Question 649397
Let the shorter leg of a right angled triangle be 'x', hypotenuse be 'h'and other leg be 'y'
Given: y= x+1
       h= x+9
According to the hypotenuse theorem,
h^2= x^2+y^2
(x+9)^2= x^2+(x+1)^2
Apply formula:(a+b)^2= a^2+b^2+2a*b
x^2+9^2+2*x*9= x^2+x^2+1^2+2*x*1
x^2+81+18x= 2x^2+1+2x
x^2-2x^2+18x-2x+81-1=0
-x^2+16x+80= 0
x^2-16x-80= 0
We get quadratic equation and need to factorise it
Factors Of 80= 2,2,2,2,5
We have to arrange the factors to get the middle term, i.e. 16
20= 2*2*5 and 4=2*2
x^2-20x+4x-80= 0
x(x-20)+4(x-20)= 0
Take (x-20) common
(x-20)(x+4)=0
x=20,-4
the leg of the triangle cannot be negative so the answer is 20
The shorter leg is 20 cm
Other leg,y=x+1
          y=20+1=21cm
Hypotenuse,h=x+9
           h=20+9=29cm