Given-that-x-1-3-x-are-lengths-of-the-sides-of-a-right-angled-triangle-pythagoras-tripple-find-the-value-of-x-

Question Number 35512 by Rio Mike last updated on 19/May/18

G i v e n t h a t (x + 1, 3, x) a r e l e n g t h s

o f t h e s i d e s o f a r i g h t a n g l e d

t r i a n g l e (p y t h a g o r a s t r i p p l e)

f i n d t h e v a l u e o f x .

Answered by Rasheed.Sindhi last updated on 20/May/18

Case - 1 : x + 1 > 3

∴ x + 1 is hypatenuse

{(x + 1)}^{2} = {(3)}^{2} + x^{2}

x^{2} + 2 x + 1 = 9 + x^{2}

2 x = 9 - 1 = 8

x = 4

x + 1 = 4 + 1 = 5

(5, 3, 4) is required pythagorean triplet .

Case - 2 : 3 > x + 1

∴ 3 is hypatenuse

{(3)}^{2} = {(x + 1)}^{2} + x^{2}

{2 x}^{2} + 2 x - 8 = 0

x^{2} + x - 4 = 0

x = \frac{- 1 \pm \sqrt{17}}{2}

x + 1 = \frac{- 1 \pm \sqrt{17}}{2} + 1 = \frac{1 \pm \sqrt{17}}{2}

(\frac{1 + \sqrt{17}}{2}, 3, \frac{- 1 + \sqrt{17}}{2}) contains irrational numbers

(\frac{1 - \sqrt{17}}{2}, 3, \frac{- 1 - \sqrt{17}}{2}) contains negative

numbers .

Hence

(5, 3, 4) is required pythagorean triplet .