A7. The swimmer crossed the river, which is $8 \mathrm{~m}$ wide. He swam straight, but due to the water current with a speed of $1 \frac{\mathrm{m}}{\mathrm{s}}$, he landed on the opposite bank of the river $6 \mathrm{~m}$ lower. How far did he swim?
(A) $8 \mathrm{~m}$
(B) $10 \mathrm{~m}$
(C) $11 \mathrm{~m}$
(D) $14 \mathrm{~m}$
(E) $20 \mathrm{~m}$
A7 The pre-impact distance is the hypotenuse of a right-angled triangle. To calculate it, we use the Pythagorean theorem: $d=\sqrt{8^{2}+6^{2}}=10 \mathrm{~m}$.