Half an equilateral triangle (30-60-90 special right triangles)

Just like the $45^\circ$45°-$45^\circ$45°-$90^\circ$90° triangle, the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle is another special type of right triangle. We can construct a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle by starting with an equilateral triangle and cutting it into two halves.

Here is an equilateral triangle:

To form a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle, we draw in an altitude from any vertex. In fact, we get two congruent $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangles by doing so:

In order to look at the relationships between the side lengths of such a triangle, let's suppose that the initial equilateral triangle $\triangle ABC$△ABC had side lengths of $2$2 units each. Since the altitude $\overline{AD}$AD bisects the side $\overline{BC}$BC, this means that the length of the short side $\overline{BD}$BD will be $1$1 unit. Here is this information on the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle $\triangle ABD$△ABD:

There is only one unknown side length left on this triangle, $AD$AD, which has been labelled $x$x on the diagram above. Since $\triangle ABD$△ABD is a right triangle, we can use the Pythagorean Theorem to find this length:

$1^2+x^2$12+x2	$=$=	$2^2$22	Using the Pythagorean Theorem
$1+x^2$1+x2	$=$=	$4$4	Simplifying
$x^2$x2	$=$=	$3$3	Subtracting $1$1 from both sides
$x$x	$=$=	$\sqrt{3}$√3	Taking the positive square root of both sides

Here is a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° with all side lengths filled in:

Note that any triangle with angle measures of $30^\circ$30°, $60^\circ$60° and $90^\circ$90° is necessarily similar to this triangle. Using what we know about similar triangles, this means that the sides of any $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle are in the ratio $1:\sqrt{3}:2$1:√3:2.

 

Practice questions

Question 1

Question 2

Question 3