Question

find the surface area of this sphere. use 3.14 for π. image of sphere with radius 2 cm surface area of a sphere sa = 4πr² sa = ? cm² round your answer to the nearest tenth.

Explanation:

Answer:

To find the surface area of the sphere, we use the formula \( SA = 4\pi r^2 \), where \( r = 2 \) cm and \( \pi \approx 3.14 \).

Substitute the values into the formula:
\( SA = 4 \times 3.14 \times (2)^2 \)
First, calculate \( 2^2 = 4 \).
Then, \( 4 \times 3.14 = 12.56 \).
Next, \( 12.56 \times 4 = 50.24 \).

\( 50.2 \) (rounded to the nearest tenth) or \( 50.24 \) (exact with the given \( \pi \)). Since the problem says to round to the nearest tenth, \( 50.2 \) cm². But if we use the exact calculation with \( 3.14 \), it's \( 50.24 \), which rounds to \( 50.2 \) (or sometimes \( 50.2 \) is acceptable, but let's check: \( 50.24 \) to the nearest tenth is \( 50.2 \) because the hundredth digit is 4, which is less than 5). Wait, actually, \( 50.24 \) rounded to the nearest tenth is \( 50.2 \)? Wait, no: the tenths place is 2, the hundredths is 4. So yes, \( 50.2 \). But maybe the problem expects the unrounded or rounded. Wait, let's recalculate:

\( 4 \times 3.14 \times 4 = 4 \times 4 \times 3.14 = 16 \times 3.14 = 50.24 \). Rounded to the nearest tenth is \( 50.2 \) (since 4 < 5, we don't round up the tenths place).

So the surface area is \( 50.2 \) cm² (or \( 50.24 \) cm² if not rounded). But the problem says "Round your answer to the nearest tenth", so \( 50.2 \).

Wait, but maybe I made a mistake. Let's check again:

\( r = 2 \) cm.

\( SA = 4 \times 3.14 \times (2)^2 = 4 \times 3.14 \times 4 = 16 \times 3.14 = 50.24 \). Rounded to the nearest tenth: look at the hundredth digit, which is 4. Since 4 < 5, we keep the tenths digit as is. So \( 50.2 \) cm².

Alternatively, if the problem allows two decimal places, it's \( 50.24 \), but the question says "to the nearest tenth", so \( 50.2 \).

So the final answer is \( \boxed{50.2} \) (or \( \boxed{50.24} \) if not rounded, but the problem specifies nearest tenth).