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Tagged: Cylinder, Measurement, surface area
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Surface Area
Cube

$$SA=6s^2$$
Rectangular Prism

$$SA=2LW+2LH+2WH$$
Triangular Prism

$$SA=2(\frac{1}{2}bh)+2sL+bL$$
Pyramid

$$SA=LW+the\:sum\:of\: the\: 4\:triangles$$
Cylinder

$$SA=2\pi r^2+2\pi rh$$
Open Cylinder

$$SA=2\pi r h$$
Open One End Cylinder

$$SA=\pi r^2+2\pi r h$$
Cone

$$SA= \pi r^2+\pi rL$$
Sphere

$$SA=4\pi r^2$$
Truncated Cone

$$SA=\pi\, R^2+\pi\,r^2+\pi\,R\,L-\pi\,r\,l$$
A sphere has a radius of 10cm. Find:
a. the surface area
b. the volumea. Surface area = $$4\times\pi\times10^2$$
= 1256.5cm2
Use the surface area formula: $$4\pi r^2$$b. Volume = $$\frac43\pi\times 10^3$$
= 4188.8cm3Use the volume formula: $$\frac43\pi r^3$$A cylinder, open at one end, has a radius of 10cm and a height of 15cm. Find the surface area.
$$SA=\pi\times10^2+2\times\pi\times10\times20$$
= 1570.80cm2Calculate the surface area of a closed hemisphere with a radius of 5cm. Answer correct to the nearest cm.
Total surface area = half of the sphere plus the circular base
SA = ½ × 4 × π × 52 + π × 52
SA = 235.619
= 236 cm2
A cylinder, open at both ends, has a radius of 10cm and a height of 15cm. Find the surface area. $$SA=2\times\pi\times10\times20$$
= 1256.64cm2$$SA=2\pi r h$$ r = 10, h = 20 a. V = 4/3 × π × 2.253
= 47.7cm3Use the volume formula for a sphere: $$V=\frac{4}{3}\pi r^3$$ and calculate, where r = ½ of the diameter
radius = ½ × 4.5 = 2.25b. 2 × 1 = 2 If it is 10cm wide, that means it can fit 2 balls across its width.
2 × 1 × 6 = 12 balls If 5cm high, can only fit one ball. so if we can fit two per row, and we need 12, so it has to be able to fit six rows along its length, 2 × 1 × 6 = 12 balls 6 × 4.5 = 27cm now if each ball is 4.5cm in diameter, then 6 balls long to fit the 12, so 6 × 4.5 -
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