Example 1 The diagonal of a cube is [latex]\sqrt[6]{3}[/latex] cm. Find its volume and surface area.
Solution:

Let the edge of the cube be a.
[latex]\sqrt{3}[/latex]a = [latex]\sqrt[6]{3}[/latex] ⇒ a = 6.
So, Volume = [latex]cm^{3}[/latex] = (6 x 6 x 6) [latex]cm^{3}[/latex] = 216 [latex]cm^{3}[/latex].
Surface area = 6[latex]a^{2}[/latex] = (6 x 6 x 6) [latex]cm^{2}[/latex] = 216 [latex]cm^{2}[/latex].

Example 2 The surface area of a cube is 1734 sq. cm. Find its volume.
Solution:

Let the edge of the cube be a. Then,
6[latex]a^{2}[/latex] = 1734 ⇒ [latex]a^{2}[/latex] = 289 ⇒ a = 17cm.
∴ Volume = [latex]a^{3}[/latex] = [latex](17)^{3} cm^{3}[/latex] = 4913 [latex]cm^{3}[/latex].

Example 1 Find the volume and surface area of a cubiod 16m long, 14 m broad and 7m high.
Solution:

Volume = (16 x 14 x 7) [latex]m^{3}[/latex] = 1568 [latex]m^{3}[/latex].
Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] [latex]cm^{2}[/latex] = 868 [latex]cm^{2}[/latex].

Example 2 Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad and 9 m high.
Solution:

Length of logest pole = length of the diagonal of the room
= [latex]\sqrt{(12)^{2} + 8^{2} + 9^{2}}[/latex] = [latex]\sqrt{289}[/latex] = 17 m.

Example 1 Find the volume, curved surface area and the total surface area of a cylinder with diameter of base 7 cm and height 40cm.
Solution:

Volume = [latex]\pi r^{2}h[/latex] = ([latex]\frac{22}{7}[/latex] x [latex]\frac{7}{2}[/latex] x [latex]\frac{7}{2}[/latex] x 40) [latex]cm^{3}[/latex] = 1540 [latex]cm^{3}[/latex].
Curved surface area = [latex]2\pi rh[/latex] = (2 x [latex]\frac{22}{7}[/latex] x [latex]\frac{7}{2}[/latex] x 40) [latex]cm^{2}[/latex] = 880 [latex]cm^{2}[/latex].
Total surface area = [latex]2\pi rh[/latex] + [latex]\pi r^{2}[/latex] = [latex]2\pi r (h + r)[/latex]
= [2 x [latex]\frac{22}{7}[/latex] x [latex]\frac{7}{2}[/latex] x (40 + 35)] c = 957 [latex]cm^{2}[/latex].

Example 2 If the capacity of a cylindrical tank is 1848 [latex]m^{3}[/latex] and the diameter of its base is 14 [latex]m[/latex], then find the depth of the tank.
Solution:

Let the depth of the tank be [latex]h[/latex] meters. Then,
[latex]\pi[/latex] x [latex](\frac{0.50}{2 \times 100})^{2} \times h[/latex] = [latex]\frac{22}{1000}[/latex] ⇒ [latex]h \pm (\frac{22}{1000} \times \frac{100 \times 100}{0.25 \times 0.25} \times \frac{7}{22})[/latex] = 112 m.

Example 1: Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.
Solution:

Here, [latex]r[/latex] = 21 cm and [latex]h[/latex] = 28 cm.
∴ Slant height, [latex]l[/latex] = [latex]\sqrt{r^{2} + h^{2}}[/latex] = [latex]\sqrt{(21)^{2} + (28)^{2}}[/latex] = [latex]\sqrt{12225}[/latex] = 35 cm.
Volume = [latex]\frac{1}{3} \pi r^{2}h[/latex] = ([latex]\frac{1}{3}[/latex] x [latex]\frac{22}{7}[/latex] x 21 x 21 x 28) [latex]\frac{1}{3}[/latex] = 12936 [latex]cm^{3}[/latex].
Curved surface area = [latex]\pi rl[/latex] = ([latex]\frac{22}{7}[/latex] x 21 x 35)[latex]cm^{2}[/latex] = 2310 [latex]cm^{2}[/latex].
Total surface area = ([latex]\pi rl[/latex] + [latex]\pi r^{2}[/latex]) = [latex](2310 + \frac{22}{7} x 21 x 21) cm^{2}[/latex] = 3696 [latex]cm^{2}[/latex].

Example 2: Find the length of canvas 1.25 [latex]m[/latex] wide required to build a conical tent of radius 7 meters and height 24 meters.
Solution:

Here, [latex]r[/latex] = 7[latex]m[/latex] and [latex]h[/latex] = 24m.
So, [latex]l[/latex] = [latex]\sqrt{r^{2} + h^{2}}[/latex] = [latex]\sqrt{7^{2} + (24)^2}[/latex] = [latex]\sqrt{625}[/latex] = 25m.
Area of canvas = [latex]\pi rl[/latex] = ([latex]\frac{22}{7}[/latex] x 7 x 25)[latex]m^{2}[/latex] = 550 [latex]m^{2}[/latex].
∴ Length of canvas = ([latex]\frac{Area}{Width}[/latex]) = ([latex]\frac{550}{1.25}[/latex])m = 440 m.

Example 1: Find the volume and surface area of a sphere of radius 10.5 cm.
Solution:

Volume = [latex]\frac{4}{3} \pi r^{3}[/latex] = ([latex]\frac{4}{3}[/latex] x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex]) [latex]cm^{3}[/latex] = 4851 [latex]cm^{3}[/latex].
Surface area = [latex]4 \pi r^{2}[/latex] = (4 x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex])[latex]cm^{2}[/latex] = 1386 [latex]cm^{2}[/latex].

Example 2: If the radius of a sphere is increased by 50 %, find the increase percent in volume and the increase percent in the surface area.
Solution:

Let original radius = R. Then, new radius = [latex]\frac{150}{100}[/latex] R = [latex]\frac{3R}{2}[/latex].
Original volume = [latex]\frac{4}{3} \pi R^{3}[/latex], New volume = [latex]\frac{4}{3} \pi (\frac{3R}{2})^{3}[/latex] = [latex]\frac{9 \pi R^{3}}{2}[/latex].
Increase % in volume = ([latex]\frac{19}{6} \pi R^{3}[/latex] x [latex]\frac{3}{4 \pi R^{3}}[/latex] x 100)% = 237.5%.
Original surface area = [latex]4 \pi R^{2}[/latex]. New surface area = [latex]4 \pi (\frac{3R}{2})^{2}[/latex] = [latex]9 \pi R^{2}[/latex]
Increase % in surface area = ([latex]\frac{5 \pi R^{2}}{4 \pi R^{2}}[/latex] x 100)% = 125%.

Example 1: Find the volume, curved surface area and the total surface area of a hemisphere of radius 10.5 cm.
Solution:

Volume = [latex]\frac{2}{3} \pi r^3[/latex] = ([latex]\frac{2}{3}[/latex] x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex]) [latex]cm^{3}[/latex] = 2425.5 [latex]cm^{3}[/latex].
Curved surface area = 2[latex]\pi r^2[/latex] = (2 x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex]) [latex]cm^{2}[/latex].
Total surface area = 3[latex]\pi r^2 [/latex] = (3 x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex])[latex]cm^{2}[/latex] = 693 [latex]cm^{2}[/latex].
Total surface area = 3[latex]\pi r^2 [/latex] = (3 x [latex]\frac{22}{7}[/latex] x [latex]\frac{21}{2}[/latex] x [latex]\frac{21}{2}[/latex])[latex]cm^{2}[/latex] = 1039.5 [latex]cm^{2}[/latex].

Example 2: A hemispherical bowl of inrenal radius 9 cm contains a liqud. That liquid is to be filled into cylidrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?
Solution:

Volume of bowl = ([latex]\frac{2}{3} \pi[/latex] x 9 x 9 x 9) [latex]cm^{3}[/latex].
Number of 1 bottle = ([latex]\pi \times \frac{3}{2} \times \frac{3}{2} \times 4[/latex])[latex]cm^{3}[/latex] = 9 [latex]\pi[/latex] [latex]cm^{3}[/latex].
Number of bottles = ([latex]\frac{486 \pi}{9 \pi}[/latex]) = 54.

Cube:

Let each edge of a cube be of length [latex]a[/latex]. Then, Volume = [latex]a^3[/latex] cubic units Surface area = 6[latex]a^2[/latex] sq. units Diagonal = [latex]\sqrt{3}a[/latex]units

Cuboid:

Let l - length, b - breadth, h - height. Then, Volume = (l x b x h) cubic units Surface area = 2(lb + bh + lh) sq. units Diagonal = [latex]\sqrt{l^2 + b^2 + h^2}[/latex] units

Cylinder:

Let radius of base = r and height( or length) = h. Then, Volume = [latex]\pi r^2 h [/latex] cubic units Curved Surface area = 2[latex]\pi r h[/latex] sq. units Total surface area = 2([latex]\pi rh + 2\pi r^2[/latex]) = 2[latex]\pi r(h + r)[/latex]sq. units

Cone:

Let radius of base = r and height = h. Then, Slant height, l = [latex]\sqrt{h^2 + r^2}[/latex] units Volume = [latex]\frac{1}{3}\pi r^2 h[/latex] cubic units Curved surface area = [latex]\pi r l[/latex] sq. units Total surface area = [latex]\pi r l + \pi r^2[/latex] sq. units

Sphere:

Let radius of the sphere be r. Then, Volume = ([latex]\frac{4}{3} \pi r^3[/latex]) cubic units Surface area = [latex]\pi r^2[/latex] sq. units

Hemisphere:

Let the radius of a hemisphere be [latex]r[/latex]. Volume = [latex]\frac{2}{3} \pi r^3[/latex] cubic units. Curved surface area = 2[latex]\pi r^2[/latex] sq. units. Total surface area = 3[latex]\pi r^2 [/latex] sq. units.

Hollow cylinder:

A solid bounded by two co-axial cylinders of the same height, is called a hollow cylinder. Let height = h, external radii = R and internal radii = r Volume = [latex]\pi R^2 h - \pi r^2 h[/latex] = [latex]\pi h(R^2 - r^2)[/latex] Curved surface area = 2[latex]\pi (R + r) h[/latex]

Frustum of a cone:

If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called frustum of the cone. Let R, r be radii of base and top of frustum of a cone, h is the height of the frustum Lateral surface area of right circular one = [latex]\pi l(R + r)[/latex] sq. units and [latex] l^2 [/latex] = [latex] h^2 + (R - r)^2[/latex] volume of frustum of a cone = [latex]\frac{\pi h}{3}(R^2 + r^2 + Rr)[/latex] cubic units Total surface area of frustum of right circular cone = Area of base + area of top + lateral surface area = [latex]\pi R^2 + \pi r^2 + \pi l(R + r)[/latex] = [latex]\pi [R^2 + r^2 + l(R + r)][/latex] sq. units

Prism:

A prism is a solid whose side faces are parallelograms and whose base are equal and parallel rectilinear figures. A prism is called a right prism, if the axis is perpendicular to the base. Volume of right prism = (Area of the base * height)cu. units Lateral surface area of a right prism = (perimeter of the base * height)sq. units Total surface area of a right prism = lateral area + 2(area of one base)sq. units

Pyramid:

It is defined as a polyhedron whose one face is a polygon and the other faces are triangles having a common vertex. Volume of a right pyramid = [latex]\frac{1}{3}(area of the base) * height [/latex]cu. units Surface area of aright pyramid = [latex]\frac{1}{2}[/latex] (perimeter of the base) x slant height sq. units Total surface area of a right pyramid = surface area + area of the base sq. units.

1. A toy is in the form of a cone mounted on a hemisphere. The diameter of the base of the cone 6 cm and its height 4 cm. Calculate the surface area of the toy?
Solution:

Given that,
Radius of the base of cone, r = 3 cm
Radius of the base of hemisphere, R = 3 cm
Therefore, surface area of hemispherical base = 2[latex]\pi R^2[/latex] = 2[latex]\pi 3^2[/latex] = 18[latex]\pi[/latex]
Similarly, surface area of cone = [latex]\pi r l[/latex]
= [latex]\pi r\sqrt{h^2 + r^2}[/latex]
= [latex]\pi * 3\sqrt{3^2 + 4^2}[/latex]
= 15[latex]\pi[/latex]
Hence, Surface area of toy = surface area of hemispherical base + surface area of cone
= 18[latex]\pi[/latex] + 15[latex]\pi[/latex] = 33[latex]\pi[/latex] = 33 x 3.14 = 103.62 sq.cm

2. The radii of the ends of a bucket of height 24 cm are 15 cm and and 5 cm. Find the capacity?
Solution:

Given that,
As the bucket is in the form of a frustum of a cone.
Here, [latex]r_{2}[/latex] = 5 cm, [latex]r_{1}[/latex] = 15 cm, [latex]h[/latex] = 24 cm
Therefore, volume of bucket (i.e. capacity) = [latex]\frac{\pi h}{3}[(r_{1})^2 + (r_{2})^2 + r_{1}r_{2}][/latex]
= [latex]\frac{22}{7 * 3}[(15)^2 + (5)^2 + 15 * 5](cm)^2[/latex]
= [latex]\frac{22 * 8}{7}[225 + 25 + 75](cm)^3[/latex]
= [latex]\frac{22 * 8 * 325}{7}(cm)^3[/latex]
= 8171.43[latex](cm)^3[/latex]
Therefore, the capacity of bucket = 8171.43[latex](cm)^3[/latex]

3. A copper sphere of daimeter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire?
Solution:

Given that,
Diameter of the copper wire = 18 cm
So, radius(r) = 9 cm
Now, consider
Volume of sphere = ([latex]\frac{4}{3} \pi r^3[/latex]) cubic units
⇒ Volume of sphere = ([latex]\frac{4}{3} \pi 9^3[/latex]) cubic units
⇒ Volume of sphere = 972 [latex]\pi (cm)^3[/latex]
Volume of wire = [latex]\pi[/latex] x 0..2 x 0.2 x 0.2 x h[latex](cm)^3[/latex]
Therefore,
972 [latex]\pi (cm)^3[/latex] = [latex]\pi * \frac{2}{10} * \frac{2}{10} * h [/latex]
⇒ h = (972 x 5 x 5)cm
⇒ h = [latex]\frac{972 x 5 x 5}{100}[/latex] m
⇒ h = 243 m
Therefore, length of wire = 243 m

4. A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Find the ratio of their volumes?
Solution:

Let R be the radius of each.
Height of hemisphere = its radius = R
Height of each = R
Ratio of volumes = [latex]\frac{1}{3} \pi R^2 * R[/latex] : [latex]\frac{2}{3} \pi R^3[/latex] : [latex]\pi R^2 * R[/latex] = 1 : 2 : 3.

5. Find the volume, curved surface area and the total surface area of a hemisphere of radius 10.5 cm?
Solution:

Volume = [latex]\frac{2}{3} \pi r^3[/latex] cubic units = [latex]\frac{2}{3} * \frac{22}{7} * \frac{21}{2} * \frac{21}{2} * \frac{21}{2}[/latex][latex](cm)^3[/latex] = 2425.5 [latex](cm)^3[/latex]
Curved surface area = 2[latex]\pi r^2[/latex] sq. units = 2 * [latex]\frac{22}{7} * \frac{21}{2} * \frac{21}{2} [/latex] sq. units = 693 sq. units
Total surface area = 3[latex]\pi r^2[/latex] sq. units = 3 * [latex]\frac{22}{7} * \frac{21}{2} * \frac{21}{2} [/latex] sq. units = 1039.5 sq. units