Adhir has drawn a right-angled triangle whose length of hypotenuse is 6 cm more than the twice of the shortest side. If the length of the third side is 2 cm less than the length of the hypotenuse, then calculate the lengths of the three sides of the right-angled triangle drawn by Adhir.

Explanation

Let the length of the smallest side of the right angled triangle be x cm.
`:.` The length of the hypotenuse =(2x+6)cm
and the length of the third side `=(2x+6-2)cm=(2x+4)cm.`
By Pythagoras' theorem we get, `(2x+6)^(2)=x^(2)+(2x+4)^(2)`
or, `(2x+6)^(2)-(2x+4)^(2)=x^(2)`
or, `(2x+6+2x+4)(2x+6-2x-4)=x^(2)`
or, `(4x+10)xx2=x^(2)`
or, `8x+20=x^(2)or,x^(2)-8x-20=0..........(1)`
Comparing (1) with `ax^(2)+bx+c=0,(ane0)` we get, `a=1,b=-8andc=-20`.
`:.` by Sreedhar Acharya's formula we get, `x=(-(-8)pmsqrt((-8)^(2)-4xx1xx(-20)))/(2xx1)`
or, `x=(8pmsqrt(64+80))/(2)`
or, `x=(8pmsqrt(144))/(2)or,x=(8pm12)/(2)`
`:.x=(8+12)/(2)` (taking +sign) and `x=(8pm12)/(2)` (taking -sign)
or, `x=(20)/(2)" "or,x=(-4)/(2)`
or, `x=10" "or,x=-2`.
But the length of a side can never be negative, `:.xne-2,i.e.,x=10`
`:.` Hypotenuse `=(2x+6)cm=(2xx10+6)cm=26cm.`
Third side `=(2x+4)cm=(2xx10+4)cm=24cm`.