00:01
In this problem, we're first asked to find the area of the parallelogram, pqrs.
00:06
The area of any parallelogram is equal to the base times the height.
00:11
In this case, the base of the parallelogram is 30, and the height is 16.
00:20
The product of these two numbers is 480, which is the area of our parallelogram.
00:28
Now are asked to find the area of the triangle, psr.
00:33
The area of any triangle is equal to one half times the base, times the height.
00:39
But notice that we can use the same values for the base and height as we did in the previous problem, problem a.
00:47
So the base is equal to 30, and the height is 16.
00:57
This gives us the area of the triangle 240.
01:07
Now we need to find the area of the triangle osr, which i will draw in blue.
01:20
We're given a hint for this problem to use the proof the property that was discussed in problem 28.
01:29
As a reminder, i will write down this property for a triangle abc with a point m that bisects the line b c, we're told that the area of the triangle amc is equal to half of the area of the triangle abc.
02:02
I have rotated the figure from problem 28 to look like our triangle psr...