**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 √15^{2} + 20^{2} = √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 = πr^{2}h = 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 = 1/3πr^{2}h = 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

πR^{2}h = 4/3 πr^{3} ⇒ 6 x 6 x h = 4/3 x 3^{3} ⇒ h = 1 inch.

Must Read 3D Geometry Articles

**Example 8:** If the length of a rectangle is increased by 10% and the breadth decreased by 20%, a square of area 484 m^{2} 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 m^{2}. So we have

1.1l x 0.8b = 484 ⇒ lb = 550 m^{2}. ∴ Area of rectangle = 550 m^{2}.

**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 =

4/3πr^{3} = 4/3π(4.5)^{3} = 121.5πcm^{3}

Wire is of cylindrical shape

∴πR^{2}H = 121.5π ⇒ (1.5) ^{2} H = 121.5

H = 54 cm

**Example 11:** Area of square pyramid is 36 m^{2}. The volume of the pyramid 288mk^{3}., what is the height of square pyramid?

**Sol:** Since area of the base = 36 m^{2},

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 = 2πr^{2}

50 = 2πr^{2}

R^{2} = 50/2π (π = 3.14) = 7.96

r = 2.82

Radius of the hemisphere is 2.82 inches

Volume of a hemisphere =2 πr^{3}/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 cm^{3}

**Example 13:** The volume of cubicle box is 343m^{3}. Find the surface area of a cubicle box.

**Sol:** Volume= a^{3} = 343m^{3}

A=7m

Surface area of cube= 6a^{2}= 6 x 7 x 7=294m^{2}

**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) a^{2} = √3 / 4 x 12 x 12 = 36√3cm^{2}

LSA = Perimeter * Altitude = 36 x 17 = 612 cm^{2}

TSA = LSA + 2 x Area of XYZ = 612+72√3 cm^{2}

Volume = Area x altitude = 36√3 x 17 = 612√3 cm^{3}