Challenge!!!
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!!
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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