(d) 40 cm
Let the length of the hypotenuse be x cm.
Then, length of base = (x – 2) cm
x - 2 length of altitude =1
⇒ Length of altitude = \(\frac12(x - 1)\) cm
Applying Pythagoras' Theorem,
(Hyp.)2 = (Base)2 + (Perp.)2
⇒ x2 = (x - 2)2 + \(\big(\frac12(x-1)\big)^2\)
⇒ x2 = x2 -4x + 4 + \(\frac14\)(x2 - 2x + 1)
⇒ 4x2 = 4(x2 - 4x + 4) + (x2 - 2x + 1)
⇒ 4x2 = 4x2 - 16x + 16 + x2 - 2x +1
x2 - 18x + 17 = 0
⇒ (x - 17)(x - 1) = 0 ⇒ x = 17, 1
x = 1 is not possible.
∴ Length of hypotenuse = 17 cm
Length of base = 15 cm
Length of altitude = \(\frac12\) x 16 cm = 8 cm
∴ Perimeter of the triangle = 17 cm + 15 cm + 8 cm = 40 cm.