Let the three sides of the triangle be a, b and c.

Let the shorter side be a, the longer side be b and the hypotenuse be c.

It is given that the shorter side of the triangle is 7 centimeters less than the longer side.

=> a = b...

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Let the three sides of the triangle be a, b and c.

Let the shorter side be a, the longer side be b and the hypotenuse be c.

It is given that the shorter side of the triangle is 7 centimeters less than the longer side.

=> a = b - 7

The hypotenuse of triangle is 2 more than the longer side of the triangle.

=> c = b + 2

Now we use the Pythagorean Theorem

a^2 + b^2 = c^2

=> (b - 7)^2 + b^2 = (b + 2)^2

=> b^2 + 49 - 14b + b^2 = b^2 + 4 + 4b

=> b^2 + 45 - 18b = 0

=> b^2 - 15b - 3b + 45 = 0

=> b(b - 15) - 3(b - 15) = 0

=> (b - 3)(b - 15) = 0

b = 3 and b = 15

b cannot be 3 as b - 7 would be negative

c = b + 2 = 17

**The length of the hypotenuse is 17 cm.**

Let the sides be x, y and the hypotenuse h.

Given that h is 2 more than the longer sides.

Let the longer side be x.

==> h = x+2...............(1)

The shorter side (y) is 7 less than the longer sides.

==> y= x-7 ..............(2)

But we know that L

h^2 = x^2 + y^2

==> (x+2)^2 = x^2 + (x-7)^2

==> x^2 + 4x +4 = x^2 + x^2 -14x + 49

==> x^2 + 4x +4 = 2x^2 -14x + 49

==> x^2 -18x + 45 = 0

==> (x-3)(x-15) = 0

==> x1= 3 ==> y= 3-7 = -4 (impossible)

==> x2=15 ==> y2= 15-7 = 8

==> h= 15+2 = 17

**Then, the length of the hypotenuse is 17 cm.**