Respuesta :

ronhagrid310

Answer:

The lengths of the sides of a right angled triangle are x + 1, 4x, and (4x + 1).

Supposing: x + 1 < 4x < 4x + 1

=> Applying Pythagorean theorem:

(x + 1)^2 + (4x)^2 = (4x + 1)^2

=> x^2 + 2x + 1 + 16x^2 = 16x^2 + 8x + 1

=> x^2 - 6x = 0

=>x(x - 6) = 0

=> x = 0 (invalid)

=> x = 6

Check:

=> a = x + 1 = 7,

=> b = 4x = 4*6 = 24,

=> c = 4x + 1 = 4*6 + 1 =25

7^2 + 24^2 = 625 = 25^2

Hope this helps!

:)

CHERLYNnotCHERRY

Answer:

The value of x is 6.

Step-by-step explanation:

You have to use Pythagorean Theorem, a² + b² = c². Make (4x+1)cm as the hypotenuse because any values substitute into x will give the highest length among these 3 expressions of length :

[tex] {a}^{2} + {b}^{2} = {c}^{2} [/tex]

Let a = side = (x+1)cm,

Let b = side = 4x cm,

Let c = hypotenuse = (4x+1)cm,

[tex] {(x + 1)}^{2} + {(4x)}^{2} = {(4x + 1)}^{2} [/tex]

[tex] {x}^{2} + 2x + 1 + 16 {x}^{2} = 16 {x}^{2} + 8x + 1[/tex]

Then you have to simplify :

[tex]17 {x}^{2} + 2x + 1 = 16 {x}^{2} + 8x + 1[/tex]

[tex]17{x}^{2} + 2x + 1 - 16 {x}^{2} - 8x - 1 = 0[/tex]

[tex] {x}^{2} - 6x = 0[/tex]

Next you have to solve it :

[tex] {x}^{2} - 6x = 0[/tex]

[tex]x(x - 6) = 0[/tex]

[tex]x = 0 \: (rejected)[/tex]

[tex]x - 6 = 0[/tex]

[tex]x = 6cm[/tex]

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