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The length of the hypotenuse of a right triangle exceeds the length of the base by 2 cm and exceeds twice the length of the attitude by 1 cm. The perimeter of the triangle is : 

(a) 18 cm 

(b) 17 cm 

(c) 25 cm 

(d) 40 cm

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Best answer

(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.

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