Question

the volume of a cone is 926 cubic cm. its height is 6.3 cm. the base radius is closest to?

Explanation:

Step1: Recall the volume formula of a cone

The volume \( V \) of a cone is given by the formula \( V=\frac{1}{3}\pi r^{2}h \), where \( r \) is the radius of the base and \( h \) is the height of the cone. We know \( V = 626\space cm^{3}\) and \( h=6.3\space cm \), and we need to solve for \( r \).

First, we can rearrange the formula to solve for \( r^{2} \):
\[
r^{2}=\frac{3V}{\pi h}
\]

Step2: Substitute the given values

Substitute \( V = 626 \), \( h = 6.3 \) and \( \pi\approx3.14 \) into the formula for \( r^{2} \):
\[
r^{2}=\frac{3\times626}{3.14\times6.3}
\]
First, calculate the numerator: \( 3\times626 = 1878 \)
Then, calculate the denominator: \( 3.14\times6.3=19.782 \)
So, \( r^{2}=\frac{1878}{19.782}\approx94.93 \)

Step3: Solve for \( r \)

Take the square root of \( r^{2} \) to find \( r \):
\[
r=\sqrt{94.93}\approx9.74\space cm
\]
If we round to a reasonable decimal place or check the closest value, we can see that the radius is closest to \( 10\space cm \) (or depending on the required precision, but if we calculate more accurately, let's re - check the calculation:

Wait, let's recalculate \( \frac{3\times626}{3.14\times6.3} \):

\( 3\times626 = 1878 \)

\( 3.14\times6.3=19.782 \)

\( 1878\div19.782\approx94.93 \)

\( \sqrt{94.93}\approx9.74 \), which is closest to \( 10\space cm \) (if we consider whole numbers) or if we have options with one decimal place, maybe \( 9.7\space cm \) or \( 9.8\space cm \), but the main calculation gives us approximately \( 9.7\space cm \) to \( 9.8\space cm \), and the closest whole number is \( 10\space cm \).

Answer:

The base radius is closest to \(\boldsymbol{10\space cm}\) (or approximately \(9.7\space cm\) depending on the required precision). If we consider the calculation more precisely, \(\sqrt{\frac{3\times626}{3.14\times6.3}}\approx9.7\space cm\), so the radius is closest to \(10\space cm\) (or \(9.7\space cm\) if we take one decimal place).