1. Question 3

2. √ √ Question 3 (8, 15) 15 8 (0, 0) a2 + b2 = c2 82 + 152 = c2
What is the distance between (0, 0) and (8, 15) on the xy-coordinate plane?
a) 7 units b) 8 units c) 17 units d) 23 units
(8, 15) 15 8
(0, 0)
a2 + b2 = c2
To solve this one, the students should pretend to draw the shape so they can see what it makes. When they draw it, it ends up being a right triangle and they can use the Pythagorean theorem to solve.
The first thing I did was draw a quick graph. I, then, drew in points where the coordinates state. It doesn’t have to be perfect for this example because it is a rough estimate. We then want to see what the distance between the two points are. In order to do this, we can draw a triangle. The bottom distance is 8 units. The height is 15 units. Now that we know the two sides of the right triangle, we can use those values for the Pythagorean Theorem.
The Pythagorean Theorem is a2 + b2 = c2. We plug in 8 for A and 15 for B. (It actually doesn’t matter what number you plug in where; 15 could be A and 8 could be B). When we square a number, it’s saying that we are mulitplying a number by itself. So 82 is 8 times 8, which is is 15 times 15, which is We add , and get In order to find out what just “c” is, we take the square root of both sides. When we do that, we find out that 17 is the distance.
Also, students should have access to a calculator for this problem
= c2
= c2 √ √
289 = c2
17 = c