CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 09

Surface Areas and Volumes of Basic Solids

Cuboid (l, b, h): total surface area 2(lb + bh + hl), volume lbh, diagonal \( \sqrt{l^2+b^2+h^2} \). Cube of edge a: surface area 6a², volume a³, diagonal a√3, lateral surface 4a².
Right circular cylinder (r, h): curved surface \( 2\pi rh \), total surface \( 2\pi r(r+h) \), volume \( \pi r^2h \). Hollow cylinder with outer radius R and inner radius r: volume of material \( \pi h(R^2-r^2) \), total surface \( 2\pi (R+r)(R-r+h) \).
Cone (r, h, slant height l with l² = r² + h²): curved surface \( \pi rl \), total surface \( \pi r(l+r) \), volume \( \tfrac13\pi r^2h \).
Sphere: surface \( 4\pi r^2 \), volume \( \tfrac43\pi r^3 \). Hemisphere: curved surface \( 2\pi r^2 \), total surface \( 3\pi r^2 \), volume \( \tfrac23\pi r^3 \). Spherical shell: volume of material \( \tfrac43\pi(R^3-r^3) \).

Melting, Recasting and Flow Problems

When a solid is melted and recast, the volume stays the same. Write the volume of the original piece(s) equal to the volume of the new shape and solve for the unknown.
Wire from a solid: the wire is a cylinder, so volume = \( \pi r^2 \times \) length. Reducing the diameter by p% makes the radius (1 − p/100) times the old one and the length increases to keep the volume.
Flow of liquid: in time t the liquid forms a cylinder of length (speed × t) and cross-section equal to that of the pipe or canal. Volume that flows = area of cross-section × speed × time. A solid dropped into a cylinder raises the level by (volume of solid) / (area of base of the cylinder).

Combination of Solids and Frustum

For a solid made of joined parts, add the volumes. For the surface area add only the surfaces that are visible: the area at a joint is not counted. For a cylinder with a cone on one end and a hemisphere on the other, total surface = cone curved + cylinder curved + hemisphere curved.
Frustum of a cone (radii R > r, height h): slant height \( l=\sqrt{h^2+(R-r)^2} \), volume \( \tfrac13\pi h(R^2+r^2+Rr) \), curved surface \( \pi l(R+r) \), total surface \( \pi l(R+r)+\pi R^2+\pi r^2 \). The cone of which a frustum is part has height \( \frac{hR}{R-r} \).
A cone cut by a plane parallel to its base gives a small cone similar to the large one: volumes are in the ratio of the cubes of the heights. If the cut is at half the height, the small cone is 1/8 of the volume and the frustum is 7/8.

Worked Examples

Example 1: a cuboid 100 × 80 × 64 cm is melted into a cube. Volume = 512000 cm³, so the edge is 80 cm and the surface area is 6 × 80² = 38400 cm².
Example 2: a wire's diameter is reduced by 5% with no change in volume. New radius = 0.95 r, so the new length is h/0.9025 and the increase is about 10.8%.
Example 3: water flows at 7 m/s through a pipe of diameter 2 cm into a tank of base radius 40 cm. In 30 minutes the volume is π × 1² × 700 × 1800 cm³. Dividing by the base area π × 40² gives a rise of 787.5 cm.