   SEARCH HOME Math Central Quandaries & Queries  Question from Brian: Seeking the square footage of a triangular property lot, dimensions are: 620 X 620 X 720. I tried the Herons formula myself, but could not complete it :( Thank you!! Brian We have two responses for you

Hi Brian,

You haven't stated the units but I assume all three lengths are in the same units.

For Heron's Formula I use a, b and c as the side lengths and s as the semi=perimeter, s = (a + b + c)/2. Thus for your example

a = 620, b = 620, c = 720 so s = (620 + 620 + 720)/2 = 980.

The area is then given by

A = sqrt[s × (s - a) × (s - b) × (s - c)]

where sqrt is the square root. Thus

A = sqrt[980 × (980 - 620) × (980 - 620) × (980 - 720)]
= sqrt[980 × (360) × (360) × (260)]
= sqrt
= 181719.78 square units.

Penny

Brian,

In order to use Heron's formula, you need to label each side a, b, and c.
Check out the following example:
A triangular lot measures 520 X 520 X 620, find the square footage.
Let side a=520, side b=520 and side c=620,
Then using Heron 's formula,
A= (s(s-a)(s-b)(s-c))1/2
s= (a + b + c)/2
s = (520 + 520 + 620)/2 = 830
A = (830(830-520)(830-520)(830-620))1/2
A = (830 × 310 × 310 × 210)1/2

Because the numbers were so large, I divided each number under the square root by 10, thus I divided by 100001/2 = 100 and hence I need to also multiply by 100

A = (830 × 310 × 310 × 210)1/2
A = 100 × (83 × 31 × 31 × 21)1/2
A = 12 9422.68 square units.

Good luck,
Melanie     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.