Interesting problem (don't think too hard)

Geometry Level 2

If rectangle ABCD has an area of 100cm $^2$ , what is the area of the area shaded in yellow?

The answer is 50.

2 solutions

Richard Desper
Oct 21, 2019

Let $X$ be the intersection of $\overline{BE}$ and $\overline{CD}$ .

The area of the upper triangle is $(1/2)(XD * BD)$ . The area of the lower triangle is $(1/2)(XD * DF)$ .

Thus the area shaded in yellow is $(1/2)(XD(BD + DF))$ .

Assuming the line $\overline{EF}$ is parallel to the line $\overline{CG}$ , the triangles $\triangle BXD$ and $\triangle BEF$ are similar. That implies that

$\frac{XD}{BD} = \frac{EF}{BD + DF}$ . If we cross-multiply and use fact $EF = CD$ , we see that $BD*CD = XD(BD+DF)$ . Thus the area in yellow is $(1/2)BD*CD = 50$ .

Barry Leung
Oct 21, 2019

Just imagine dragging the tip at H to E, and then drag it from E to C. Then you will have half of the rectangle, which has an area of 50 cm $^2$ .

A few more words would be useful. Something like "the area of triangle CDH = the area of triangle CDE" followed by "the area of triangle BDE = the area of triangle BDC", along with a few words about why. Your two sliding actions have different meanings, as the horizontal sliding preserves the areas of both triangles while the vertical sliding does not. (My initial reading kept the point of intersection fixed, which would have kept the upper triangle fixed while the area of the lower triangle dropped to zero.)

Richard Desper - 1 year, 7 months ago

Interesting problem (don't think too hard)

The answer is 50.

2 solutions

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