MHB Find area of a triangle in coordinate plane

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To find the area of triangle QRS in the coordinate plane, the area of the surrounding rectangle is calculated first, yielding 20. The areas of three right triangles within the rectangle, which are not part of triangle QRS, are then determined to be 6, 2, and 5, totaling 13. By subtracting this total from the rectangle's area, the area of triangle QRS is found to be 7. It is noted that triangle QRS is not a right triangle, as it does not satisfy the Pythagorean Theorem. The discussion concludes with the confirmation that the area of triangle QRS is indeed 7.
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Hello, and welcome to MHB! (Wave)

$$\triangle QRS$$ is not a right triangle, but I suspect what is intended is for you to find the area of the rectangle, and then subtract away the areas of the 3 right triangles that are within the rectangle, but not part of $$\triangle QRS$$.

Can you proceed?
 
Um does the position on the grid mean anything. i honestly only know 1/2BH formula and its not giving me the right answer. I honestly don't know any part of how to do this
 
I figured it out! 7 is the answer. Thanks!
 
Very good! For those who are interested, MarkFL could see that the central triangle was NOT a right triangle because it does not satisfy the "Pythagorean Theorem". The two legs have length 5 and 2\sqrt{2} while the hypotenuse has length $\sqrt{29}$. $29\ne 25+ 8$.

The bounding 4 by 5 rectangle has area 20. The right triangle at the upper left corner has area 3(4)/2= 6. The right triangle at the lower left has area 2(2)/2= 2. And the right triangle at the right has area 2(5)/2= 5. Those have total area 6+ 2+ 5=13 so the area of the central triangle is 20- 13= 7 as funnigen said.
 

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