# Examples on Surface Area and Volume

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### Volume Problems

We live in a three-dimensional world surrounded by objects of various shapes. Each of these products can be estimated regarding length, broadness, and tallness. For instance, the house that we live in has measurements. The surface area is the external aspect of the figure, while volume is the internal part. This article contains fathomed models on surface region and volume.

### Definition of Surface Area Volume Problems

The total region of the outside of a three dimensional solid can be named as surface territory. It is denoted by "S" and estimated in square units. The amount of a three-dimensional shape or the space taken by it is called volume. Volume is indicated by "V" and estimated in cubic units.

The picture above represents Surface Area Volume Problems

## Formulae for Surface Area Volume Problems

 Shapes Curved Surface Area Total Surface Area Volume Cube 4 (l2) 6 (l2) l3Â Cuboid 2h ( l+ b) 2(lb + lh + bh) lh Cylinder 2Ï€rh 2Ï€r (r + h) Ï€ (r2)h Cone Ï€rl = Ï€(h2+r2)1/2 Ï€r (r + l) â…“ Ï€(r2)h Hollow Cylinder 2 Ï€h (R + r) 2 Ï€h (R + r) + 2Ï€h (R2 â€“ r2) Ï€ (R2)h - Ï€ (r2)h Frustum of Cone Ï€l (R + r) Ï€R2 + Ï€r2 + Ï€l (R + r) â…“ Ï€h [R2 + r2 + Rr] Sphere 4 Ï€r2 4 Ï€r2 4/3 Ï€r3 Pyramid Â½ (perimeter of base) * Slant Height Curved Surface Area + Area of the Base â…“ Area of Base * Height Prism Perimeter is of Base * Height Curved Surface Area + 2 (Area of the End Surface) Area of the Base * Height

### Examples on Surface Area Volume Problems

Model 1: A customary hexagonal crystal has its border of the base as 600 cm and tallness 200 cm. Discover the weight of petroleum that it can hold if the thickness is 0.8 g/cc. (Take âˆš3 = 1.73)Â

Side of hexagon = (Perimeter)/(Number of Sides) = 600/6 = 100 cm.Â

Territory of ordinary hexagon = (3âˆš3)/2 x 100 x 100 = 25,950 sq.cm.Â

Volume = Base Area Ã— stature = 25950 x 200 = 5,190,000 cu.cmÂ

Weight of petroleum = Volume x Density = 5,190,000 cc Ã— 0.8 g/cc = 4,152,000 gm. = 4,152 kg.Â

Model 2: Find the volume of the most prominent right round cone that can be removed of a 3D shape of edge 42 cm.Â

The cone base will be hover recorded in the face of the 3D shape, and its tallness will be equivalent to the edge of the 3D square. Range of cone = 21 cm. Stature = 42 cm.Â

Volume of cone = 1/3Ï€r2h = 1/3x 22/7 x 21 x 21 x 42 = 19,404 cu. cm.Â

Model 3: An iron bundle of the distance across 6 inches is dropped into a round, hollow vessel of measurement 1ft loaded up with water. Discover the ascent at the water level.Â

Radius of the vessel = 12 inch/2 = 6 inch.Â

Volume of water that has risen = Volume of circleÂ

Ï€R2h = 4/3 Ï€r3 â‡’ 6 x 6 x h = 4/3 x 33 â‡’ h = 1 inch.Â

Model 4: The measurement of a circle is 9 cm. It is softened and brought into a wire of width 3 cm. Discover the length of the wire.Â

Vol. of circle = 4/3Ï€r3 = 4/3Ï€(4.5)3 = 121.5Ï€cm3Â

A wire is of round and hollow shapeÂ

âˆ´Ï€R2H = 121.5Ï€ â‡’ (1.5) 2 H = 121.5Â

H = 54 cmÂ

Model 5: The area of a square pyramid is 36 m2. The volume of the pyramid is 288mk3. What is the tallness of the square pyramid?Â

Since the zone of the base = 36 m2,Â

Volume of the pyramid = (1/3) * (base zone) * (tallness)Â

288 = (1/3) * (36) * (H)Â

H=24mÂ

Model 6: A crystal with symmetrical three-sided base (side= 12 cm) and Altitude (H)= 17cm. Locate the Lateral surface Area, Total surface territory, and Volume?Â

Suppose the triangle XYZ is symmetrical.Â

Edge (P) =12+12+12 = 36cmÂ

The zone of XYZ = (âˆš3/4) a2 = âˆš3/4 x 12 x 12 = 36âˆš3cm2Â

CSA = Perimeter * Altitude = 36 x 17 = 612 cm2Â

TSA = CSA + 2 x Area of XYZ = 612+72âˆš3 cm2Â

Volume = Area x elevation = 36âˆš3 x 17 = 612âˆš3 cm3