movies back home.

`d((-4,-6)(0,0))=\sqrt[]{16+36}=\sqrt[]{52}`

Now, we're already on saturday. From home to the park we have

`d((0,0)(7,-5))=\sqrt[]{49+25}=\sqrt[]{74}`

From the park to school

`d((7,-5)(4,3))=\sqrt[]{9+64}=\sqrt[]{73}`

From school back home:

`d((4,3)(0,0))=\sqrt[]{16+9}=\sqrt[]{25}=5`

Now we have all the individual routes, to get the total amount he walked during friday and saturday, we just need to add all of them.

`\begin{gathered} \sqrt[]{145}+\sqrt[]{80}+\sqrt[]{52}+\sqrt[]{74}+\sqrt[]{73}+5=50.3432980521\ldots \\ \approx50.34 \end{gathered}`

The total distance is 50.34 miles.