Hello again. I have a problem that occured over a year ago. But, I cannot solve it until now. The problem is: Given that the rectangle with length x and width y. This rectangle was folded based on its diagonal to form two triangles. Find the area ratio between red and yellow triangles! I've already tried to find it, but still don't get the solution. Maybe any theorems or formulas that I can use to solve this problems? Thanks

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TopNewestFirst off, I think the graph is innacurate, because the botom angle is produced by a reflection of the top-right angle, which is 90 degrees. The bottom angle is not 90 degrees though. Let's not let that impede us though.

Here's my method of solving it. It is based on the double-angle formula for tangent. The slope of AC is \(-\frac{y}{x}\). Call the point on the bottom E, and the intersection of AE and DC point F. The duplication of \(\angle CAB\) onto \(\angle EAC\) produces \(\angle EAB\), which the slope of AE is derived from.

My patience with LaTex is not good enough to write out all the algebra, but the slope of AE is \(-\frac{2xy}{x^{2}-y^{2}}\).

Next, DF is the run of line AE when the rise is \(-y\) Using the slope equation for AE, the run is \(\frac{x^{2}-y^{2}}{2x}\). That is the base for the red triangle. The base for the yellow triangle is \(x-\frac{x^{2}-y^{2}}{2x}=\frac{x^{2}+y^{2}}{2x}\).

SInce we have found the two bases, and the height for the two is both y , the ratio of the red triangle to the yellow is\(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\).

\(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\) is the answer. – Clarence Chen · 3 years, 7 months ago

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Is it supposed to be x and y, or specifically 7 and 12 ? – Kathleen Kasper · 3 years, 7 months ago

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– Leonardo Chandra · 3 years, 7 months ago

yes, it's only tell me about the variables, x and y, no other specifics number. The picture were drawn not in the real measure, only approximation.Log in to reply

– Tim Ye · 3 years, 7 months ago

It states 'length x and width y' in the actual problem. I believe the diagram was posted so that we could see which triangles he was talking about.Log in to reply