Theorems

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Solution: Since we know that 8 + 11 must be greater than <math>S_3</math>, we have <math>8\ + 11 > S_3</math>,
Solution: Since we know that 8 + 11 must be greater than <math>S_3</math>, we have <math>8\ + 11 > S_3</math>,
so <math>S_3</math> must be less than 19. But what is the smallest value that <math>S_3</math> can be?  
so <math>S_3</math> must be less than 19. But what is the smallest value that <math>S_3</math> can be?  
-
Well, we know that <math>S_3\ + 11 must be > 8 </math>, but 11 is already greater than 8 so that doesn't help us.  
+
Well, we know that <math>S_3\ + 11\ must be > 8 </math>, but 11 is already greater than 8 so that doesn't help us.  
-
However we also know that <math>S_3\ + 8 must be > 11 </math>. Using algebra we subtract 8 from both sides of our equation we get <math>S_3\ + 8 {\color{Red}-8} must be > 11 {\color{Red}-8} </math>
+
However we also know that <math>S_3\ + 8\ must be > 11 </math>. Using algebra we subtract 8 from both sides of our equation we get <math>S_3\ + 8 {\color{Red}-8}\ must be > 11 {\color{Red}-8} </math>
[[Category:Geometry]]
[[Category:Geometry]]

Revision as of 23:08, 18 October 2009

The Distance Formula

d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

Example: Find the distance between point A(5,-7) and B(-6,-2) and leave your answer in radical form.

Solution: Let A be point 1 and B be point 2. We have x_1\ =5,y_1 = -7, x_2 = -6, y_2 = -2

Substituting we have d=\sqrt{(\ -6 -5)^2 + (-2 - -7)^2} = \sqrt{(\ -11)^2 + (-2 +7)^2} = \sqrt{(\ -11)^2 + (+5)^2} = \sqrt{\ 121 + 25} = \sqrt{146}

Try some distance problems at Distance & Midpoint 9-2 ME Worksheet

Sum of the Sides of a Triangle

Thm: The sum of two sides of any triangle must be greater than the third side

Example: Given two sides of a triangle are 8 inches and 11 inches, find all the possible lengths of the third side (also known as S3)

Solution: Since we know that 8 + 11 must be greater than S3, we have 8\ + 11 > S_3, so S3 must be less than 19. But what is the smallest value that S3 can be? Well, we know that S_3\ + 11\ must be > 8 , but 11 is already greater than 8 so that doesn't help us. However we also know that S_3\ + 8\ must be > 11 . Using algebra we subtract 8 from both sides of our equation we get S_3\ + 8 {\color{Red}-8}\ must be > 11 {\color{Red}-8}

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