 # Examples of Larger Constructions

You’re going to construct a triangle $△ABC$ and a quadrilateral $\square ABCD$ with the following specifications:

1.
$\angle A$ is $60$°, the line $AB$ is $5\phantom{\rule{0.17em}{0ex}}\text{cm}$ long, and $\angle B$ is $90$°. Construct the triangle $△ABC$.

What kind of triangle is this? How large is $\angle C$?

2.
The area of the triangle $△ABC$ in Exercise 1 is half the area of the rectangle $\square ABCD$.

How long are the sides in the rectangle? How long is the diagonal?

Example 1

### $△ABC$

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 $5\phantom{\rule{0.17em}{0ex}}\text{cm}$ along the line to point $B$.

• Construct $60$° in $A$.

• Construct $90$° in $B$.

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

• This is a right triangle.

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

Example 2

### $\square ABCD$

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

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

• Construct a normal on $AB$ in $A$. 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 $2\cdot 5\phantom{\rule{0.17em}{0ex}}\text{cm}=10\phantom{\rule{0.17em}{0ex}}\text{cm}$.

• To find the height of the triangle, you use the Pythagorean theorem: $\begin{array}{llll}\hfill {a}^{2}+{b}^{2}& ={c}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill B{C}^{2}+{5}^{2}& =1{0}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill B{C}^{2}+25& =100\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill BC& =75\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{B{C}^{2}}& =\sqrt{75}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill BC& =8.67\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You’ve found that the sides are $5\phantom{\rule{0.17em}{0ex}}\text{cm}$ and $8.67\phantom{\rule{0.17em}{0ex}}\text{cm}$, and the diagonal is $10\phantom{\rule{0.17em}{0ex}}\text{cm}$. 