Mensuration problems

=Problem 1= Exercise 16.1 Problem 7. 1)Craft teacher of a school taught the students to prepare cylindrical pen holders out of card board. In a class of strength 42, if each child prepared a pen holder of radius 5 cm and height 14 cm, how much cardboard was consumed?

Case 1)Pen stand which is open on top and closed base. Then we have to calculate Curved surface area of cylinder and area of one circle.
 * Interpretation;

visualising the cylinder into rectangle and circle
 * skills
 * Assumptions:

Concepts used
 * 1) basics of circles -radius
 * 2) area of circles
 * 3) addition and multiplication of fractions
 * 4) unit of area

knowledge to be used
 * 1) knowledge of Polygons -Rectangles
 * 2) knowledge of measurements
 * 3) Formula of CSA of cylinder
 * 4) substitution
 * 5) computing
 * 6) Value of Л

Concepts to be taught solution: r=5cm, h=14cm, Л=$${\frac{22}{7}}$$ Curved surface area of cylinder and area of one circle= CSA of cylinder+area of circle =$$2Лrh+Лr^2$$ =$$2X{\frac{22}{7}}X5X14+{\frac{22}{7}}X{5^2}$$ =440+78.5 =518.5cm
 * 1) basics of circles -radius
 * 2) area of circles
 * 3) relation between circumference of circle and length of rectangle in a cylinder
 * 4) addition and multiplication of fractions
 * 5) unit of area

Cardboard required to prepare 42 penstands= 42X518.5 =$$21777{cm^2}$$ =$$2.1777{m^2}$$

Case 2:- Hollow pen stand: Pen stand which is open both the sides i.e., top and base. Then we have to calculate Curved surface area of cylinder. Case 3:-Pen stand which is closed both the sides i.e., top and base. Then we have to calculate total surface area of cylinder.

Activity Making use of 2200 cmcardboard sheet how many hollow cylinders of radius 7 cm and height 5 cm can be prepared.

=Problem 2= Solved problems on cone Example 1

cm, if the material of which it is made weighs 10gm/
 * 1) Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51,

Calculating the weight of solid cone by calculating its volume. density of a material is given. it is related to volume and weight.
 * Statement of the problem:What is the weight of a solid cone whose base is of diameter 14 cm and vertical height 51cm, if the material of which it is made weighs 10gm/
 * Interpretation;

Concepts used
 * Assumptions:
 * 1) basics of circles -radius
 * 2) area of circles
 * 3)  multiplication of fractions
 * 4) unit of volume,density and weight
 * 5) density

knowledge to be used Concepts to be taught
 * 1) radius=diameter/2
 * 2) density=mass/volume
 * 3) knowledge of measurements
 * 4) Formula of volume of cones
 * 5) substitution
 * 6) computing
 * 7) Value of Л
 * 1) density
 * 2) units conversion

solution: d=14cm   r=7cm, h=51cm Volume of cone= =$$\frac{1}{3}{Л}{r^2}h$$

=$$\frac{1}{3}{X}{\frac{22}{7}}{X}{7^2}{X}51$$ =$$2618{cm^3}$$ To calculate weight of the solid cone we have to use density of the given material. density=$$10gm/{cm^3}$$ =$${\frac{10Kg}{1000cm^3}}$$ Weight of solid cone=volumeXdensity =$$2618{cm^3}{X}{\frac{10Kg}{1000cm^3}}$$ =26.18Kg

=Problem 3= problem on combination of solids example 01

of the conical portion is 4 times the radius of its base, find the radius of the cream cones. Solution Solution of the conical portion is 4 times the radius of its base. .
 * 1) A Cylindrical container of radius 6cm and height 15cm is filled with ice cream. The whole ice cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height
 * 1) Statement of the problem
 * 2) A Cylindrical container of radius 6cm and height 15cm is filled with ice cream. The whole ice cream has to be filled in 10 equal cones with hemispherical tops. and the height

Assumptions
 * 1) Student should know volume of cone,and volume of Hemisphere
 * 2) Student should know the value of л=22/7
 * 3) Student should know the difference between radius and height
 * 4) Student should know the proper substitution simplification

Concepts to be taught
 * 1) Let us consider the volume of a cone having hemispherical top =2л
 * 2) Volume of Cylindrical container is equated to volume of cone having hemispherical top

Gaps
 * 1) Comparision between the volumes of cylinder and (cone+Hemisphere)

Skills
 * 1) To imagine a cone, Hemisphere,and cylinder
 * 2) To imagine a cone having Hemispherical top

Algorthem Part 1: To derive the volume of a cone with hemispherical top =$${\frac{1}{3}}л4r +{\frac{2}{3}}л{r^3}$$(on simplification) =$$2л{r^3}$$
 * 1) volume of a cone with hemispherical top =$${\frac{1}{3}}л{r^2}h+{\frac{2}{3}}л{r^3}$$

Part 2 : To calculate the volume of 10 cone with hemispherical top =$$10X2л{r^3}$$ =$$20л{r^3}$$ To calculate the volume of ice-cream in cylindrical container =$$л{r^2}h$$ = л X6X6X15 =540лc

Part 3 : To apply condition given in the problem volume of 10 cone with hemispherical top= volume of cylindrical container $$20л{r^3}$$ =540л $${r^3}$$=$${\frac{540л}{20л}}$$ =27 				r=3cm Hence the radius of cream cones=3cm