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[tex]\pink{\sf Third \: side \: of \: the \: triangle = 45}[/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 (51)^{2} = (24)^{2} + P^{2}[/tex]
[tex] \sf : \implies 51 \times 51 = 24 \times 24 + P^{2} [/tex]
[tex] \sf : \implies 2601 = 576 + P^{2}[/tex]
[tex] \sf : \implies P^{2} = 2601 - 576[/tex]
[tex] \sf : \implies P^{2} = 2025[/tex]
By squaring both sides :
[tex] \sf \sqrt{P^{2}} = \sqrt{2025}[/tex]
[tex] \sf : \implies P^{2} = \sqrt{2025}[/tex]
[tex] \sf : \implies P^{2} = \sqrt{(45)^{2}}[/tex]
[tex] \sf : \implies P^{2} = 45 [/tex]
[tex]\pink{\sf \therefore \: Third \: side \: of \: the \: triangle \: is \: 45}[/tex]
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