Answer:
Sheridan's Work is correct
Step-by-step explanation:
we know that
The lengths side of a right triangle must satisfy the Pythagoras Theorem
[tex]c^{2}=a^{2}+b^{2}[/tex]
where
a and b are the legs
c is the hypotenuse (the greater side)
In this problem
Let
[tex]a=7\ cm\\c=13\ cm[/tex]
substitute
[tex]13^{2}=7^{2}+b^{2}[/tex]
Solve for b
[tex]169=49+b^{2}[/tex]
[tex]b^{2}=169-49[/tex]
[tex]b^{2}=120[/tex]
[tex]b=\sqrt{120}\ cm[/tex]
[tex]b=10.95\ cm[/tex]
we have that
Jayden's Work
[tex]a^{2}+b^{2}=c^{2}[/tex]
[tex]a=7\ cm\\b=13\ cm[/tex]
substitute and solve for c
[tex]7^{2}+13^{2}=c^{2}[/tex]
[tex]49+169=c^{2}[/tex]
[tex]218=c^{2}[/tex]
[tex]c=\sqrt{218}\ cm[/tex]
[tex]c=14.76\ cm[/tex]
Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle
Sheridan's Work
[tex]a^{2}+b^{2}=c^{2}[/tex]
[tex]a=7\ cm\\c=13\ cm[/tex]
substitute
[tex]7^{2}+b^{2}=13^{2}[/tex]
Solve for b
[tex]49+b^{2}=169[/tex]
[tex]b^{2}=169-49[/tex]
[tex]b^{2}=120[/tex]
[tex]b=\sqrt{120}\ cm[/tex]
[tex]b=10.95\ cm[/tex]
therefore
Sheridan's Work is correct
