Respuesta :

Answer:

let the third side be x

Using pythagoras theorem we get,

(58)^2 = (42)^2 + (x)^2

3364=1764+x^2

x^2=3364-1764

x^2= 1600

x=√(1,600)

x=40

Answer :

[tex]\pink{\sf Third \: side \: of \: the \: triangle = 40}[/tex]

Solution :

As, the given triangle is a right angled triangle,

Hence, We can use the Pythagoras' Theorem,

[tex]\star\:{\boxed{\sf{\pink {H^{2} = B^{2} + P^{2}}}}}[/tex]

Here,

  • H = Hypotenuse of triangle
  • B = Base of triangle
  • P = Perpendicular of triangle

In given triangle,

  • Base = 42
  • Hypotenuse = 58
  • Perpendicular = ?

Now, by Pythagoras' theorem,

[tex]\star\:{\boxed{\sf{\pink {H^{2} = B^{2} + P^{2}}}}}[/tex]

[tex] \sf : \implies (58)^{2} = (42)^{2} + P^{2}[/tex]

[tex] \sf : \implies 58 \times 58 = 42 \times 42 + P^{2} [/tex]

[tex] \sf : \implies 3364 = 1764 + P^{2}[/tex]

[tex] \sf : \implies P^{2} = 3364 - 1764[/tex]

[tex] \sf : \implies P^{2} = 1600[/tex]

By squaring both sides :

[tex] \sf \sqrt{P^{2}} = \sqrt{1600}[/tex]

[tex] \sf : \implies P^{2} = \sqrt{1600}[/tex]

[tex] \sf : \implies P^{2} = \sqrt{(40)^{2}}[/tex]

[tex] \sf : \implies P^{2} = 40 [/tex]

[tex]\pink{\sf \therefore \: Third \: side \: of \: the \: triangle \: is \: 40}[/tex]

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