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

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Answer 1 · 2018-03-12T11:41:29

Area of the triangle is #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]