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Quadrilateral $ABCD$ has right angles only at vertices $A$ and $D$. The numbers show the areas of two of the triangles. What is the area of $ABCD$?

The rectangle $DABC'$ will have an area of $30$. I am not sure how to proceed after that. Can anyone help?


Let $E$ be the intersection of $AC$ and $BD$. Observe that $\Delta ABE$ and $\Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $\overline{DE}$ to $\overline{BE}$ by comparing areas. Namely, $\frac{\overline{DB}}{\overline{BE}} = \frac{A(\Delta ADB)}{A(\Delta ABE)} = \frac{15}{5} = 3$, so $\overline{DE} = 2 \overline{BE}$. Then using the similarity observed above, we have $\overline{CD} = 2\overline{AB}$. In particular, $A(\Delta BCD) = 2 A(\Delta ABD) = 30$; summing these two triangles gives a total are of $45$.

  • $\begingroup$ Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles. $\endgroup$ – Hari Dec 8 '18 at 2:04
  • $\begingroup$ Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with. $\endgroup$ – platty Dec 8 '18 at 3:17

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