Solving right-angled triangle with one side

Question

My teacher challenged us to get the sides of right angled triangle with one side.

ABC is a right-angled triangle at B. AB is 6 cm long. Find AC and BC.

Answer 1

I assume your teacher meant integer sides (otherwise it has an infinite solutions!!).

So,we can see that one side is $6$ units.

We know,$AB^2+BC^2=AC^2$.So,the three are Pythagorean triples.

We can generate Pythagorean triples in the following way.

$(m^2-n^2),2mn,(m^2+n^2)$ are three numbers which are Pythagorean triples for any integers $m,n$.

So,here we let,$2mn=6\implies mn=3$ Now,as $3$ is a prime it can be factorized only as $3\times1$.So,$m=3,n=1$.

Substituting these values into $(m^2-n^2)$ and $(m^2+n^2)$ we get $8,10$ respectively.So,the sides are $8,6,10$.

Hope this helps!!

Answer 2

It is possible if the given side AB (= 6cm) is the base of a right - angled isosceles triangle for which the other two sides are of three multiplied by square root of two (measured in cm.)

Answer 3

I suspect that you have been doing Pythagoras theorem on right angled triangles. First think about of the sides were string that is elastic. As you move C left or right along the horizontal your triangle still keeps its 90 degree angle and AB stays 6 cm. So just those two conditions have infinitely many triangles that satisfy them. Now since we have the 90 degree angle we can use Pythagoras to get length from A to C multiplied by itself is equal to the length from B to C multiplied by itself plus the length from B to A multiplied by itself. In formula form if we let the length from A to C be x and the length from B to C be y we have $ x^2 = y^2+6^2$ and so $x^2 -y^2 = 36$. Now you can write the left hand side as $(x+y)(x-y)$. So this tells you that since $x$ and $y$ are both whole numbers bigger than zero (since they are lengths) then $x+y$ must also be a whole number bigger than zero AND it also divides into 36 a whole number of times. So you can now get a short list of the only numbers $(x+y)$ can be. Then for each number on your list there is only certain pairs of whole numbers which when added together give you this number. These pairs are the only POSSILE values for $ x$ and $y$. However for each pair you have one more thing to check which is when you work out $x-y$ for that pair then it multiplied by ... has to equal ...

Hope this helps