Hey Everyone! I'm not sure if Mr. H want's me to write here, {please let me know if this is bad haha} but my name is Chantelle and I am going to the U of Regina and have been invited to your Blog!!!
I came across a question in one of my math classes that I feel will be right up your alley with Pythagorean Theorem! So here is a little challenge for you!
The question is: The perimeter of an isosceles triangle {2 sides are equal and the two angles at the base are equal} ABC with AB=BC is 128 inches. The altitude {height of the triangle which is perpendicular (90 degree angle) to the base ( and in the case of an isosceles triangle, hits at the midpoint of the base)} BD is 48 inches. What is the area of the triangle??
So, you will need a couple formulas other than just pythagorus. However, I think you can do it!
This is from my Math 308 class, which is a 4th year math class!!!
Good Luck!!
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Hi Chantelle! first let me say welcome to our blog! I hope you enjoy it here! :) The question you asked us is a little confusing... well for me I can't speak for other people.. hahaha anyways, maybe you should put a picture up so that we can see what your asking...???!!! :)
- Camille 8-17
March 18, 2009 at 6:43 PM
That is a VERY good idea!! Now, I am just learning how to blog. I have created a picture on paint. How do I get it up on here?
March 18, 2009 at 6:49 PM
There's an Add Image icon when you're editing a post. It's a blue square with a mountain on it, I think. Hopefully you're enjoying blogger.
-linda
March 18, 2009 at 9:48 PM
Hello Chantelle and welcome to our online classroom! I was very interested by your math problem and I believe the answer is the area=5695.632 inches. That's if you that the base=237.318 (118.659x2). Please tell me if I'm right. :)
March 19, 2009 at 9:25 PM
Hi Chantelle! Welcome to our blog and have fun around on the blog!
March 19, 2009 at 11:34 PM
Nicky D, That is a little big.
March 20, 2009 at 9:07 PM