Ming had 13 unsharpened pencils bundled in the shape of a cylinder. Of the diameter is 0.5 cm and the length is 19 cm, what is the surface area of Ming's bundle of pencils? Use 3.14 for π and round to the nearest tenth
It IS NOT 30.3
To find the total surface area of the bundle of pencils, we need to calculate the lateral surface area of the cylinder (the curved surface) as well as the surface area of the two circular ends.
Lateral surface area of cylinder = (2 * π * radius * height)
= 2 * 3.14 * 0.25 cm (radius = 0.5 cm / 2) * 19 cm
= 6.28 * 0.25 * 19
= 29.86 cm²
Surface area of two circular ends = 2 * π * radius^2
= 2 * 3.14 * 0.25^2
= 2 * 3.14 * 0.0625
= 0.3925 cm²
Total surface area of the bundle of pencils = Lateral surface area + Surface area of two ends
= 29.86 + 0.3925
= 30.2525 cm²
Rounded to the nearest tenth, the surface area of Ming's bundle of pencils is 30.3 cm², which is approximately equal to 30.3.
thats not right
I apologize for the error in my previous calculation. Let's correct it:
Lateral surface area of cylinder = 2 * π * radius * height
= 2 * 3.14 * 0.25 cm (radius = 0.5 cm / 2) * 19 cm
= 1.57 * 19
= 29.83 cm²
Surface area of two circular ends = 2 * π * radius^2
= 2 * 3.14 * 0.25^2
= 2 * 3.14 * 0.0625
= 0.3925 cm²
Total surface area of the bundle of pencils = Lateral surface area + Surface area of two ends
= 29.83 + 0.3925
= 30.2225 cm²
Rounded to the nearest tenth, the surface area of Ming's bundle of pencils is 30.2 cm².