Chapter 3:Pair of Linear Equations in Two Variables
Chapter 5: Arithmetic Progression
Chapter 8: Introduction to Trigonometry
Chapter 9:Some Applications of Trigonometry
Chapter 12:Area Related to Circles
Surface Areas and Volumes
Radius r1 of the upper end = 4/2 = 2 cm
Radius (r2) of lower the end = 2/2 = 1 cm
Height = 14 cm
Capacity of glass = Volume of frustum of cone
Capacity of glass = (1/3)×π×h(r12+r22+r1r2)
= (1/3)× π × (14)(22+12+ (2)(1))
∴ The capacity of the glass = 102×(2/3) cm3
Given,
Slant height l = 4 cm
Circumference of upper end of the frustum = 18 cm
∴ 2πr1 = 18
Or, r1 = 9/π
circumference of lower end of the frustum = 6 cm
∴ 2πr2 = 6
Or, r2 = 6/π
Now, Coverd surface area of frustum = π(r1+r2) × l
= π(9/π+6/π) × 4
= 12×4 = 48 cm2
Given,
the radius of lower end,(r1) = 10 cm
radius of upper end, (r2) = 4 cm
Slant height l of frustum = 15 cm
Now,
The area of material to be used for making the fez = Coverd surface area of frustum + Area of upper circular end
Coverd surface area of frustum = π(r1+r2)×l
= 210π
And, Area of upper end = πr22
= 16π
∴ The area of material used = 710 × (2/7) cm2
Given,
r1 = 20 cm,
r2 = 8 cm and
h = 16 cm
∴ Volume of the frustum = (1/3)×π×h(r12+r22+r1r2)
the rate of milk = Rs. 20/litre
So, Cost of milk = 20×volume of the frustum
= Rs. 209
slant height will be
So, CSA of the container = π(r1+r2)×l
l= 1758.4 cm2
the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)
= 1758.4+201 = 1959.4 cm2
∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157
Consider AEG
Radius r1 of upper end of frustum =Height h3 of container = 10 cm
Now,
Volume of the frustum =On Solving this we get,
Volume of the frustum =Now,
Let the length of wire is l.
Volume of wire = Area of cross-section x Length
= (πr2)xl
= π (Now, Volume of frustum = Volume of wire
we get,
l = 7964.44 m