Question

The hypotenuse of a right angled triangle exceeds twice one side by 1 and exceeds the other by 2.find the hypotenuse.

Answer 1

Answer:

c = 2a + 1

c = b + 2

c - 1 = 2a

a = (1/2) * (c - 1)

c - 2 = b

a^2 + b^2 = c^2

(1/2)^2 * (c - 1)^2 + (c - 2)^2 = c^2

(1/4) * (c^2 - 2c + 1) + c^2 - 4c + 4 = c^2

(1/4) * (c^2 - 2c + 1) - 4c + 4 = 0

c^2 - 2c + 1 - 16c + 16 = 0

c^2 - 18c + 17 = 0

c = (18 +/- sqrt(324 - 68)) / 2

c = (18 +/- sqrt(256)) / 2

c = (18 +/- 16) / 2

c = 9 +/- 8

c = 1 , 17

c = 17

b = 15

a = 8

Answer 2

Answer:

let the hypotenuse be 'x' cm

let the base b

x=2b+1 ==       b=(x-1)/2

let the height be h

x=h+2 ==     h=x-2

By using pythagoras theorem

substitute the values of base and height with the values shown above:-

we get the values of x as 1 or 17

x=1 is not possible

so the hypotenuse is 17 units.

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