Question

Show algebraically that Heron's formula can also be written as...

Answer 1

\[area= (\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}) / 4\]

Answer 2

FYI, Heron's formula is \[\sqrt{s(s-a)(s-b)(s-c)}\] when s=(a+b+c)/2

Answer 3

it is the same right?

Answer 4

oh that is what you have to show
ok start with \[\sqrt{s(s-a)(s-b)(s-c)}\] and then replace \(s\) by \(\frac{a+b+c}{2}\) and it will pop right out

Answer 5

where does the /2 go?

Answer 6

you have it four places and \(2^4=16\)

Answer 7

then when you take the square root you will get your \(4\) in the denominator

Answer 8

Oh... But, \[\sqrt{(a+b+c)(b+c)(a+c)(a+b)}\] How does that work?

Answer 9

That's not the same...

Answer 10

\[\frac{a+b+c}{2}-a=\frac{b+c-a}{2}\]

Answer 11

for example

Answer 12

what happened to a?

Answer 13

lets go slow

Answer 14

\[s=\frac{a+b+c}{2}\]
\[s-a=\frac{a+b+c}{2}-a=\frac{a+b+c-2a}{2}=\frac{b+c-a}{2}\]

Answer 15

so, you multiply something by 2?

Answer 16

similarly \[s-b=\frac{a+b+c}{2}-b=\frac{a+b+c-2b}{2}=\frac{a+c-b}{2}\]

Answer 17

yeah, to put it over the same denominator
just like
\[\frac{7}{2}-3=\frac{7-2\times 3}{2}\]

Answer 18

huh. Got it! Thanks.

Answer 19

yw