How do you use Heron's formula to find the area of a triangle with sides of lengths #15 #, #16 #, and #14 #?

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Dean R. Share
Apr 30, 2018

Answer:

Know more than the other students! Archimedes' Theorem is a modern form of Heron's Formula. We set #a=15, b=16, c=14# in

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 149175#

# A = (15 sqrt(663))/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 - (a^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 fractions till the end, 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.

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1s2s2p Share
Mar 8, 2018

Answer:

#"Area"~~96.56#

Explanation:

Heron's formula says that:
#"Area"=sqrt(S(S-A)(S-B)(S-C))# given #S=(A+B+C)/2#

For this triangle, #S=(14+15+16)/2=22.5#

#"Area"=sqrt(22.5(22.25-14)(22.5-15)(22.5-16))=sqrt(149175/16)=(15sqrt(663))/4~~96.56#

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