CBSE Class 10 Mathematics Surface Areas and Volumes Notes Set 12

Formulas for Standard Solids

For a cone, l = √(r² + h²). The diagonal of a cuboid is √(l² + b² + h²) and of a cube a√3.

Frustum of a Cone

When a cone is cut by a plane parallel to its base, the lower portion is a frustum. With radii R and r (R > r), height h and slant height l:
Slant height l = √(h² + (R − r)²).
Volume = (π h/3)(R² + Rr + r²).
Curved surface area = π(R + r)l.
Total surface area = π(R + r)l + πR² + πr².
The small cone cut off is similar to the whole cone, so the ratio of the volumes is the cube of the ratio of the heights (or radii).

Combination and Conversion of Solids

Combination: when two solids are joined (for example a cone on a hemisphere or a cylinder with hemispherical ends), the total volume is the sum of the volumes, but the total surface area is the sum of the visible curved and flat surfaces only; the joined faces are not counted.
Conversion: when a solid is melted and recast into another shape, the volume is unchanged. If n small solids are formed from a large solid, n × (volume of each) = volume of the large one.
For a vessel in which a solid is dropped, the rise in water level = volume of the solid ÷ area of the base of the vessel.
Flowing water: volume flowed = cross-sectional area × speed × time.

Percentage Change Facts

1. If all linear dimensions are multiplied by k, areas are multiplied by k² and volumes by k³.
2. If the radius of a cylinder is multiplied by a and its height by b, the volume is multiplied by a²b.
3. If a wire's radius becomes 1/n of its original, then for the same volume its length becomes n² times.

Worked Examples

Example 1. Find the volume of a cone of radius 7 cm and height 24 cm (π = 22/7).
Solution: V = (1/3)(22/7)(49)(24) = 1232 cm³. The slant height is 25 cm.

Example 2. A solid metallic sphere of radius 3 cm is melted and recast into a cylinder of radius 1 cm. Find the height of the cylinder.
Solution: (4/3)π(27) = π(1)²h, so h = 36 cm.

Example 3. A toy is a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height is 15.5 cm. Find its volume (π = 22/7).
Solution: The cone height is 15.5 − 3.5 = 12 cm. Volume = (1/3)(22/7)(12.25)(12) + (2/3)(22/7)(42.875) = 154 + 89.83 = 243.83 cm³.

Example 4. The radii of the ends of a bucket 16 cm deep are 20 cm and 8 cm. Find its capacity (π = 3.14).
Solution: V = (3.14 × 16/3)(400 + 160 + 64) = 3.14 × 16 × 208 = 10449.92 cm³ ≈ 10.45 litres.

Example 5. Water flows through a pipe of internal diameter 2 cm at 0.7 m per second into a cylindrical tank of base radius 40 cm. By how much does the water level rise in 30 minutes?
Solution: Length of water in one second = 70 cm, so the volume in 1800 s is π(1²)(70)(1800) = 126000π cm³. Rise = 126000π/(π × 40²) = 78.75 cm.