Answer:
Let the required side (=BA) be x meter
According to the problem, (using the Pythagorus Theorem)
[tex] {x}^{2} + ({4.1)}^{2} = {(4.7)}^{2} \\ {x}^{2} = {(4.7)}^{2} - {(4.1)}^{2} \\ {x}^{2} = (4.7 + 4.1)(4.7 - 4.1) \\ {x}^{2} = 8.8 \times 0.6 \\ {x}^{2} = 5.28 \\ x = \sqrt{5.28} \\ \boxed{x = 2.3 \: meter}[/tex]
2.3 m is the right answer.
