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Help with this problem please!!!!?
a right angled triangle is provided all of whose sides are integers.it is required to prove that the area of the triangle is an integer which is divisible by 6.
1 Answer
- 8 years agoFavorite Answer
Name the height 'a', and the base 'b'
Name the hypotenuse (opposite the right-angle) 'h'
Pythagoras's Theorem: a² + b² = h²
Area of Triangle = ½base x height = ½ba
Rearranging Pythagoras's Theorem for h: h = √(a² + b²)
In this equation, in order for h to be an integer, a must be a multiple of 4 and b must be a multiple of 3 (or the other way round) - this comes from knowledge of '3,4,5 right-angled triangles'.
So (where z is any integer (whole number)): Area of Triangle = ½ba = ½(3z)(4z) = 6z²
If z is an integer, z² must also be an integer.
Show this is divisible by six by writing it as:
6z² = 6(z²) - where z² is an integer
Source(s): You might want to look up '3,4,5' triangles, if you don't know about them
