If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Class 10 math (India)

### Course: Class 10 math (India)>Unit 8

Lesson 3: Trigonometric ratios of some specific angles

# Special right triangles intro (part 1)

Introduction to 45-45-90 Triangles. Created by Sal Khan.

## Want to join the conversation?

• At he explains how to rationalize. What I don't understand is when do I cross out and when do I multiply.

Like in Sal's example:
c/sqrt2 x sqrt2/sqrt2.

Wouldn't the denominator from the first fraction just be crossed out the with nominator of the second franction which would end up again with c/sqrt2?
• Yes, sqrt2/sqrt2 is the same thing as 1 and the whole expression would just turn back to c/sqrt2, but what Sal wanted to do is to have the irrational number be the numerator of the fraction (in this case, the irrational number is sqrt2).
When you multiply c/sqrt2 x sqrt2/sqrt2, the numerator of the product would be c sqrt2, and the denominator would be 2 (since sqrt2^2 = 2). This way, we would have the irrational on top and the rational on the bottom.
Both c/sqrt2 and c sqrt2/2 are equivalent, but some people prefer the second option because if you were to add fractions, having a 2 as a denominator is a whole lot easier to work with than having sqrt2.
• When you are trying to solve for the hypotenuse in a 90-45-45 triangle with only the length of one side (either a or b) given, is it possible to just substitute in the side lengths into the Pythagorean theorem?
For example, you are given a 90-45-45 triangle where the length of a = 5. To figure out the length of c, can you do 5² + 5² = c²?
• Yes, but no matter what the side is, the hypotenuse will always be x√2 length, so it would be 5√2, this should be easier than the Pythagorean theorem and get to the exact answer much quicker.
• Can (sqrt(2)/2)*C also be expressed as sqrt(0.5*C)?
• Close, but not quite.
√0.5 = ½ √2
Thus, ½ C√2 = C√0.5
So, you are correct, except that the C must be outside the square root on both expressions.
• Are there many ways to do this because the way my teacher taught me is way easier?
• There are different ways to get to most answers in Math, but you want to be sure that you understand the underlying concept and aren't just using a trick to get the answer quickly (note: it's fine to use a trick, as long as you understand why it works!). Can't say more than that without knowing what you were taught.
• how do you get 16 over square root of 2 out of 8? I got very confused
• At around he shows that when you want the hypotenuse, c, in a 45-45-90 triangle, then you can solve for it with the expression b = c/sqrt(2), so to solve for c you get that c = b*sqrt(2)
• Is a triangle with side lengths √125, √80 and √45 possible?
• Yes, I don't see why not! As long as the sum of the two shorter sides is greater than the longest side, you have a triangle: (√80 + √45) > √125
• okay this stuff is hard im not even gonna try
• Everything can be hard when you are first learning it, but if you practice it some, it will quickly become less hard until it is all of the sudden easy.
• At how do we come to the conclusion that c√2 / 2 = c / √2

Can somebody also please point me to a lesson plan where we learned this?
I am currently working on the world of math mastery challenge and started again at Kindergarten. I don't see how I could have missed a lesson on how to work with radicals in fractions. Right now this just blows my mind. :( I'd really like to learn, instead I feel a fear of radicals creeping up inside me.

I did find the lesson plans on radicals in general and simplifying those. I had no problems there. Your help is much appreciated.
• With fractions, you can divide or multiply both the numerator and denominator by the same number and it will still be the same. For example, 2/3 is the same as 4/6 and is the same as 3/4.5.
At , they simply divided both sides of the fraction by the square root of 2. Since the square root of 2 times the square root of 2 = 2, dividing 2 by the square root of 2 must equal the square root of 2. (And in the numerator, anything divided by itself is 1.) Hope this helps- keep exploring!
• I have a question.
What if the x side is squared like 2 squared root of 2
how will I find the hypotenuse?