Big Ideas Math Integrated I, 2016
BI
Big Ideas Math Integrated I, 2016 View details
4. Perimeter and Area in the Coordinate Plane
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Exercise 29 Page 412

Practice makes perfect
a Let's create a diagram showing where the car, the waterfall, and the lookout point are.

We traditionally write a map with north up, south down, west to the left and east to the right.

The car is at the origin. The waterfall is 4 miles to the east of the car. If one unit equals one mile then the waterfall is 4 units to the right away from the car. That is the point (4,0) and, to indicate that it marks the waterfall, we write it as W(4,0). The lookout point is located north of the car. Again, we let it be a point in the diagram, a point 2 units up from the origin. It is the point (0,2) and we write it as L(0,2). We will also put a point in the diagram at the origin where the car is parked. We name that point C(0,0). The diagram now look like this.

We have now mapped out the route.

b The route we follow is in the form of a triangle. We find the distance we travel by calculating the triangle's perimeter. The perimeter we find by calculating the length of each side and add them together. The length of the side CW, that is the distance from the car to the waterfall, we already know since it is given in the exercise. It is 4 miles. Let's now continue and calculate the length of the side WL. We then make use of the Distance Formula.
WL = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
WL = sqrt(( 0- 4)^2 + ( 2- 0)^2)
WL=sqrt((- 4)^2+(2)^2)
WL=sqrt(16+4)
WL=sqrt(20)
The last leg of the hike went from the lookout point back to the car. That distance is given in the exercise and it is 2 miles. The total distance we hiked, D, was D&=CW+WL+LC &=4+sqrt(20)+2 &≈ 10.47 miles.
c As we altered the plans and added the wishing well to hike we also need to update the diagram. The wishing well is located 3 miles north of the lookout point, at y=5, and 2 miles west of it, at x=- 2. The coordinate for the wishing well is (- 2, 5) and we will write it

X(- 2,5).

Let's mark it in the diagram and rewrite the route so that it goes from the car, to the waterfall, up to the wishing well, down to the lookout point and back to the car.
To find the distance of the new route we must find the length of each side and add them together. Like before we already know the lengths of CW, which is 4 miles, and LC, which is 2 miles. Let's now calculate the distance from the waterfall to the wishing well. We again make use of the Distance Formula.
WX = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
WX = sqrt(( - 2- 4)^2 + ( 5- 0)^2)
WX=sqrt((- 6)^2+5^2)
WX=sqrt(36+25)
WX=sqrt(61)
We go on and calculate the distance from the wishing well to the lookout point since that is also a leg of the hike which has been altered from the original plan.
XL = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
XL = sqrt(( 0-( - 2))^2 + ( 2- 5)^2)
XL=sqrt((0+2)^2+(2-5)^2)
XL=sqrt(2^2+(- 3)^2)
XL=sqrt(4+9)
XL=sqrt(13)
Let's now calculate the total distance, D, traveled.
D=CW+WX+XL+LC
D=4+sqrt(61)+sqrt(13)+2
D=17.41580...
D≈ 17.42
The total distance traveled if we follow the new route is ≈ 17.42 miles.