Question

The length of the hypotenuse of a right triangle exceeds the length of its base by 2cm and exceeds twice the length of the altitude by 1cm. Find the length of each side of the triangle.

Answer 1

Let hypotenuse be a

Given, base = a-2               ---------------- (1)

Given, altitude = (a-1)/2.       ---------------- (2)


By Pythagoras theorem

a^2 = (a-2) + ((a-1)/2)^2

a^2 = (a^2-2a+1)/4 + a^2-4a+4 

4a^2 = a^2-2a+1+4(a^2-4a+4)

4a^2 = a^2-2a+1+4a^2-16a+16

a^2 - 18a + 17 =0

(a-17)(a-1) = 0

a = 17

Substitute a = 17 in (2) we get

and lenght of each side = 17 - 1/2

                                         = 16/2

                                         = 8


Hope this helps!

Answer 2

Answer:

Step-by-step explanation:

Solution :-

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.