# Solution to the [[Four Houses in a Semi-Circle]] Puzzle +-- {.image} [[FourHousesinaSemiCircle.jpeg:pic]] > These houses are made from squares and equilateral triangles. What’s their total area? =-- ## Solution by Properties of an [[Equilateral Triangle]] and [[Pythagoras' Theorem]] +-- {.image} [[FourHousesinaSemiCircleLabelled.jpeg:pic]] =-- As the squares and equilateral triangles all share edges, all those edges are the same length. Let this length be $x$. In the diagram above, $O$ is the centre of the circle and $A$ is such that angle $O \hat{A} B$ is a [[right-angle]]. Since $B$ is the apex of an [[equilateral triangle]], $B A$ [[bisects]] the base of that triangle, and hence also of the square below it. Line segment $O B$ is a [[radius]] of the circle so has length $2$. Line segment $O A$ comprises one-and-a-half of the sides of the squares, so has length $\frac{3}{2} x$. Then line segment $A B$ is the height of one square and the height of one equilateral triangle, so from the lengths in an [[equilateral triangle]] is has length $x + \frac{\sqrt{3}}{2} x$. Applying [[Pythagoras' theorem]] to triangle $O A B$ then: $$ \begin{aligned} 2^2 &= \left(\frac{3}{2} x\right)^2 + \left(x + \frac{\sqrt{3}}{2} x\right)^2 \\ &= \frac{9}{4} x^2 + x^2 + \sqrt{3} x^2 + \frac{3}{4} x^2 \\ &= 4x^2 + \sqrt{3} x^2 \end{aligned} $$ From the properties of an [[equilateral triangle]], the area of one with side length $x$ is $\frac{\sqrt{3}}{4} x^2$. So the total area of the squares and triangles is $4 x^2 + \sqrt{3} x^2 = 4$.