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\(\displaystyle \triangle QRS\) 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?

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- Jan 30, 2018

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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.