contestada

2x +1
d A right-angled triangle is drawn so
that the hypotenuse is twice the
shortest side plus 1 cm, and the other
side is twice the shortest side less 1 cm.
Find the length of the hypotenuse.
2x-1

Respuesta :

x = 8

If the hypotenuse = 2x + 1 and the shortest side = x and the other side = 2x - 1

Then we can substitute rose values into the Pythagorus Theorem (a squared plus b squared equals c squared which is the hypotenuse) to the equation in the diagram attached.

Then you just expand the brackets and simplify to get x. Hope that helps.

Answer:

Hypotenuse = 17cm

Step-by-step explanation:

From the question we have that,

  • It is a right-angled triangle
  • The shorter side's length is x
  • The hypotenuse's length is 2x + 1
  • The other side's length is 2x - 1

Since this is a right-angled triangle, we can apply the Pythagoras formula:

[tex]H^{2} = C_{1}^{2} + C_{2}^{2}[/tex]

This says that the square of the hypotenuse is the sum of the square of the other sides.

From this, we can replace the variables with the actual values:

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

We can then expand the exponents to have:

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

We can proceed to isolate x:

[tex]4x^{2} + 4x + 1 = 4x^{2} - 4x + 1 + x^{2}\\\\4x^{2} + 4x + 1 = 5x^{2} - 4x + 1\\\\4x^{2} + 4x + 1 - 5x^{2} = - 4x + 1\\\\- x^{2} + 4x + 1 = - 4x + 1\\\\- x^{2} + 4x + 1 + 4x = 1\\\\- x^{2} + 8x + 1 = 1\\\\- x^{2} + 8x = 1 - 1\\\\- x^{2} + 8x = 0\\\\[/tex]

We can now multiply the whole equation by -1 just to have a positive [tex]x^{2}[/tex]:

[tex]x^{2} - 8x = 0[/tex]

And we can now solve it using Bhaskara:

[tex]x = \frac{-b\ \pm \sqrt{b^{2} - 4ac }}{2a}[/tex]

Replacing out values, we've got:

[tex]a = 1\\b = -8\\c = 0\\x = \frac{8\ \pm \sqrt{(-8)^{2} - 4*1*0 }}{2*1}\\x = \frac{8\ \pm \sqrt{(-8)^{2} - 0 }}{2}\\x = \frac{8\ \pm \ 8}{2}\\x = 0\ or\ x = 8[/tex]

Since x can't be 0 because we're talking about a drawn triangle, x can only be 8.

We can then calculate the hypotenuse based on this value:

[tex]Hypotenuse = 2x + 1\\Hypotenuse = 2*8 + 1\\Hypotenuse = 17[/tex]

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