The diagram shows two identical triangular pieces of paper A and B. The side lengths of each triangle are in the proportion 3, 4 and 5. Each triangle is folded along a line through a vertex, so that the two sides meeting at this vertex coincide. The regions not covered by the folded parts have respective areas SA and SB. If SA+SB=39, find the area of the original triangular piece of paper A.

Figure for above question

I understand that the triangles which cover each other, become congruent triangles. But, I am stuck here. I do not know how to proceed. Please advise.

  • $\begingroup$ Yes, I will edit the question now. $\endgroup$
    – Math Tise
    Jun 12 '18 at 16:04
  • $\begingroup$ question edited. $\endgroup$
    – Math Tise
    Jun 12 '18 at 16:05

The big triangle is 3-4-5

The yellow triangle is a right triangle. And it shares an angle with the big triangle, so it must be similar to the big triangle. Lets call that angle $\alpha$

The leg adjacent to $\alpha$ in $A = 4$

The leg adjacent to $\alpha$ in the yellow triangle $= 5-3 = 2$

Each side is half as long as the corresponding side in A.

  • $\begingroup$ Considering the sides of the big triangles as 3x, 4x and 5x. So, the corresponding sides of the smaller triangles are 1.5x, 2x and 2.5x. Now, SA+SB=0.5*(2x)(1.5x)+0.5*(2x)(1.5x)=39 => x = sqrt(13). So, area of big triangle A = 0.5*(4x)(3x) = 0.5(4)(3)(13) = 6*13 = 78. So, area of original piece of triangle A = 78. Is this correct? Please advise. $\endgroup$
    – Math Tise
    Jun 15 '18 at 5:09
  • 1
    $\begingroup$ In $SA$ the side lengths are $\frac {3}{2}x, \frac {4}{2}x, \frac {5}{2}x.$ In SB they are $\frac {3}{3}x, \frac {4}{3}x, \frac {5}{3}x$ The areas of each then are $\frac {6}{4}x^2, \frac {6}{9}x^2$ and $\frac 64 x^2 + \frac 69 x^2 = 39$ or $\frac {13}{6} x^2 = 39 \implies x^2 = 18$ and the big triangle has area $6x^2 = 108$ $\endgroup$
    – Doug M
    Jun 15 '18 at 16:20
  • $\begingroup$ Thank You very much for the Answer. However, please explain why the sides in SB are 3x/3, 4x/3 and 5x/3 ? $\endgroup$
    – Math Tise
    Jun 15 '18 at 17:47
  • $\begingroup$ ... I understood now ...thanks again Doug :-) $\endgroup$
    – Math Tise
    Jun 15 '18 at 17:52

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