Thursday, March 17, 2011

Square root practice.

The square root sign is not available with the editor for Blogger (not an extended ASCII symbol is the nerdy explanation), so I will be writing sqrt(n) to indicate the square root of n.

The Pythagorean Theorem tells us that a² + b² = c², where a and b are the short sides of a right triangle, called the "legs", and c is the long side opposite the right angle, known as the "hypotenuse". This means since the square of the hypotenuse is equal to a² + b², the hypotneuse itself is equal to sqrt(a² + b²). What this means is we can find the square root of any number that is the sum of two perfect squares. Here are some examples.

If the legs are 1 and 1, then 1² + 1² = 2 = c², so c = sqrt(2)
If the legs are 1 and 2, then 1² + 2² = 5 = c², so c = sqrt(5)
If the legs are 1 and 3, then 1² + 3² = 10 = c², so c = sqrt(10)

We can also re-arrange the equation to a² = c² - b², which means we can find the square root of any number that can be written as the difference of two squares.

If the hypotenuse is 2 and one of the legs is 1, then 2² - 1² = 3 = a², so a = sqrt(3)
If the hypotenuse is 3 and one of the legs is 1, then 3² - 1² = 8 = a², so a = sqrt(8)
If the hypotenuse is 5 and one of the legs is 2, then 5² - 2² = 21 = a², so a = sqrt(21)

Here are some practice problems.

1) Make a right triangle that will have two whole number side lengths and one side with length of sqrt(7).

2) Make a right triangle that will have two whole number side lengths and one side with length of sqrt(18).

3) Make TWO different right triangles that will have two whole number side lengths and one side with length of sqrt(20).

4) Make TWO different right triangles that will have two whole number side lengths and one side with length of sqrt(50).

Answers in the comments.

4 comments:

Prof. Hubbard said...

1) Make a right triangle that will have two whole number side lengths and one side with length of sqrt(7).

Answer: 4² - 3² = 7, so we can make a right triangle with legs of length 3 and sqrt(7) and hypotenuse 4.




2) Make a right triangle that will have two whole number side lengths and one side with length of sqrt(18).

Answer: 3² + 3² = 18, so we can make a right triangle with legs of length 3 and 3 and hypotenuse of sqrt(18).



3) Make TWO different right triangles that will have two whole number side lengths and one side with length of sqrt(20).


Answer: 4² + 2² = 20, so we can make a right triangle with legs of length 4 and 2 and hypotenuse of sqrt(20).

Answer: 6² - 4² = 20, so we can make a right triangle with hypotenuse of length 6 and legs of length 4 and sqrt(20).




4) Make TWO different right triangles that will have two whole number side lengths and one side with length of sqrt(50).

Answer: 7² + 1² = 50, so we can make a right triangle with legs of length 7 and 1 and hypotenuse of sqrt(50).

Answer: 5² + 5² = 50, so we can make a right triangle with legs of length 5 and 5 and hypotenuse of sqrt(50).

Erica said...

but how are you finding the numbers (legs)? Is their an easy way to figure out what those numbers are?

Prof. Hubbard said...

What we were doing here is guess and check, but we've learned in class a way to find any odd number 2a+1 as the difference between (a+1)^2 - a^2.

Erica said...

I get it now....I get it. :)