Warm-Up

1.) Find the value of *x* above.

2.) Evaluate each (simplify).

a.) √25

b.) 3√4

c.)√9√16

d.) √36/√9

Consider that each of the above examples shows that you can multiply and divide two different ways with square roots: √9•√16=√9•16 (keep in mind the root should completely be above the 9 and 16 on the right side of the equation).

So we have these rules of radicals (square roots):

√m•√n = √n•m

√m/√n = √m/n

There is another rule that says we are not supposed to have radicals in the denominator.

See the top example on this website: http://www.purplemath.com/modules/radicals5.htm

**Special Right Triangles**

Recall: Equilateral, Isosceles, Scalene. Of these, one might consider “Equilateral” or “Isosceles” to be more special, especially with the cases right triangles they give us.

Consider an Isosceles Right Triangle. (45-45-90)

Let the congruent sides each equal length 1. Use the Pythagorean Theorem to solve for the hypotenuse. You should get √2.

So one example of our 45-45-90 (Isosceles Right Triangle) looks like this:

Any other 45-45-90 is similar to the one above, and you can use ratios and scale factors to find missing sides of any similar triangle.

Consider an equilateral triangle, with one of its angles bisected.

Isolate one of the smaller triangles created.

Let the smallest side (two tick marks) equal 1. Doing the same on the equilateral triangle diagram shows us that the one tick mark side must equal 2.

This gives us two sides in the triangle, and since the third angle is 90º (by the Triangle Sum Theorem), we can use the Pythagorean Theorem to solve for the third side.

You should get √3 for the missing medium length side.

This triangle is called a 30-60-90. All other triangles that have the same angles, are similar to it by AA, and so we can use the known ratios of *this triangle’s sides* to solve for unknown sides in similar triangles.

We then practiced problems on this worksheet:

Special Right Triangles- Front