Ram's Tricky Cardboard Challenge: What % Remains?

[himanshu121]

Ram has a rectangular piece of card board out of which he clipped away the largest possible square and was left with a piece similar in shape to the original piece. The area of the piece remaining with Ram forms approx what percent of the original piece

 

[HallsofIvy]

This is a "golden rectangle" problem isn't it?

Let the original rectangle have length l and width w. We an assume that w< l. The largest square that can be cut from that is "w by w" leaving a rectangle of sides w and l-w.

Now which is larger, w or l-w? If we started with a skinny rectangle, in which l was not only larger than w but larger than 2w, l-w is still larger than w and saying that this new rectangle is similar to the orginal says that l/w= (l-w)/w (writing longer side over shorter for both rectangles). But that's the same as l= l-w which is impossible.

Thus, in the new rectangle, the longer side (corresponding to l in the original rectangle) has length w and the shorter l-w. Since the new rectangle is similar to the original, we have w/(l-w)= l/w or
w²= l²- lw. Dividing both sides by l², (w/l)²= 1- w/l. We can think of this a quadratic equation and solve for w/l.

The area of the new rectangle is w(l-w)= lw- w². The area of the original rectangle was lw. The new rectangle has area
(lw- w²)/lw= 1- (w/l).

 

[himanshu121]

Thanks Halls I knew i have to apply the prop u quoted but never applied it was just an odd time