Surface Area of Combination of Solids

Class 10 NCERT Mathematics Ch Surface Area and Volumes

1. Class X ANJALI SINGH INTEGRATED B.SC. B.ED. School of Education, LPU

2. WHAT WILL BE THE SURFACE AREA OF THE CONTAINER OF THIS TRUCK?

3. • Describe combination of solids with appropriate examples. • Enlist examples of combination of solids from their daily life. • Analyse a given solid for finding its surface area. • Predict the formulae for finding the surface area of a given combination of solids. • Solve problems based on surface area of combination of solids. • Search for more applications of surface area of combination of solids from the book/internet. Appreciate the role of surface area of combination of solids.

4. SURFACE AREA OF A COMBINATION OF SOLIDS A combination of a solid is that figure which is formed by combining two or more different solids.   The total surface area of a combined solid is the sum of the total surface areas of the individual solids that make up the combined solid, excluding the overlapping parts from each figure.

5. SURFACE AREA OF A COMBINATION OF SOLIDS To determine the surface area of a solid, we take the sum of the area of all the surfaces of a 3-dimensional solid object. The surface area is the area that describes the material that will be used to cover a geometric solid. When we determine the surface areas of a geometric solid we take the sum of the area for each geometric form within the solid.

6. Find the total surface area of the given figure. Question

7. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm

8. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion,  r =3.5/2 ​ =7/4 ​cm

9. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion, r =3.5/2 ​ =7/4 ​cm Height of the conical portion,  h= (5−3.5/2) =13/4 ​ cm

10. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion,  r =3.5/2 ​ =7/4 ​cm Height of the conical portion,  h= (5−3.5/2) =13/4 ​ cm Slant height of the conical part,  l= r 2 +h 2 ​  

11. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion,  r =3.5/2 ​ =7/4 ​cm Height of the conical portion,  h= (5−3.5/2) =13/4 ​ cm Slant height of the conical part,  l= r 2 +h 2 l= (7/4​) 2 +( 13/4 ​) 2 ​ = 218/16 ​​ =3.69 cm ≈3.7 cm ​  

12. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion,  r =3.5/2 ​ =7/4 ​cm Height of the conical portion,  h= (5−3.5/2) =13/4 ​ cm Slant height of the conical part,  l= r 2 +h 2 l= (7/4​) 2 +( 13/4 ​) 2 ​ = 218/16 ​​ =3.69 cm ≈3.7 cm Total surface area of the figure will be,   2πr 2 + πrl =πr (2r+l) = 22/7​×7/4 ​(2×7/4​+3.7) ​  

13. We have, Radius of hemispherical portion, r =3.5/2 ​ =7/4 ​ cm Radius of the conical portion,  r =3.5/2 ​ =7/4 ​cm Height of the conical portion,  h= (5−3.5/2) =13/4 ​ cm Slant height of the conical part,  l= r 2 +h 2 l= (7/4​) 2 +( 13/4 ​) 2 ​ = 218/16 ​​ =3.69 cm ≈3.7 cm Total surface area of the figure will be,   2πr 2 + πrl =πr (2r+l) = 22/7​×7/4 ​(2×7/4​+3.7) = 39.6 cm 2 ​  

14. If two solid hemispheres of same base radius are joined together along their bases, then curved surface area of this new solid is ___________ .

15. PRACTICE!!! Given Total height of given figure = 14 cm Height of cylindrical portion= 8 cm Diameter of cylindrical portion= 4 cm. (Take = 22/7)   4 cm 14 cm 8 cm Calculate the total surface area of the given figure.

16. https://www.google.com/ imgres?imgurl =https%3A%2F%2Fspectrumscanindia.com%2Fimages%2Fbigstory%2Fh-1.jpg&imgrefurl=https%3A%2F%2Fspectrumscanindia.com%2Fbigstory&tbnid=DmHxxh0gpoFFBM&vet=12ahUKEwjvteHmne72AhWAZWwGHYSvCxoQMygregUIARClAg..i&docid=qJUp151rO6MmiM&w=1200&h=464&q=bharat%20petroleum%20truck&hl= en&ved =2ahUKEwjvteHmne72AhWAZWwGHYSvCxoQMygregUIARClAg https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.brainkart.com%2Farticle%2FVolume-and-Surface-Area-of-Combined-Solids_39428%2F&psig=AOvVaw0Juidkgfn_2XA3DxAU10Ce&ust=1648743440366000&source=images&cd=vfe&ved=0CAsQjRxqFwoTCND-koye7vYCFQAAAAAdAAAAABAD ABH Rohini, January 2021, MATHEMATICS Textbook for Class: https://ncert.nic.in/textbook/pdf/iemh113. pdf https://byjus.com/ncert-solutions-class-10-maths/ https://www.learncbse.in/ncert-class-10-math-solutions/