The length of the hypotenuse of a right triangle exceeds the length of

1.	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
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.

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

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.