Pleaseeeee answer correctly !!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!

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