Geometry questions sometimes involve various graphs. For example:
Find the area of the following square:
We see that the square has points at (3, 1) and (1, 2). Using the Pythagorean Theorem, we can find the length of that side as Thus, the side is units long and so the square has an area of 5.
Now, these problems do not differ substantially from what we have before. Essentially, graphing problems give you the same information in a slightly different way. So instead of telling you how long the sides of a particular shape are, they might give you the coordinates of the endpoints. But of course, we know how to infer the length of that side, for we know the Pythagorean Theorem:
A rectangle has a side whose vertices are at (1, 3) and (2, 2). Find the length of that side.
We can apply the Pythagorean Theorem to get that which is the length of that side.
In this vein, a more realistic problem might look like:
A triangle has vertices at (1, 4), (2,3), and (0,0). Find its perimeter.
This problem simply involves repeated uses of the Pythagorean Theorem. Draw in the following additional triangles:
And we can apply the Pythagorean Theorem to get the following lengths:
So the answer is
Similarly, graphing problems can involve angles:
A triangle has vertices at and . Find the degree of angle .
We know that is a right isosceles triangle since and Thus, we know that
In short, if you have a geometry problem that involves certain points or a diagram, you can generally use those points or that diagram to infer aspects like the lengths of certain sides or the degrees of various angles, and then use that information as you normally would.
1. A rectangle (shown below) has vertices at (0,1), (2, 0), and (2,5). Where is its fourth vertex?
To answer this question, we need to remember that the opposite sides of a rectangle are the same length and point in the same direction. So the bottom side of the rectangle, going from (0, 1) to (2, 0) travels in the same direction and for the same distance as the top side of the rectangle going from (2, 5) to our mystery point.
And we know how far the bottom side changes: it gains 2 units in the x-direction and loses 1 unit in the y-direction. So too for the top side. So (2,5) must become (4, 4) and that is the coordinate of our missing point.
A similar route would be to note that these are congruent triangles:
and so, again, the change over the course of the longer side of the rectangle must be the same. So in going from (0, 1) to (2, 5), there is a gain of 2 units in the x-direction and 4 units in the y-direction. So (2,0) must undergo the same change, giving us (4,4).
2. A circle has its center at point and its circumference includes the point . Find its radius.
Recall that the radius of a circle is just the distance from the center to the circumference (at any point on the circumference, since they are all equally far from the center). So we just need to find:
which, by the Pythagorean Theorem, is just
3. A triangle has vertices at (1, 4), (2,3), and (0,0). Find its area.
The key observation is that we can take the area of the surrounding rectangle and subtract the area of bits outside of our desired triangle, like so:
Thus, we get that the surrounding rectangle has an area of 8 and that the triangles in red have an area of Thus, our triangle has an area of
4. The square in the following diagram has points at (2, 1) and (1, 4) as labelled. What is the area of the square?
To find the area of the square, we just need to find the length of any side. We are given, in effect, a diagonal of the square. We know that if a square has sides of length , then its diagonal will be units long. And we can use the Pythagorean Theorem for the above to find the length of the diagonal:
so we get that the diagonal is units long. Dividing by we get that the sides must be units long and so the area of the square is 5.