# Solution to the Two Rectangles and a Semi-Circle Puzzle +-- {.image} [[TwoRectanglesandaSemiCircle.png:pic]] > Two rectangles and a semicircle. What's the shaded area? =-- ## Solution by [[Dissection]], [[Angle in a Semi-Circle]], and [[Similar Triangles]] +-- {.image} [[TwoRectanglesandaSemiCircleLabelled.png:pic]] =-- With the points labelled as above, first note that there is a [[dissection]] argument to show that the two rectangles have the same area. Triangles $I B C$ and $G K E$ are [[congruent]], so the green rectangle has the same area as [[parallelogram]] $G I C E$. Then triangles $G I F$ and $E C D$ are also [[congruent]], so the parallelogram has the same area as the red rectangle. Alternatively, this can be seen by a composition of two [[shears]] centred at $C$. The first is vertical and takes $B K$ to $I G$. The second is parallel to $I C$ and takes $G E$ to $F D$. Triangle $A I C$ is [[right-angled triangle|right-angled]] since the [[angle in a semi-circle]] is $90^\circ$. It is [[similar]] to triangle $C D E$ since each of angles $D \hat{C} E$ and $I \hat{C} A$ makes a [[right-angle]] when added to angle $E \hat{C} I$. Since $E C$ is a radius of the semi-circle and $A C$ a diameter, the length scale factor is $2$. Therefore the length of $I C$ is twice that of $D C$, so is of length $6$. The area of the red rectangle, and hence also of the green, is then $6 \times 3 = 18$. ## Solution by [[Invariance Principle]] +-- {.image} [[TwoRectanglesandaSemiCircleSpecial.png:pic]] =-- In this special case, the red rectangle coincides with the green and so they have the same area. Also, the height of the red rectangle is the radius of the semi-circle, so its diameter is $6$. The area of the rectangle is therefore $3 \times 6 = 18$.