Question

what is the greatest possible perimeter of a right angled triangle with integer side lengths if one of its side is 12 ?

Answer 1

Let the 2 other sides be x and y

So, x2=y2+122x2=y2+122

So, x2−y2=122x2−y2=122

(x−y)(x+y)=72∗2(x−y)(x+y)=72∗2……….(I have used this method just to Find maximum of x+y to get maximum perimeter)

On comparing, Let x−y=2x−y=2

And x+y=72x+y=72

So, x=37x=37

y=35y=35

So, perimeter max . will be p=x+y+12=84p=x+y+12=84 units.

Note :

If x is not bound to be an integer the best solution could have been this :

Let the 2 other sides be x and y

So, x2=y2+122x2=y2+122

i.e. x=sqrt(y2+144)x=sqrt(y2+144)

Now, perimeter = x+y+12x+y+12

So, p=sqrt(y2+144)+y+12p=sqrt(y2+144)+y+12

To find maxima,, just differentiate the above Equation

i.e. dp/dy=2y/(sqrt(y2+144))+1dp/dy=2y/(sqrt(y2+144))+1

For maxima or minima, dp/dy=0dp/dy=0

So, −2y=sqrt(y2+144)−2y=sqrt(y2+144)

Squaring both sides, we get :

4y2=y2+1444y2=y2+144

3y2=1443y2=144

y2=48y2=48

y=4∗sqrt(3)y=4∗sqrt(3)

So, x=sqrt(192)x=sqrt(192)

P=sqrt(192)+4∗sqrt(3)+12