# Mensuration-40

### From Homeworkwiki

**A rocket is in the form of a circular cylinder, closed at the lower end with a cone of the same base radius**

**attached to the top. The cylinder is of radius 5/2 m and height 21 m and the cone is of slant height 8 m.**

**Calculate in m ^{2}, the total surface area of the rocket. Also find the volume of the rocket.**

**Solution:** Curved surface area of the cone = Π r l = 22 / 7 x 5 / 2 x 8

= 22 / 7 x 5 / 2 x 8 = 440 / 7 = 62-6/7 m^{2}

Curved surface area of the cylinder = 2 Π r h = 2 x 22 / 7 x 5 / 2 x 21 = 330 m^{2}

Base area of cylinder = Π r^{2} = 22 / 7 x 5 / 2 x 5 / 2 = 19-9/14 m^{2}

Total surface area of rocket = (62-6/7 + 330 + 19-9/14) = **412-1/2 m ^{2}**

Volume of cylinder = Π r^{2} h = 22 / 7 x 5 / 2 x 5 / 2 x 21

= 825 / 2 = 412-1/2 m^{2}

If h be the height of the cone, then

h^{2} = 8^{2} - (5 / 2)^{2} = 64 - 25 / 4 = 231 / 4

=> h = √231 / 2 = 15.20 / 2 = 7.6 m

Volume of cone = 1 / 3 Π r^{2} h = 1 / 3 x 22 / 7 x 25 / 4 x 7.6 m^{3} = 1045 / 21 = 49-16/21 m^{3}

Total volume of the cone = Volume of the cylindrical portion + Volume of the conical portion

= 412-1/2 + 49-16/21 = **462-11/42 m ^{3}**