Find area of triangle with sides $\sqrt{b^2+a^2}$, $\sqrt{c^2+a^2}$ and $\sqrt{a^2+c^2}$

Question

The sides of a triangle are given by $\sqrt{b^2+a^2}$, $\sqrt{c^2+a^2}$ and $\sqrt{a^2+c^2}$.

Please help me in finding the area of the triangle!

It'll be in square root with a,b and c

Answer 1

Use the formula by Heron: $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ The variables are not identical with your variables! For your work: The area is given by $$A=\frac{1}{2}\sqrt{a^2c^2+b^2c^2+a^2b^2}$$

Answer 2

Use this variant of Heron's formula

$$4\triangle=\sqrt{4(a^2b^2+b^2c^2+c^2a^2)-(a^2+b^2+c^2)^2}$$

where $a,b,c$ are the sides of the triangle

Answer 3

Take the advantage of the unique structure of the sides to use the cosine rule, $$\cos\theta=\frac{a^2}{uv}$$

where the angle $\theta$ is between the sides $u=\sqrt{b^2+a^2}$ and $v=\sqrt{a^2+c^2}$. Then,

$$\sin^2\theta= 1-\cos^2\theta = \frac{a^2b^2+c^2a^2+b^2c^2}{u^2v^2}$$

Hence, the area

$$A = \frac12uv\sin\theta= \frac12 \sqrt{a^2b^2+c^2a^2+b^2c^2}$$