Ok so the question goes about this
A rectangular painting is twice as long as it is wide. A frame surrounds the painting. The frame is a constant 2 inches wide and has a area of 196 square inches. Find the dimensions of the painting.
I would have liked to make a visual representation of what i wanted but i don't know how to do that so i am just going to write what i inferred from the question
the painting's length is 2*W (w=width) and the width is just w. the frames width is 2 and the area of the frame is 196. How do i go on with the problem. if its just asking for the dimension of the painting wouldn't that mean that i could just multiply the width of the painting and the length of the painting. I know this is wrong because it gives me other numbers. Can anyone explain me how to solve this.
Best Answer
If the width of the painting is $w$, then its length is $2w$. Then, since the frame is 2 inches thick on each side of the painting, its width is $w + 4$ and its length is $2w + 4$. So the area of the rectangle bounded by the frame (which includes both the area of the frame and the area of the painting) is $(w + 4)(2w + 4)$.
The area of the painting is $w \times 2w = 2w^2$, and the area of the frame is 196 square inches, so the area of the two combined is $2w^2 + 196$. However, that is the same area as the other one I described, so we can then equate the two, i.e. $(w + 4)(2w + 4) = 2w^2 + 196$. You can then solve that equation for $w$, and hence find all of the relevant dimensions.