$△ABC$

Make an auxiliary figure, which is a small figure of what you’re going to construct. Mark the angles, lengths, and any other information.

Begin the construction:

• Draw a long line.

• Mark point $A$ and measure $5\text{cm}$ along the line to point $B$.

• Construct a $60$° angle at $A$.

• Construct a $90$° angle at $B$.

• Name the intersection between $\angle A$ and $\angle B$, $C$.

• This is a right triangle.

• $\angle C=180\text{°}-90\text{°}-60\text{°}=30\text{°}$.

$\square ABCD$

Continue the construction by constructing the quadrilateral $\square ABCD$:

• Construct a normal on the line $BC$ at $C$.

• Construct a normal on $AB$ at $A$. Name the intersection between the two normals $D$. You have now constructed the rectangle $\square ABCD$.

• Because the rectangle is made up of two “$30$°-$60$°-$90$°” triangles, the hypotenuse is twice the length of the shortest leg. That means the diagonal is $2\cdot 5\text{cm}=10\text{cm}$.

• To find the height of the triangle, you use the Pythagorean theorem:

  $$\begin{array}{llll}\hfill a^2+b^2& =c^2& \hfill & \\ \hfill B{C}^2+5^2& =10^2& \hfill & \\ \hfill B{C}^2+25& =100& \hfill & \\ \hfill BC& =75& \hfill & \\ \hfill \sqrt{B{C}^2}& =\sqrt{75}& \hfill & \\ \hfill BC& \approx 8.67\text{cm}& \hfill & \end{array}$$

  You’ve found that the sides are $5\text{cm}$ and $8.67\text{cm}$, and the diagonal is $10\text{cm}$.