Using Pythagoras’ theorem on each of the right-angled triangles VAB, VBC and VCA, we obtain the following values for AB, BC and CA.

AB = √(a² + b²), BC = √(b² + c²), CA = √(c² + a²).

To find the area of triangle ABC we need to know the sine of one of the angles; let’s choose the angle at A.

Using the cosine rule we have

cos A = (√(c² + a²)²+ √(a² + b²)²– √(b² + c²)²)/(2√(c² + a²) √(a² + b²))

=(2a² + b² + c² – (b²+ c²))/( 2√(c² + a²) √(a² + b²))

= a²/(√(c² + a²) √(a² + b²))

Now we can find the value of sin A using 1 – cos² A.

sin² A = 1 – [a²/(√(c² + a²) √(a² + b²))]²

= 1 - a⁴/((c² + a²) (a² + b²))

= ((c² + a²) (a² + b²) – a⁴)/((c² + a²) (a² + b²))

= (a²b² + b²c²+ c²a²) /((c² + a²) (a² + b²))

So, sin A = √(a²b² + b²c²+ c²a²)/ (√(c² + a²) √(a² + b²))

Therefore, the area of triangle ABC is

½ x AB x AC x sin A

= ½ √(a² + b²)) x √(c² + a²)x√(a²b² + b²c²+ c²a²)/ (√(c² + a²) √(a² + b²))

= ½ √(a²b² + b²c²+ c²a²)

We have found that the set of values a = 2, b = 4 and c = 6 give an integer area (equal to 14).

We’d be interested to know if you find any more!

Challenge 310: A Particular Pyramid