From a right cylindrical solid of radius 7cm and height 12cm , a right cone, half the height of the cylinder , is removed. The base of the cone is on one of the plane faces of the cylinder, their centres being the same. If the radius of the base of the cone is half of the radius of the base of the cylinder, find :-
•) The volume of the solid left
•) The total surface area of the solid left.
Answers
Solution:-According to question
height of the cone ,which will be removed=12/2=6
radius of that cone=7/2=3.5
total volume of cylinder=π×7×7×12
=>π×84×7=588π
volume of cone ,which have to remove=π×3.5×3.5×6/3
=>π×3.5×7=24.5π
volume of solid left=(588π-24.5π)=22(84-3.5)
=22×80.5=1771cm^3
total surface area of remaining solid=curved surface area of cylinder+area of one plane+curved surface area or come+π(r+R)(R-r)
=2πrh+πr^2+πr/2 ×l+π(R+r)(R-r)
=πr(2h+r+l/2)+π(R+r)(R-r)
=22×7/7(2×12+7+l/2)+π(R+r)(R-r)
l is the slant height of cone
l={(3.5)^3+36)}^1/2
l=(48.25)^1/2
l=6.8(approx)
surface area after removing cone =22×(34+7+3.4)+22/7×10.5×3.5
=22×44.4+115.5
=976.8+115.5cm^2
=1092.3cm^2
Answer:
Solution:-According to question
height of the cone ,which will be removed=12/2=6
radius of that cone=7/2=3.5
total volume of cylinder=π×7×7×12
=>π×84×7=588π
volume of cone ,which have to remove=π×3.5×3.5×6/3
=>π×3.5×7=24.5π
volume of solid left=(588π-24.5π)=22(84-3.5)
=22×80.5=1771cm^3
total surface area of remaining solid=curved surface area of cylinder+area of one plane+curved surface area or come+π(r+R)(R-r)
=2πrh+πr^2+πr/2 ×l+π(R+r)(R-r)
=πr(2h+r+l/2)+π(R+r)(R-r)
=22×7/7(2×12+7+l/2)+π(R+r)(R-r)
l is the slant height of cone
l={(3.5)^3+36)}^1/2
l=(48.25)^1/2
l=6.8(approx)
surface area after removing cone =22×(34+7+3.4)+22/7×10.5×3.5
=22×44.4+115.5
=976.8+115.5cm^2
=1092.3cm^2