Heron's Formula - Examples

1. s=\frac{1}{2}(a+b+c)=\frac{1}{2}(6 \,cm +3 \,cm +5\,cm) =\frac{14 \,cm}{2}=7 \,cm

2. A=\sqrt{s(s-a)(s-b)(s-c)}= \sqrt{7(7-6)(7-3)(7-5)\,cm^4}= \sqrt{56} \,cm^2 \approx 7,5 cm^2

1. s=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 +4 +5) =\frac{12}{2}=6

2. A=\sqrt{s(s-a)(s-b)(s-c)}= \sqrt{6(6-3)(6-4)(6-5)}= \sqrt{36} =6

• The triangle 3-4-5 is known to be a right triangle? with base а=3 , height b=4 and hypotenuse c=5
• Using the standard formula for the area of a triangle: A=\frac{1}{2}ah =\frac{1}{2}ab=\frac{1}{2}\,3 \cdot 4=6
• This means that we got the same answer with Heron's formula and with the standard formula for area. (Good thing, huh?)

1. s=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 +5 +7) =\frac{15}{2}=7,5

2. A=\sqrt{s(s-a)(s-b)(s-c)}= \sqrt{7,5(7,5-3)(7,5-5)(7,5-7)}= \sqrt{42,19} =6,5

3. On the other hand, A=\frac{1}{2}\,c \cdot h_c .

    Substituting, we have: 6,5=\frac{1}{2}\,7 \cdot h_c

    so that h_c= \frac{2 \cdot 6,5}{7} = 1,86 .

From here - if we want - with Pythagoras' theorem? we can find the length of the left and right part of the base, ...

h_c= 1,86