| Line 21: | Line 21: | ||
It's tempting to set up an underground Kiwi for ENGR 195. I suspect, however, that openly helping one another solve problems would be academically dishonest compared to the currently clandestine sharing of secrets.--[[User:Jmason|Jmason]] 11:14, 4 October 2008 (UTC) | It's tempting to set up an underground Kiwi for ENGR 195. I suspect, however, that openly helping one another solve problems would be academically dishonest compared to the currently clandestine sharing of secrets.--[[User:Jmason|Jmason]] 11:14, 4 October 2008 (UTC) | ||
| + | |||
| + | *I agree, it is tempting. But it would be academically dishonest, I think. I really don't want to try and then find out it is. [[User:Jhunsber|Jhunsber]] | ||
Revision as of 10:10, 4 October 2008
Upon further review of the problem (so about ten minutes after I posted and had walked out the door), I realized that my equation is really just a special case of Hero's Formula. This is because the semi perimeter of the triangle minus one of the sides is equal to the radius of the circle on the opposite side of the side. So
$ S=\frac{2(a+b+c)}{2}=a+b+c $
Hero's Formula:
$ A=\sqrt{(S)(S-(a+b))(S-(a+c))(S-(b+c))} $
Now do simple subtraction to find $ S-(a+b) $
$ S-(a+b)=a+b+c-a-b=c $
In a similar manner we find $ S-(a+c)=b $ and $ S-(b+c)=c $
Substituting:
$ A=\sqrt{(S)(a)(b)(c)}=\sqrt{abc(a+b+c)} $ Jhunsber
It's tempting to set up an underground Kiwi for ENGR 195. I suspect, however, that openly helping one another solve problems would be academically dishonest compared to the currently clandestine sharing of secrets.--Jmason 11:14, 4 October 2008 (UTC)
- I agree, it is tempting. But it would be academically dishonest, I think. I really don't want to try and then find out it is. Jhunsber
