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A right triangle has sides whose lengths are consecutive integers Find the lengths of the sides?
Wiki User
∙ 13y ago
The clue in the question here is that you are dealing with a right triangle. If a triangle has a right angle then Pythagoras' theorem can be used to describe the relationship between the lengths of the three sides.
Pythagoras' theorem is:
a2 + b2 = c2
a and b are used to represent the lengths of the two shorter sides and c the length of the longest side (this side is called the hypotenuse and it will be the one opposite the right angle).
If the sides are three consecutive integers they can be described as n, n+1 and n+2.
n+2 must be the hypotenuse as this will be the biggest value (the longest side, or c).
If we substitute these in to Pythagoras' theorem we get:
n2 + (n+1)2 = (n+2)2
This is an equation with one variable (one letter which we do not know). When there is only one thing we do not know in the equation we can solve it by rearranging. So, we can find n, which will be the length of the shortest side, using the method below.
n2 + n2 + 2n +1 = n2 + 4n + 4 (by expanding the squared brackets)
2n2 + 2n +1 = n2 + 4n + 4 (by simplifying)
n2 - 2n - 3 = 0 (by doing the same to both sides)
(n - 3)(n + 1) = 0 (by factorising, you can check this by multiplying the brackets back out)
Two things multiplied together will only give you zero if one of them is zero.
n - 3 = 0 and n + 1 = 0 will both give solutions which work.
n = 3 or n = -1 (by same on both sides)
However, n is the length of the shortest side so cannot be negative.
n = 3
n +1 = 4
n + 2 = 5
So, the lengths of the three sides must be 3, 4 and 5
.
This is a special set of numbers called a Pythagorean triple. They are called this because they fit Pythagoras' theorem and they are all integers. It is a good idea to know some of these for GCSE Mathematics exams. They quite often appear on the non-calculator paper. if you had known this triple already you would not need all the algebra! Another common Pythagorean triple is 8, 15, 17. You may be able to find some others.
Wiki User
∙ 13y ago
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