Two pictures have equal area. Their frames, which have a thickness of \(1\) inch, also have equal area.

Prove that the two frames are identical.

Two pictures have equal area. Their frames, which have a thickness of \(1\) inch, also have equal area.

Prove that the two frames are identical.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestFirst, cut out all the square \(1\) inch corners. Then if \(a,b\) and \(c,d\) are the dimensions of the pictures, we then see that the perimeters and areas have to be the same. The rest follows. – Michael Mendrin · 3 years ago

Log in to reply

– Daniel Liu · 3 years ago

Prove that if the perimeters and areas are equal, then the rectangles must be congruent.Log in to reply

I'll try to think of a more clever proof, if nobody else improves on this.

Okay, maybe I'll post a graphic a bit later to illustrate this. Given 2 rectangles of equal areas. Overlap them so that they share a common corner. Then, excluding the union of the 2 rectangles, we have 2 smaller rectangles, the areas of which have to be the same. If the perimeters of both rectangles are the same, then both of the smaller rectangles have to have a side of the same length. This means that the other sides they have must also be of the same length. This means that the 2 rectangles are congruent.

Picture Frames

Log in to reply

– Daniel Liu · 3 years ago

Technically \(a-b=\pm(c-d)\) which gives \(a=c,b=d\) or \(a=d,b=c\).Log in to reply

– Michael Mendrin · 3 years ago

yeah something like thatLog in to reply

I recommend downloading Geogebra, it's free the quality is very good. – Daniel Liu · 3 years ago

Log in to reply

– Michael Mendrin · 3 years ago

I used colored pencils and then I put the picture into my Underwood typewriter and clacked on some letters and then I scanned it and put it here. Just now downloaded and tried out Geogebra. Maybe I'll get the hang of it one day.Log in to reply