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 $$45$$.
• 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. – platty Dec 8 '18 at 3:17