हिंदी

A Hemi-spherical Dome of a Building Needs to Be Painted. If the Circumference of the Base of the Dome is 17.6 Cm, Find the Cost of Painting It, Given the Cost of Painting is Rs. 5 per L00 `Cm^2` - Mathematics

Advertisements
Advertisements

प्रश्न

A hemi-spherical dome of a building needs to be painted. If the circumference of the base of
the dome is 17.6 cm, find the cost of painting it, given the cost of painting is Rs. 5 per l00
`cm^2`

संक्षेप में उत्तर

उत्तर १

Given that only the rounded surface of the dome to be painted, we would need to find the
curved surface area of the hemisphere to know the extent of painting that needs to be done.
Now, circumference of the dome =17.6m.
Therefore, 17.6=2πr.

`2× 22/7r= 17.6m.`

So, the radius of the dome = 17.6× `7/(2× 22)`

m=2.8m

The curved surface area of the dome = `2πr^2`

=`2× 22/7× 2.8× 2.8cm^2`

= `49.28m^2`

Now, cost of painting 100`cm^2` is Rs. 5.

So, cost of painting `1m^2`= Rs.500

Therefore, cost of painting the whole dome
=Rs. 500×49.28
=Rs. 24640 .

shaalaa.com

उत्तर २

In the given problem, a hemispherical dome of the building needs to be painted. So, we need to find the surface area of the dome.

Here, we are given the circumference of the hemispherical dome as 17.6 m and as we know that circumference of the hemisphere is given by 2πr. So, we get

       `2πr = 17.6`

`2(22/7)r = 17.6`

            `r  = 17.6 (7/22)(1/2)`

                 = 2.8

So, now we find the surface area of the hemispherical dome.

` "surface area" = 2  π r^2`

                     ` =2(22/7)(2.8)^2`

                       = 49.28

So, the curved surface area of the dome is 49.28 m2

Since the rate of the painting is given in cm2, we have to convert the surface area from m2 to cm2.

So, we get

Curved surface area = `(49.28)(10000)cm^2`

= 492800 cm2

Now, the rate of painting per 100 cm2 = Rs 5

The rate of painting per 1 cm2 = `5/100`

So, the cost of painting the dome = `(5/100) (492800)`

=24640 

Therefore, the cost of painting the hemispherical dome of the building is Rs 24640 .

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Surface Areas and Volume of a Sphere - Exercise 21.1 [पृष्ठ ८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 9
अध्याय 21 Surface Areas and Volume of a Sphere
Exercise 21.1 | Q 10 | पृष्ठ ८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.


A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.


The diameter of the moon is approximately one fourth of the diameter of the earth. Find the
ratio of their surface areas.


A solid is in the form of a cone standing on a hemi-sphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find, in terms of π, the volume of the solid. 


If a sphere of radius 2r has the same volume as that of a cone with circular base of radius r, then find the height of the cone.


If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of the cube is


The ratio between the volume of a sphere and volume of a circumscribing right circular cylinder is 


The total area of a solid metallic sphere is 1256 cm2. It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate: the number of cones recasted [π = 3.14]


From a rectangular solid of metal 42 cm by 30 cm by 20 cm, a conical cavity of diameter 14 cm and depth 24 cm is drilled out. Find: the volume of remaining solid 


A manufacturing company prepares spherical ball bearings, each of radius 7 mm and mass 4 gm. These ball bearings are packed into boxes. Each box can have maximum of 2156 cm3 of ball bearings. Find the:

  1. maximum number of ball bearings that each box can have.
  2. mass of each box of ball bearings in kg.
    (use π = `22/7`)

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×