In Fig., a tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre. (use π = 22/7)


In the given figure,

Given diameter 3m
For conical portion, 
radius = 3/2 = 1.5m and length is 2.8m
∴ S1 = curved surface area of conical portion
∴ S1 = πrl
   = π x 1.5 x 2.1
= 6.3π m2
Area of canvas used for making tent = S1+S2
 =4.2π +6.3π
= 10.5π
=10.5 x (22/7)
=33m2
Thus, total cost of the canvas at the rate of Rs. 500 per m2 = 500 x 33 =16500
Rs. 16500 is total cost.
4302 Views

In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.



Side of square = 28 cm and radius of each circle = 28/2 cm
Area of the shaded region
= Area of the square + Area of the two circles − Area of the two quadrants
space equals space left parenthesis 28 right parenthesis squared space plus space 2 space straight x space straight pi space straight x open parentheses 28 over 2 close parentheses squared space minus 2 space straight x 1 fourth space straight x space straight pi space straight x space open parentheses 28 over 2 close parentheses squared
space equals space left parenthesis 28 right parenthesis squared space plus space 3 over 2 space straight x space straight pi space straight x open parentheses 28 over 2 close parentheses squared
space equals space left parenthesis 28 right parenthesis squared space left parenthesis 1 plus 3 over 2 space straight x 22 over 7 space straight x 1 half space straight x 1 half right parenthesis
space equals space left parenthesis 28 right parenthesis squared space open parentheses 1 plus 33 over 28 close parentheses
space equals space left parenthesis 28 right parenthesis squared space straight x space 61 over 28
space equals space 28 space straight x 61
space equals space 1708 space cm squared

6743 Views

From each end of a solid metal cylinder, metal was scooped out in the hemispherical form of the same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. ( π = 22/7)

Height of the solid metal cylinder, h = 10 cm
Radius of the solid metal cylinder, r = 4.2 cm

therefore,

Radius of each hemisphere = radius of the solid metal cylinder, r = 4.2 cm

Now, 

Volume of the rest of the cylinder = Volume of cylinder - 2 x volume of each hemisphere

 space equals space πr squared straight h space minus space 2 space straight x space 2 over 3 space πr cubed
space equals space πr squared space open parentheses straight h space minus 4 over 3 straight r close parentheses
equals straight pi space straight x space left parenthesis 4.2 space right parenthesis squared space open parentheses 10 space minus space 4 over 3 space straight x space 4.2 close parentheses
space equals straight pi space straight x space left parenthesis 4.2 right parenthesis squared space straight x space left parenthesis 4.4 right parenthesis space cm cubed

Thickness of the cylindrical wire = 1.4 cm
Therefore, 
Radius of the cylindrical wire, R = 1.4/2 = 0.7 cm

Let the length of the wire be H cm.

It is given that the rest of the cylinder is melted and converted into a cylindrical wire.

therefore,

Volume of the cylindrical wire = Volume of the rest of the cylinder

⇒ π x 0.7 x 0.7 x H = π x (4.2)2 x (4.4)

rightwards double arrow space straight H space equals space fraction numerator 4.2 space straight x space 4.2 space straight x space 4.4 over denominator 0.7 space straight x space 0.7 end fraction space equals space 158.4 space cm

Hence, the length of the wire is 158.4 cm



3312 Views

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.


Let the height and radius of the given cone be H and R respectively.
The cone is divided into two parts by drawing a plane through the mid point of its
axis and parallel to the base.
Upper part is a smaller cone and the bottom part is the frustum of the cone.
⇒ OC = CA = h/2
Let the radius of smaller cone be r cm.
In ΔOCD and ΔOAB,
∠OCD = ∠OAB = 90°
∠COD = ∠AOB (common)
∴ ΔOCD ∼ ΔOAB (AA similarity)

⇒ OA/OC = AB/CD = OB/OD
⇒ h / h/2 = R/r
⇒ R = 2r
the radius and height of the cone OCD are r cm and h/2 cm
therefore the volume of the cone OCD = 1/3 x π x r2 x h/2 = 1/6 πr2h
Volume of the cone OAD = 1/3 x π x R2 x h = 1/3 x π x 4r2 x h
The volume of the frustum = Volume of the cone OAD - Volume of the cone OCD
= (1/3 x π x 4r2 x h) – (1/3 x π x r2 x h/2)
= 7/6 πr2h
Ratio of the volume of the two parts = Volume of the cone OCD : volume of the frustum
= 1/6 πr2h : 7/6 πr2h
= 1 : 7

9459 Views

In Figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).


Area of the square lawn PQRS = 42 m x 42 m

Let OP = OS = xm
So, x2 + x2  = (42)2

⇒ 2x2 = 42 x 42

⇒ x2 = 21 x 42

Now,
area of sector POS=

 90 over 360 space x space πx squared space equals space 1 fourth space straight x space left parenthesis πx right parenthesis squared space.... space left parenthesis space straight i space right parenthesis

equals 1 fourth space space straight x space 22 over 7 space straight x space 21 space straight x space 42 space straight m squared space.... space left parenthesis ii right parenthesis
Also comma space
Area space of space increment POS space equals space 1 fourth space straight x space Area space of space square space lawn space PQRS
space equals space 1 fourth space straight x space left parenthesis space 42 space straight x space 42 right parenthesis squared space straight m squared space space space space space left parenthesis angle POQ space equals space 90 degree right parenthesis space.. space left parenthesis iii right parenthesis

So comma space

Area space of space flower space bed space PSP space equals space Area space of space Sector space POS space minus space Area space of space increment POS
space equals space 1 fourth space straight x space 22 over 7 space straight x space 21 space straight x 42 space minus space 1 fourth space straight x space 42 space straight x space 42 space space left square bracket frome space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis right square bracket

space equals space 1 fourth space straight x space 21 space straight x space 42 space straight x open parentheses 22 over 7 space minus space 2 close parentheses

equals space 1 fourth space straight x space 21 space straight x space 42 space straight x space open parentheses 8 over 7 close parentheses straight m squared
Therefore space area space of space the space two space flower space beds space equals space 2 space straight x space 1 fourth space straight x space 21 space straight x space 42 space straight x open parentheses space 8 over 7 close parentheses space
space equals space 504 space straight m squared
Hence comma space the space total space area space of space the space two space flower space beds space 504 space straight m squared

3460 Views

In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of the hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park. Write your views on recycling of water. 


Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, Vc = πr2H = π×(1)2×5=5π m

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
the height of standing water in the park = h
Volume of water in the park = lbh = 25×20×h

Now water from the tank is used to irrigate the park. So,
Volume of cylindrical tank = Volume of water in the park

⇒5π=25×20×h
⇒5π/25×20=h
⇒h=π/100 m
⇒h=0.0314 m

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

7864 Views