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

2 Answers
1s2s2p
Mar 8, 2018

"Area"~~96.56Area≈96.56

Explanation:

Heron's formula says that:
"Area"=sqrt(S(S-A)(S-B)(S-C))Area=√S(S−A)(S−B)(S−C) given S=(A+B+C)/2S=A+B+C2

For this triangle, S=(14+15+16)/2=22.5S=14+15+162=22.5

"Area"=sqrt(22.5(22.25-14)(22.5-15)(22.5-16))=sqrt(149175/16)=(15sqrt(663))/4~~96.56Area=√22.5(22.25−14)(22.5−15)(22.5−16)=√14917516=15√6634≈96.56

Dean R.
Apr 30, 2018

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

16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2 = 14917516A2=4a2b2−(c2−a2−b2)2=149175

A = (15 sqrt(663))/4A=15√6634

Explanation:

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

Given a triangle with sides a,b,ca,b,c and area AA, we have

16A^2 = 4a^2b^2 - (a^2-a^2-b^2)^2 16A2=4a2b2−(a2−a2−b2)2

= (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) =(a2+b2+c2)2−2(a4+b4+c4)

= (a+b+c)(-a+b+c)(a-b+c)(a+b-c)=(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,ca,b,c however we see fit.

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