Big Ideas Math: Modeling Real Life, Grade 8
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Big Ideas Math: Modeling Real Life, Grade 8 View details
2. The Pythagorean Theorem
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Exercise 31 Page 388

Practice makes perfect
We want to draw the given situation on a coordinate plane. We know that we run 20 feet forward and then 15 feet to our right. Our friend faces the other direction and starts running at the same moment. We also know that our friend runs 16 feet forward and then 12 feet to her right. We can add this information to a diagram.
your path and your friend's path
We know that our friend stops and throws her snowball at us. Let's draw the path of the snowball on the diagram!
snowball path

Note that this diagram is only one of the possible solutions. We could have started drawing at any point on the coordinate plane or moved in any direction. This creates infinitely many possibilities for drawing a graph illustrating the given situation.

We want to know how far our friend throws her snowball. To do so, we will find the distance between us and our friend after running. We can do it in three steps.

  1. Calculate the distance from our current position to the initial position.
  2. Calculate the distance from the initial position to our friend's current position.
  3. Add the distances.

We will do this one step at a time.

Distance From Our Current Position to the Initial Position

Let's start by looking at the diagram from Part A!

The point that represents us is situated 15 units from the y-axis and 20 units from the x-axis in the first quadrant and the point that represents the friend is 12 units from the x-axis and 16 units from the y-axis in the third quadrant

We can see that our path forward and our path to the right are the legs of a right triangle. Notice that the path between our current position and the initial position is the hypotenuse of the triangle. This means that we can use the Pythagorean Theorem to calculate the distance between ourselves and the initial position! a^2+b^2=c^2 In the formula, a and b are the legs and c is the hypotenuse of a right triangle. Let's identify a, b, and c on the diagram.

triangle with our path
We have that a= 20 feet and b= 15 feet. Let's substitute these values into the formula.
a^2+b^2=c^2
20^2+ 15^2=c^2
â–Ľ
Solve for c
400+225=c^2
625=c^2
sqrt(625)=c
c=sqrt(625)
c=25
The distance between our current position and the initial position is 25 feet.

Distance From the Initial Position to Our Friend's Current Position

Now let's look at our friend's path on the diagram!

triangle with our path and distance to initial position

Our friend's path also created a right triangle where the path from the initial position to their current position is the hypotenuse. This means that we can use the Pythagorean Theorem again to find the distance. Let's identify the legs a and b, and the hypotenuse c on the diagram!

triangle with our friend's path
We can see that a= 12 feet and b= 16 feet. Let's substitute these values into the formula.
a^2+b^2=c^2
12^2+ 16^2=c^2
â–Ľ
Solve for c
144+256=c^2
400=c^2
sqrt(400)=c
c=sqrt(400)
c=20
The distance from the initial position to our friend's current position is 20 feet.

Distance Between Ourselves and Our Friend

Finally, we can calculate the distance between ourselves and our friend. To do so, we will add the distances that we found. 25+20=45 Our friend threw her snowball 45 feet.