## Tuesday, December 8, 2015

### Geometry Problem 1170: Area of a Triangle in terms of the three sides, a-b-c

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1170.

1. heron theorem

2. Draw altitude BD = h and let AD = p

So h^2 = c^2 -p^2 = a^2 -(b-p)^2

Simplifying p = (b^2 + c^2-a^2)/2b and
h^2 = c^2 -(b^2+c^2-a^2)^2/4b^2

So K^2 = h^2b^2/4 = {4b^2 c^2 - (b^2 + c^2- a^2) }/16 which simplifies to

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

Sumith Peiris
Moratuwa
Sri Lanka

3. K^2 = (p)(p-a)(p-b)(p-c) ; where p = (a+b+c)/2 ...

K^2 = (a+b+c)/2 * (b+c-a)/2 * (a+c-b)/2 * (a+b-c)/2

K^2 = 1/16 * [(a+b)^2 - c^2] * -[(b-a)^2 - c^2]

K^2 = 1/16 * -{(a+b)^2(a-b)^2 - c^2[(a+b)^2 + (a-b)^2] + c^4}

K^2 = 1/16 * -{(a^2-b^2)^2 - c^2(2*a^2 + 2*b^2) + c^4}

K^2 = 1/16 * -{a^4 - 2*a^2*b^2 + b^4 - 2*c^2a^2 - 2*c^2*b^2 + c^4}

K^2 = [-a^4 + 2*a^2*b^2 - b^4 + 2*c^2a^2 + 2*c^2*b^2 - c^4]/16

K = [2*a^2*b^2 + 2*c^2a^2 + 2*c^2*b^2 -a^4 - b^4 - c^4]^(1/2) /4