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
\(c^{2}=a^{2}+b^{2}\)
where
a and b are the legs
c is the hypotenuse (the greater side)
In this problem
Let
\(a=7\ cm\\c=13\ cm\)
substitute
\(13^{2}=7^{2}+b^{2}\)
Solve for b
\(169=49+b^{2}\)
\(b^{2}=169-49\)
\(b^{2}=120\)
\(b=\sqrt{120}\ cm\)
\(b=10.95\ cm\)
we have that
Jayden's Work
\(a^{2}+b^{2}=c^{2}\)
\(a=7\ cm\\b=13\ cm\)
substitute and solve for c
\(7^{2}+13^{2}=c^{2}\)
\(49+169=c^{2}\)
\(218=c^{2}\)
\(c=\sqrt{218}\ cm\)
\(c=14.76\ cm\)
Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle
Sheridan's Work
\(a^{2}+b^{2}=c^{2}\)
\(a=7\ cm\\c=13\ cm\)
substitute
\(7^{2}+b^{2}=13^{2}\)
Solve for b
\(49+b^{2}=169\)
\(b^{2}=169-49\)
\(b^{2}=120\)
\(b=\sqrt{120}\ cm\)
\(b=10.95\ cm\)
therefore
Sheridan's Work is correct
