Question

2. the hypotenuse of a right angled triangle
exceds twice one side by l and exceeds the other by 2
Find the hypotenuse.
​

Answer 1

Answer:

17

Step-by-step explanation:

Let the hypotenuse be 'a'.

In question,

Hypotenuse = twice the first side + 1

→ a - 1 = twice the first side

→ (a - 1)/2 = first side.

Also, hypotenuse = second side + 2

→ a = 2nd side + 2

→ a - 2 = 2nd side

Using Pythagoras theorem:

→ a² = (a - 2)² + {(a - 1)/2}²

→ a² = (a² + 2² - 2*2*a) + {(a² + 1 - 2a)/4}

→ a² = {4(a² + 4 - 4a) + (a² + 1 - 2a)}/4

→ 4a² = 4a² + 16 - 16a + a² + 1 - 2a

→ 4a² = 5a² - 18a + 17

→ 5a² - 4a² - 18a + 17 = 0

→ a² - 18a + 17 = 0

→ a² - 17a - a + 17 = 0

→ a(a - 17) - a(a - 17) = 0

→ (a - 1)(a - 17) = 0

→ a = 1 or 17

a cant 1, as, if a = 1, then 2nd side = 1 - 2 = -1, which is not possible.

So, a = 17

Answer 2

Answer:

Step-by-step explanation:

Let the altitude of triangle be x cm.

Hypotenuse of triangle = 2x + 1

Base of triangle = 2x - 1.

According to the Question,

Using Pythagoras theorem,

⇒ (2x + 1)² = x² + (2x - 1)²

⇒ 4x² + 1 + 4x = x² + 4x² + 1 - 4x

⇒ x² - 8x = 0

⇒ x(x - 8) = 0

⇒ x = 0, 8 (Neglecting 0)

⇒ x = 8 cm

Altitude of triangle = x = 8 cm

Hypotenuse of triangle = 2x + 1 = 2(8) + 1 = 16 + 1 = 17 cm

Base of triangle = 2x - 1 = 2(18) - 1 = 16 - 1 = 15 cm.