EMAT 6690

Day 5

Special Right Triangles ... 30, 60, 90 ... 45, 45, 90 ... Constructions

30, 60, 90 Right Triangles

Construction

To construct a 30, 60, 90 triangle start with segment AB. Now construct an equilateral triangle ABC by constructing circle A with a radius congruent to AB and circle B also with a radius of AB. The intersection of these two circles, point C, is the third vertex of the constructed equilateral triangle. Now construct a line perpendicular to segment AB that passes through point C. Triangle CDB is a 30, 60, 90 triangle.

Consider triangle ABC shown below.

The hypotenuse is twice as long as the shorter leg and the longer leg is the square root of 3 times longer than the shorter leg.

Since point C is the midpoint of segment AD, it is easy to see that the hypotenuse AB of the 30, 60, 90 triangle ABC is twice as long as the shorter leg AC. By using the pythagorean theorem we can show the longer leg is the square root of 3 times longer than the shorter leg.

45, 45, 90 Triangles

Construction

To construct a 45, 45, 90 triangle start with segment AB. Construct a line perpendicular to AB passing through point A. Now construct circle A with its center at point A and its radius congruent to segment AB. The intersection of the constructed perpendicular line and constructed circle is point C. Point C is the third vertex of the constructed 45, 45, 90 triangle ABC.