10 Math Surfacearea&volume

Finish Line & Beyond SURFACE AREAS AND VOLUMES Cone and Hemisphere: Total Surface Area Total Surface Area of the toy = Curved Surface Area of hemisphere + Curved Surface Area of cone = 2 Πr²+ Πr  Total Volume: = Volume of Hemisphere + Volume of Cone =

2 1 Πr³ + Πr²h 3 3

Example: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere . The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour.

Solution: Total Surface Area of the toy = Curved Surface Area of hemisphere + Curved Surface Area of cone = 2 Πr²+ Πr  = 2×

22 3.5 3.5 22 × × × 1.75 × 3.7 + 7 2 2 7

= 39.6 cm² (  can be calculated using Pythagoras Formula) Cube & Hemisphere Total Surface Area: = Surface Area of Cube + Curved Surface Area of Hemisphere- Area of Base of Hemisphere (If it is smaller than cube’s surface) = Surface Area of Cube + Curved Surface Area of Hemisphere + Area of Base of Hemisphere – Area of one Face of Cube (If circular base is bigger than cube’s surface) Total Volume: = Volume of Cube + Volume of Hemisphere = Side³+

2 Πr³ 3

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Example: A decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.

Solution : The required answer will be the sum of areas of cube’s surface and hemisphere’s surface minus the area of circular base of the hemisphere. Total Surface Area of Cube = 6 Side² =6 × 5²= 150 cm² Surface area of hemisphere= 2Πr² = 2×

22 × 2.1 × 2.1 = 2 × 138.6 7

As area of the circular base needs to be deducted, Total Surface Area = 150+13.86= 163.86 cm² Cone & Cylinder: Total surface area will be as follows: Curved Surface Area of Cylinder + Curved Surface Area of Cone =2Πrh + Πr  There can be a chance of the circumference of the cone being either greater than or smaller than the circumference of the cylinder. If both cases area of the exposed circular portion can be calculated as follows: Area of Bigger Circle – Area of Smaller Circle. Total Volume:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond = Volume of Cylinder + Volume of Cone = 2Πrh+

1 Πr²h 3

Conversion of Solid from One Shape to Another When a solid of a certain shape is converted to a solid of another shape its volume remains unchanged. Example: A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere. Solution: Volume of cone = Volume of Sphere Now let us assume that the radius of Cone is r and the radius of sphere is R.

1 4 Πr²h = ΠR³ 3 3 1 4 Or, Π6² × 24 = ΠR³ 3 3 Or, 6 × 6 × 6 × 2 × 2 = 2 × 2 × R³ Then,

Or, R = 6 cm

Example: Selvi’s house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use π = 3.14) Solution: The volume of water in the overhead tank = Volume of Water taken from tank So, Πr²h =  × b × h Or, 3.14 × 3600 × 95 = 157 × 144 × H Or, H=47.5 cm So, height of the water left in tank = 95-47.5=47.5 cm As now water is left to half its original level so cylindrical tank’s capacity is half of the capacity of the sump. Example: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. Solution: Let us assume that height of the rod is h and radius is r and the length of the wire is H and radius is R Then, Πr²h = ΠR²H Or, Π 0.5² × 8 = ΠR²1800 cm Or, R²= 0.5 × 0.5 × 8 ÷ 1800 cm Or, R²= 5 × 5 × 80 ÷ 18000 mm

5× 5× 5× 2× 2× 2× 2 1 = 3 × 3 × 2 × 10 × 10 × 10 9 1 Or, R = mm 9 =

Frustum:

Anything in the shape of a bucket is called as frustum. Imagine a bigger cone whose top has been chopped off and you get a figure of frustum. Volume of Frustum:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Volume of Bigger Cone – Volume of Smaller Cone Let us assume that the radius of the bigger cone is R, height is H and lateral side is L. Then the radius of the smaller cone is r, height is h and lateral side is  Then Volume of Frustum=

1 1 1 ΠR²H - Πr²h = Π(R²H-r²h) 3 3 3

When height of the frustum is given in stead of height of the bigger cone then heights of both cones can be calculated using ratios of similar triangles. As bigger and smaller cones are always similar triangles so ratio of bigger radius to smaller radius will be same as ratio of height or slant height of bigger cone to those of smaller cones. Similarly Curved Surface area can be calculated as follows: Curved Surface Area of Bigger Cone – Curved Surface area of Smaller Cone ΠRL-Πr  = Π(RL-r  )

www.excellup.com ©2009 send your queries to [email protected]