How do you use Heron's formula to find the area of a triangle with sides of lengths #6 #, #6 #, and #7 #?

2 Answers
Nam D.
Apr 24, 2018

#17.06 \ "units"^2#

Explanation:

Heron's formula states for any triangle, the area is given by:

#A=sqrt(s(s-a)(s-b)(s-c))#

  • #s# is the semiperimeter of the triangle, given by #s=(a+b+c)/2#.

  • #a,b,c# are the sides of the triangle

Here, #s=(6+6+7)/2=19/2=9.5# units.

And so, the area is:

#A=sqrt(9.5(9.5-6)(9.5-6)(9.5-7))#

#=sqrt(9.5*3.5*3.5*2.5)#

#=sqrt(290.9375)#

#~~17.06#

Dean R.
Apr 30, 2018

Archimedes' Theorem is a modern form of Heron's Formula. We set #a=c=6, b=7# in

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 4655#

# A = {7 sqrt(95)}/4 #

Explanation:

Archimedes' Theorem is usually superior to Heron's Formula.

Given a triangle with sides #a,b,c# and area #A#, we have

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 #

# = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) #

#= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)#

No semiperimeter, no multiple square roots when given coordinates.

We're free to choose whichever form is convenient and assign #a,b,c# however we see fit. For an isosceles triangle, #a=c# gives nice cancellation in the first form. Let's set #a=c=6, b=7# in

#16 A^2 = 4(6^2)(7^2) - 7^4 = 4655#

# A = \sqrt{4744/16} = {7 sqrt(95)}/4 #

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