Find area of a triangle in coordinate plane

[funnijen]

View attachment 8941

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

\(\displaystyle \triangle QRS\) 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 \(\displaystyle \triangle QRS\).

Can you proceed?

 

[funnijen]

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

 

[funnijen]

I figured it out! 7 is the answer. Thanks!

 

[HOI]

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.