How do you find the area of a triangle ABC, if $a = 23$ $a = 23$ , $b = 46$ $b = 46$ , and $c = 41$ $c = 41$ ?

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Answer 1 · 2018-03-12T11:41:29

Area of the triangle is $470.914$ $470.914$ sq.unit.

In triangle $DeltaABC ; a= 23 , b=46 , c=41$

The semi perimeter of triangle is $s=(a+b+c)/2$ or

$s= (23+46+41)/2 =55$ unit

Area of the triangle is $A_t=sqrt(s(s-a)(s-b)(s-c)$ or

$A_t=sqrt(55(55-23)(55-46)(55-41)$ or

$A_t=sqrt(55*32*9*14)=sqrt(221760)~~470.914$ sq.unit

Area of the triangle is $470.914$ sq.unit. [Ans]

How do you find the area of a triangle ABC, if a = 23a=23a = 23, b = 46b=46b = 46 , and c = 41c=41c = 41?