Heron’sFormula

Heron’s Formula is used to determine the area of any triangle when only the lengths of the three

sides are known.

2

cbas

where

c)b)(sa)(s(s)(sk

The ‘s’ represents the

semi-perimeter.

The semi-perimeter is nothing more

than half the perimeter.

2

cbas

Add the lengths of the three known

sides together and then divide by two.

Next substitute the known values into Heron’s Formula then solve for the

area.

A

B

C

a

b

c

Side a=5

Side b=8

Side c=7

First determine ‘s’, which is the

semi-perimeter.

2

cbas

2

785s

2

20s

10s

Now we know:

Side a: = 5

Side b: = 8

Side c: = 7

Semi-perimeter: s = 10

Replace the letters in Heron’s Formula with their respective known values. (Substitute)

c)b)(sa)(s(s)(sk

7)8)(105)(10(10)(10k

c)b)(sa)(s(s)(sk

(3)(10)(5)(2)k

7)8)(105)(10(10)(10k

c)b)(sa)(s(s)(sk

300k

(3)(10)(5)(2)k

7)8)(105)(10(10)(10k

c)b)(sa)(s(s)(sk

320508.17k

300k

(3)(10)(5)(2)k

7)8)(105)(10(10)(10k

c)b)(sa)(s(s)(sk

3.17k

320508.17k

300k

(3)(10)(5)(2)k

7)8)(105)(10(10)(10k

c)b)(sa)(s(s)(sk

Another problem solved via

trigonometry!

Try this one on your own!

Side a = 5

Side b = 6

Side c = 7

Did you getArea = k = 14.7

Now for the solution of this

problem.

First determine ‘s’, which is the

semi-perimeter.

2

cbas

2

765s

2

18s

9s

Now we know:

Side a: = 5

Side b: = 6

Side c: = 7

Semi-perimeter: s = 9

c)b)(sa)(s(s)(sk

7)6)(95)(9(9)(9k

c)b)(sa)(s(s)(sk

2)(9)(4)(3)(k

7)6)(95)(9(9)(9k

c)b)(sa)(s(s)(sk

216k

2)(9)(4)(3)(k

7)6)(95)(9(9)(9k

c)b)(sa)(s(s)(sk

696938.14k

216k

2)(9)(4)(3)(k

7)6)(95)(9(9)(9k

c)b)(sa)(s(s)(sk

7.14k

696938.14k

216k

2)(9)(4)(3)(k

7)6)(95)(9(9)(9k

c)b)(sa)(s(s)(sk

This was createdby

Frank J. Antoniazzi Jr.on

15 April 2002