Question

A map of an obstacle course is shown in the graph. The running path for the course is shaped like a right triangle where each unit is equal to 1 meter. graph of a right triangle with points at negative 4 comma 0 labeled Obstacle 1, negative 4 comma 3 labeled Starting Point, and 0 comma 0 labeled Obstacle 2 Part A: Find the distance in meters from the starting point to obstacle 2. Show every step of your work. (3 points) Part B: How many meters is one full lap around the course? Show every step of your work. (1 point)

Answer 1

1. **Distance to Obstacle 2:** Using the distance formula, the distance from the starting point (-4, 3) to Obstacle 2 (0, 0) is found to be 5 meters.

2. **Full Lap Distance:** Applying the Pythagorean theorem to the right triangle formed by the course's running path, the distance for one full lap around the obstacle course is determined to be 5 meters.

Given the information provided, let's first find the distance from the starting point to obstacle 2, which forms one leg of the right triangle.

Part A:

The distance formula in a coordinate plane between two points
`\((x_1, y_1)\) and \((x_2, y_2)\)` is given by:

`\[d = √((x_2 - x_1)^2 + (y_2 - y_1)^2)\]`

For the distance from the starting point (-4, 3) to obstacle 2 (0, 0):

`\[d = √((0 - (-4))^2 + (0 - 3)^2)\]`

`\[d = √(4^2 + (-3)^2)\]`

`\[d = √(16 + 9)\]`

`\[d = √(25)\]`

`\[d = 5 \text{ meters}\]`

So, the distance from the starting point to obstacle 2 is 5 meters.

Part B:

To find the distance of one full lap around the course, we need to consider the hypotenuse of the right triangle. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides a and b.

`\[c^2 = a^2 + b^2\]`

In this case, the legs of the triangle are 3 meters and 4 meters (as obtained in Part A).

`\[c^2 = 3^2 + 4^2\]`

`\[c^2 = 9 + 16\]`

`\[c^2 = 25\]`

`\[c = √(25)\]`

`\[c = 5 \text{ meters}\]`

So, one full lap around the course is 5 meters.