3D Geometry: Mensuration - Solved Examples

Example 1. The diagonals of a rhombus are 40 cm and 30 cm. What is the length of each side of the rhombus?

Sol Diagonals are 40 cm and 30 cm. ∴ Half of the diagonals are 20 cm and 15 cm.
∴ Side of rhombus is √15² + 20² = √225 + 400 = √625 = 25cm.

Example 2. A rectangle 7 cm x 5 cm is rotated about its smaller edge as axis. Find the curved surface area and volume of solid generated.

Sol When the rectangle is rotated about its smaller side as axis, then the radius will be the larger side and the height will be the smaller side.
Curved Surface Area = 2πrh = 2 x 22/7 x 7 x 5 = 220 sq. cm.
Volume of solid = πr²h = 22/7 x 7 x 7 x 5 = 770 cu. cm.

Example 3: A semi-circle has a diameter of 49 m. Its perimeter (in m) equals

Sol: Here diameter = 49 m, Radius = Diameter/2 = 49/2m
Perimeter = D + πr = 49 + (22/7) x (49/2) =126 m.

Example 4: The perpendicular sides of the base of a right triangular metallic prism are 4 cm and 7 cm. It weighs 868 g. Find its height if density of metal is 12.4.g/cc.

Sol: Volume = Weight / Density = 868 / 12.4 = 70cc. Volume = 1/3 x Base area x height.
70 = 1/3 x 1/2 x 4 x 7 x height ⇒ height = 15 cm.

Example 5: A regular hexagonal prism has its perimeter of base as 600 cm and height 200 cm. Find weight of petrol that it can hold if density is 0.8 g/cc. (Take √3 = 1.73)

Sol: Side of hexagon = (Perimeter) / (Number of Sides) = 600/6 = 100 cm.
Area of regular hexagon = (3√3) / 2 x 100 x 100 = 25,950 sq.cm.
Volume = Base Area × height = 25950 x 200 = 5,190,000 cu.cm
Weight of petrol = Volume x Density = 5,190,000 cc × 0.8 g/cc = 4,152,000 gm. = 4,152 kg.

Example 6: Find the volume of the largest right circular cone that can be cut out of a cube of edge 42 cm.

Sol: The base of the cone will be circle inscribed in a face of the cube and its height will be equal to an edge of the cube. Radius of cone = 21 cm. Height = 42 cm.
Volume of cone = 1/3πr²h = 1/3x 22/7 x 21 x 21 x 42 = 19,404 cu. cm.

Example 7: An iron ball of diameter 6 inch is dropped into a cylindrical vessel of diameter 1ft filled with water. Find the rise in water level.

Sol: Radius of vessel = 12 inch/2 = 6 inch. Volume of water that has risen = Volume of sphere
πR²h = 4/3 πr³ ⇒ 6 x 6 x h = 4/3 x 3³ ⇒ h = 1 inch.

Example 8: If the length of a rectangle is increased by 10% and the breadth decreased by 20%, a square of area 484 m² is obtained. What is the area of the rectangle in square meters?

Sol: Let the length of the rectangle is 'l' and its breadth is 'b'. Now the length is increased by 10% and its breadth is decreased by 20% and the resulting area is 484 m². So we have
1.1l x 0.8b = 484 ⇒ lb = 550 m². ∴ Area of rectangle = 550 m².

Example 9: The radius of a circle is decreased by 20%. What will be the percentage decrease in its surface area?

Sol: New radius is 0.8 times the previous radius. Therefore the new surface area will be 0.8 x 0.8 = 0.64 times the previous surface area. So, the percentage decrease is 36%.

Example 10: The diameter of a sphere is 9 cm. It is melted and drawn into a wire of diameter 3 cm. Find the length of the wire.

Sol: Vol. of sphere =
4/3πr³ = 4/3π(4.5)³ = 121.5πcm³
Wire is of cylindrical shape
∴πR²H = 121.5π ⇒ (1.5) ² H = 121.5
H = 54 cm
 

Example 11: Area of square pyramid is 36 m². The volume of the pyramid 288mk³., what is the height of square pyramid?

Sol: Since area of the base = 36 m²,
Volume of the pyramid = (1/3) * (base area) * (height)
288 = (1/3) * (36) * (H) ;
H = 24m

Example 12: Find the volume of the hemisphere whose surface area is 50 sq. inches.

Sol: SA = 50 sq. inches
Surface area of the hemisphere = 2πr²
50 = 2πr²
R² = 50/2π (π = 3.14) = 7.96
r = 2.82
Radius of the hemisphere is 2.82 inches
Volume of a hemisphere =2 πr³/3 cubic units
= 2/3 x 3.14 x (2.82)³
= 2/3 x 3.14 x 22.43
= 2/3 x 70.43
= 46.96
The volume of the hemisphere is 46.96 cm³

Example 13: The volume of cubicle box is 343m³. Find the surface area of a cubicle box.

Sol: Volume= a³ = 343m³
A=7m
Surface area of cube= 6a²= 6 x 7 x 7=294m²

Example 14: A prism with equilateral triangular base (side= 12 cm) and Altitude (H)= 17cm. Find the Lateral surface Area, Total surface area and Volume?

Sol: So, the triangle XYZ is an equilateral triangle.
Perimeter (P) =12+12+12 = 36cm
The area of XYZ = (√3 / 4) a² = √3 / 4 x 12 x 12 = 36√3cm²
LSA = Perimeter * Altitude = 36 x 17 = 612 cm²
TSA = LSA + 2 x Area of XYZ = 612+72√3 cm²
Volume = Area x altitude = 36√3 x 17 = 612√3 cm³