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Area With Determinants - Problem 3
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I have another problem here. I have 4 points R, O, F and L, and I want to find the area of the quadrilateral they form. It turns out that this quadrilateral is not a parallelogram. So we’re going to have to get creative in our method for finding the area. Let me show you the plotted points. Here’s point R, O, F, and L.
My strategy here, is going to be to divide this quadrilateral into two triangles. We know how to find the area of triangles. I’m going to find the area of each of these triangles; ROL and RLF, and those will add up to the area of quadrilateral ROFL.
Area of ROFL equals the area of triangle ROF plus the area of triangle RLF, those two triangles. Now let’s remember that, the area of a triangle is going to be ½ the absolute value of a determinant formed by two vectors, that make up two sides, that share an initial point of the triangle. Let’s try RO and RF.
Vector RO and vector RF. I need to come up with components for these guys. RO goes from point R to point O, and so it’s going to be 0 minus 2, -2, 0 minus 6, -6. And RF is from R to point F, 7 minus 2, 5, 1 minus 6, -5. So my determinant is going to be <-2,-6> <5,-5>.
Plus I need to find the area of RLF and so I’m going to need two more vectors. RL, I already have RF. So I can use this very same RF, <5, -5>. RL is going to be from point R to point L. so 7 minus 2, 5, 4 minus 6, -2. And so here I’m going to have ½ the absolute value of the determinant, 5 -2, 5 -5. This is going to be ½ the absolute value of 10 minus -30, 10 plus 30, 40. That’s going to 20 plus ½ the absolute value of -25 minus -10. -25 plus 10, -15. Absolute value of -15 is 15, times ½ is 7.5. So the total area is 27.5.
Now let’s go back to our picture. What we found was triangle ROF has an area of 20 and RLF has an area of 7.5, for a total area of 27.5.
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