source: #alcumus tags: #special-right-triangles #area #rectangles course: [[Geometry]] status: #solved ### Problem In the diagram, $A B C D$ is a rectangle with $A B=12$ and $B C=18$. Rectangle $A E F G$ is formed by rotating $A B C D$ about $A$ through an angle of $30^{\circ}$. What is the total area of the shaded regions? ![[alc-overlp-rect.svg]] #### Solution Perhaps there is a way to partition the shaded area, but it might be easier to calculate the non-shaded area and then subtract it. In the process, we will use a few [[Special Right Triangles#30-60-90]] ##### Finding an unknown length First, name some additional points: Let $M$ be the point of intersection of $\overline{EF}$ and $\overline{DC}$ Let $N$ be the point of intersection of $\overleftrightarrow{EF}$ and $\overleftrightarrow{AD}$ Let $x$ be the length of $\overline{DM}$ In this extended diagram, label all instances of $30\degree$ and $60\degree$ angles, showing $\overline{DN}$ has a length of $x\sqrt3$ since $\triangle DMN$ is 30-60-90. ($m\angle DNM=30\degree$ because it is [[‡ Alternate Interior]] to $\angle GAD$) By the [[◊ Segment Addition Postulate]] $\begin{aligned} m\overline{AN}&=m\overline{AD}+m\overline{DN}\\ &=18+x\sqrt{3} \end{aligned} $ ![[alc-ovrlp-rect-1.svg]] Since $\overline{AN}$ is the hypotenuse of the 30-60-90 $\triangle AEN$, it should be twice the length of the short leg $\overline{AE}$. This will be enough to solve for $x$ $\begin{aligned} m\overline{AN}&=2\cdot m\overline{AE}\\ 18+x\sqrt{3}&=2\cdot12 \\ 18+x\sqrt{3}&=24 \\ x\sqrt{3}&=6 \\ x&=2\sqrt3 \end{aligned}$ In preparation for finding other areas, find $\triangle ANE$ and $\triangle DEM$ ##### Area of $\triangle ANE$ $\frac{bh}{2}=\frac{12\cdot12\sqrt{3}}{2}=72\sqrt{3}$ ##### Area of $\triangle DEM$ $\frac{bh}{2}=\frac{2\sqrt{3}\cdot6}{2}=6\sqrt{3}$ ##### Area of the non-shaded region $ADME$ $\begin{aligned} \text{area of }ADME &= \triangle ANE - \triangle DEM \\ &=72\sqrt{3} - 6\sqrt{3}\\ &=66\sqrt{3} \end{aligned}$ ##### Area of the shaded region If the rectangles did not overlap, there would be nothing to subtract; just add their areas. But the overlapping(non-shaded) area is a part of both of them so it should be subtracted twice. $\begin{aligned} \text{area of shaded region} &= ABCD + AEFG - 2\cdot ADME \\ &=2\cdot12\cdot18 - 2\cdot66\sqrt{3}\\ &=\boxed{432-132\sqrt{3}} \end{aligned}$