43. Television screens and computer monitors are measured along the diagonal. Ray just bought a 46-cm computer monitor. The width of the monitor is $35 \mathrm{~cm}$. a) What is the height of the monitor, to the nearest centimetre? b) The monitor has a $2-\mathrm{cm}$ frame around it. What is the length of the diagonal of the monitor, including the frame, to the nearest centimetre?

Step 1 ：Given the width \(w = 35\) cm and the diagonal \(d = 46\) cm of the monitor, we can use the Pythagorean theorem to find the height \(h\): \(w^2 + h^2 = d^2\).

Step 2 ：Substitute the given values: \(35^2 + h^2 = 46^2\).

Step 3 ：Solve for \(h^2\): \(h^2 = 46^2 - 35^2\).

Step 4 ：Calculate \(h^2\): \(h^2 = 1681\).

Step 5 ：Find the height \(h\) by taking the square root: \(h = \sqrt{1681}\).

Step 6 ：Calculate the height: \(h = 29.8496231131986\) cm.

Step 7 ：Round the height to the nearest centimetre: \(h = \boxed{30}\) cm.

Step 8 ：Now, we need to find the length of the diagonal of the monitor including the frame. The frame adds \(2\) cm to each side, so the new width is \(35 + 2 + 2 = 39\) cm and the new height is \(30 + 2 + 2 = 34\) cm.

Step 9 ：Use the Pythagorean theorem again to find the new diagonal \(d'\): \(39^2 + 34^2 = d'^2\).

Step 10 ：Calculate \(d'^2\): \(d'^2 = 2677\).

Step 11 ：Find the new diagonal \(d'\) by taking the square root: \(d' = \sqrt{2677}\).

Step 12 ：Calculate the new diagonal: \(d' = 51.64103973493939\) cm.

Step 13 ：Round the new diagonal to the nearest centimetre: \(d' = \boxed{52}\) cm.