Heron's formula when side lengths include radicals

We can use the fourth form of Heron's formula to get an easy answer.

Let $a=2, b=\sqrt{2}, c=\sqrt{3}-1$. We can use the formula $$A=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$$ We have $a^2 = 4, b^2 = 2, c^2 =4-2\sqrt{3}$Thus, $$A=\frac{1}{4}\sqrt{32-(2+2\sqrt{3})^2}$$ $$=\frac{1}{2}\sqrt{8-(1+\sqrt{3})^2}$$ $$=\frac{1}{2}\sqrt{4-2\sqrt{3}}$$ $$=\frac{\sqrt{3}-1}{2}$$


A 7th grader who does not know Heron's formula can still solve this.

Consider the right triangles with two common vertices $A$ and $B$:

  • triangle $ABC$ with $AB=1$, $AC=1$, $BC=\sqrt{2}$; its area is $1/2$.
  • triangle $ABD$ with $AB=1$, $AD=\sqrt{3}$, $BD=2$; its area is $\sqrt{3}/2$.

Drawing a sketch of these triangles, with point $C$ between $A$ and $D$, we see that the sought-for area is the difference of the above areas, $ABD$ minus $ABC$; i.e. it's the area of triangle $BCD$, which is $\sqrt{3}/2-1/2$.

Tags:

Geometry

Area

Triangles

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