Easy and simple triangle area problem

Using Heron's Formula,

If a,b,c are the three sides of a triangle, then we define s(semi-perimeter) as $s=\dfrac{a+b+c}{2}$ , then

So here as per given information, let $a=10,b=8,c=4$

$Area=\sqrt{s(s-a)(s-b)(s-c)}$

$\quad \quad =\sqrt{231}$

So $answer=\boxed{2+3+1=6}$

We can solve it by herons formulae.

by heros formulae

We can find the area of a triangle by using the Heron's formula . If all the sides of a triangle are give: Area = square root of s(s-a)(s-b)(s-c) Where s= a+b+c/2 here: a=10 ; b=8 ; c=4 So Area = square root of 231 So the Answer is 6.

we can easily find the area of any triangle by using Heron's Formula.... Area= square root of s(s-a)(s-b)(s-c).. where s is the (a+b+c)/2.......... by replaing the values we get 231 under the square root......... which is a 3 digit number... so the sum of digits is 2+3+1=6.........

let a=8,b=4,c=10; using cosine rule cosC=a^2+b^2-c^2/2ab=-5/16; now sinC=√ (1-cos^2C)=√(231)/16; Area=1/2 a b*sinC=√(231); therefore, a+b+c=2+3+1=6