Area Example 2 (Triangle)
If you are given the measurements of the base and height of a triangle you can use this formula.
If you don't have those measurements and the triangle is on a graph you can solve it like this:
First you have to find the side lengths of the triangle.
You can find the side lengths of a triangle by finding the rise over run between two of the points.
From point A to B you have to go over 6 and down 1, so you would plug this number into the area formula of a triangle.
a=½*bh
a=½*(1)(6)
AB=3
Continue this process for the other two sides.
BC:
a=½*(4)(5)
BC=10
CA:
a=½*(1)(5)
CA=2.5
Next find the total area surrounding the triangle. In this example this area is represented by a blue box.
Using the formula that we learned in the previous example find the area of the square.
L*W
5*6=30
Now take the total area of the shape (30) and subtract the smaller areas from it.
30-2.5-10-3=14.5
This means that the area of the triangle is 14.5 units ^2.
If you don't have those measurements and the triangle is on a graph you can solve it like this:
First you have to find the side lengths of the triangle.
You can find the side lengths of a triangle by finding the rise over run between two of the points.
From point A to B you have to go over 6 and down 1, so you would plug this number into the area formula of a triangle.
a=½*bh
a=½*(1)(6)
AB=3
Continue this process for the other two sides.
BC:
a=½*(4)(5)
BC=10
CA:
a=½*(1)(5)
CA=2.5
Next find the total area surrounding the triangle. In this example this area is represented by a blue box.
Using the formula that we learned in the previous example find the area of the square.
L*W
5*6=30
Now take the total area of the shape (30) and subtract the smaller areas from it.
30-2.5-10-3=14.5
This means that the area of the triangle is 14.5 units ^2.