tag:blogger.com,1999:blog-6933544261975483399.post1503085115464644476..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1170: Area of a Triangle in terms of the three sides, a-b-cAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-10668957160051206442015-12-09T08:38:54.019-08:002015-12-09T08:38:54.019-08:00K^2 = (p)(p-a)(p-b)(p-c) ; where p = (a+b+c)/2 .....K^2 = (p)(p-a)(p-b)(p-c) ; where p = (a+b+c)/2 ...<br /> <br />K^2 = (a+b+c)/2 * (b+c-a)/2 * (a+c-b)/2 * (a+b-c)/2<br /><br />K^2 = 1/16 * [(a+b)^2 - c^2] * -[(b-a)^2 - c^2]<br /><br />K^2 = 1/16 * -{(a+b)^2(a-b)^2 - c^2[(a+b)^2 + (a-b)^2] + c^4}<br /><br />K^2 = 1/16 * -{(a^2-b^2)^2 - c^2(2*a^2 + 2*b^2) + c^4}<br /><br />K^2 = 1/16 * -{a^4 - 2*a^2*b^2 + b^4 - 2*c^2a^2 - 2*c^2*b^2 + c^4}<br /><br />K^2 = [-a^4 + 2*a^2*b^2 - b^4 + 2*c^2a^2 + 2*c^2*b^2 - c^4]/16<br /><br />K = [2*a^2*b^2 + 2*c^2a^2 + 2*c^2*b^2 -a^4 - b^4 - c^4]^(1/2) /4Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75447313547140654062015-12-08T10:51:53.465-08:002015-12-08T10:51:53.465-08:00Draw altitude BD = h and let AD = p
So h^2 = c^2 ...Draw altitude BD = h and let AD = p<br /><br />So h^2 = c^2 -p^2 = a^2 -(b-p)^2<br /><br />Simplifying p = (b^2 + c^2-a^2)/2b and<br />h^2 = c^2 -(b^2+c^2-a^2)^2/4b^2<br /><br />So K^2 = h^2b^2/4 = {4b^2 c^2 - (b^2 + c^2- a^2) }/16 which simplifies to <br /><br />K^2 ={ 2b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b ^4 - c^2}/16 from which upon taking the square root the result follows <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87232729791316688272015-12-08T10:45:20.143-08:002015-12-08T10:45:20.143-08:00heron theoremheron theoremAnonymousnoreply@blogger.com