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.

enter image description here

  • $\begingroup$ Hint: Don't try to calculate the area of the painting. Just calculate the area of the frame. Since this depends upon the dimensions of the painting, you'll be able to solve for them. $\endgroup$ – John Douma Oct 8 '15 at 22:56
  • $\begingroup$ Thanks for the hint so than the area is L*W=A so than 2 * L = 196 than L=98? this does not sound good. and also could you explain by what you said that the dimensions of the paintings depend on the frame? $\endgroup$ – MATH ASKER Oct 8 '15 at 23:02
  • $\begingroup$ Umm. When they say the frame is 2 inches wide, they don't mean from top to bottom (or from side to side). They mean the distance from the painting to the outside the frame is 2 inches in all directions. $\endgroup$ – Paul Sinclair Oct 8 '15 at 23:02
  • $\begingroup$ Oh so well how would i go on solving it...even a hint would help... $\endgroup$ – MATH ASKER Oct 8 '15 at 23:06
  • $\begingroup$ Break the frame up into rectangles and add the areas of those rectangles to get the area of the frame. You will find that $l$ and $w$ make their way into the expression for the area of the frame. $\endgroup$ – John Douma Oct 8 '15 at 23:12

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.


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