Special Triangles

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Recall that a triangle has three angles that add up to $180^\circ$ . A right triangle is a triangle that has one $90^\circ$ angle; the side opposite this angle is called the hypotenuse. In this activity, we will look at two special cases of right triangles: when the remaining two angles are each $45^\circ$ , and when the remaining two angles are $30^\circ$ and $60^\circ$ . Our goal is to determine the lengths of the legs of these triangles when the hypotenuse is 1.

For the $45^\circ -45^\circ -90^\circ$ triangle, (the isosceles right triangle), there are two legs of length $a$ and the hypotenuse of length 1.

Use the Pythagorean Theorem to write an equation relating the lengths of the sides of the triangle: $\answer {a}^2+\answer {a}^2=\answer {1}^2$ .

In a right triangle with sides lengths $a$ and $b$ and hypotenuse $c$ , the Pythagorean Theorem states that $a^2+b^2=c^2$ .

Solve the equation for $a = \answer {\frac {\sqrt {2}}{2}}$ .

How can you write $a^2+a^2$ as one term?

In summary, a $45^\circ -45^\circ -90^\circ$ triangle has:

base $=\answer {\frac {\sqrt {2}}{2}}$

height $=\answer {\frac {\sqrt {2}}{2}}$

hypotenouse $=\answer {1}$

To find the lengths of the legs of the $30^\circ -60^\circ -90^\circ$ triangle, begin with an equilateral triangle, all of whose sides are length 1. Recall that in an equilateral triangle, all three angles are equal to $60^\circ$ . One possible approach to solving for the side lengths is the following:

From the top vertex, draw a line segment perpendicular to the bottom side, cutting the original triangle into two congruent triangles. In an equilateral triangle, this line segment is called a perpendicuar bisector, a median, and an altitude.

The length of each half of the bottom side is $b=\answer {\frac {1}{2}}$ .

Find all of the angles in the new triangles determined by the perpendicular bisector you drew in the previous problem.

Smallest $\measuredangle = \answer {30}^\circ$

Second smallest $\measuredangle = \answer {60}^\circ$

Largest $\measuredangle = \answer {90}^\circ$

The angles add up to $180^\circ$ .

Label the length of the altitude $h$ , and use the Pythagorean theorem to write an equation involving $h$ :

In a right triangle with sides lengths $a$ and $b$ and hypotenuse $c$ , the Pythagorean Theorem states that $a^2+b^2=c^2$ .

Solve the equation for $h$ in the expression you wrote above:

$h = \answer {\frac {\sqrt {3}}{2}}$

In summary, a $30^\circ -60^\circ -90^\circ$ triangle has:

base $=\answer {\frac {1}{2}}$

height $=\answer {\frac {\sqrt {3}}{2}}$

hypotenouse $=\answer {1}$

Draw the 45-45-90 triangle in as many orientations as possible, keeping the legs either horizontal or vertical. How many are there? $\answer {4}$

You can rotate and reflect the triangle.

Draw the 30-60-90 triangle in as many orientations as possible, keeping the legs either horizontal or vertical. How many are there? $\answer {8}$

Do you expect this number to be the same as in the 45-45-90 case?