Volume and surface areas Problems deals with parameters like volume, surface area, diagonal of cube, cuboid, cylinder, cone, sphere, hemisphere, hollow cylinder, frustum of a cone, prism, pyramid.

A cube is a three dimensional figure and has six square faces which meet each other at right angles. It has eight vertices and twelve edges.Let each edge of a cube be of length \(a\). Then,

- Volume = \(a^3\) cubic units

Surface area = 6\(a^2\) sq. units

Diagonal = \(\sqrt{3}a\)units

**Example 1**

The diagonal of a cube is \(\sqrt[6]{3}\) cm. Find its volume and surface area.

**Solution**:

- Let the edge of the cube be a.

\(\sqrt{3}\)a = \(\sqrt[6]{3}\) ⇒ a = 6.

So, Volume = \(cm^{3}\) = (6 x 6 x 6) \(cm^{3}\) = 216 \(cm^{3}\).

Surface area = 6\(a^{2}\) = (6 x 6 x 6) \(cm^{2}\) = 216 \(cm^{2}\).

**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\(a^{2}\) = 1734 ⇒ \(a^{2}\) = 289 ⇒ a = 17cm.

∴ Volume = \(a^{3}\) = \((17)^{3} cm^{3}\) = 4913 \(cm^{3}\).

A solid body having six rectangular faces, is called cuboid.

(or) A parallelopiped whose faces are rectangles is called rectangular parallelopiped or 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 = \(\sqrt{l^2 + b^2 + h^2}\) units

**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) \(m^{3}\) = 1568 \(m^{3}\).

Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] \(cm^{2}\) = 868 \(cm^{2}\).

**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

= \(\sqrt{(12)^{2} + 8^{2} + 9^{2}}\) = \(\sqrt{289}\) = 17 m.

A solid geometrical figure with straight parallel sides and a circular or oval cross section is called cylinder. Let radius of base = r and height( or length) = h. Then,

- Volume = \(\pi r^2 h \) cubic units

Curved Surface area = 2\(\pi r h\) sq. units

Total surface area = 2(\(\pi rh + 2\pi r^2\)) = 2\(\pi r(h + r)\)sq. units

**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 = \(\pi r^{2}h\) = (\(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 40) \(cm^{3}\) = 1540 \(cm^{3}\).

Curved surface area = \(2\pi rh\) = (2 x \(\frac{22}{7}\) x \(\frac{7}{2}\) x 40) \(cm^{2}\) = 880 \(cm^{2}\).

Total surface area = \(2\pi rh\) + \(\pi r^{2}\) = \(2\pi r (h + r)\)

= [2 x \(\frac{22}{7}\) x \(\frac{7}{2}\) x (40 + 35)] c = 957 \(cm^{2}\).

**Example 2**

If the capacity of a cylindrical tank is 1848 \(m^{3}\) and the diameter of its base is 14 \(m\), then find the depth of the tank.

**Solution**:

- Let the depth of the tank be \(h\) meters. Then,

\(\pi\) x \((\frac{0.50}{2 \times 100})^{2} \times h\) = \(\frac{22}{1000}\) ⇒ \(h \pm (\frac{22}{1000} \times \frac{100 \times 100}{0.25 \times 0.25} \times \frac{7}{22})\) = 112 m.

A solid (3-dimensional) object with a circular flat base joined to a curved side that ends in an apex point is called cone. Let radius of base = r and height = h. Then,

- Slant height, l = \(\sqrt{h^2 + r^2}\) units

Volume = \(\frac{1}{3}\pi r^2 h\) cubic units

Curved surface area = \(\pi r l\) sq. units

Total surface area = \(\pi r l + \pi r^2\) sq. units

**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, \(r\) = 21 cm and \(h\) = 28 cm.

∴ Slant height, \(l\) = \(\sqrt{r^{2} + h^{2}}\) = \(\sqrt{(21)^{2} + (28)^{2}}\) = \(\sqrt{12225}\) = 35 cm.

Volume = \(\frac{1}{3} \pi r^{2}h\) = (\(\frac{1}{3}\) x \(\frac{22}{7}\) x 21 x 21 x 28) \(\frac{1}{3}\) = 12936 \(cm^{3}\).

Curved surface area = \(\pi rl\) = (\(\frac{22}{7}\) x 21 x 35)\(cm^{2}\) = 2310 \(cm^{2}\).

Total surface area = (\(\pi rl\) + \(\pi r^{2}\)) = \((2310 + \frac{22}{7} x 21 x 21) cm^{2}\) = 3696 \(cm^{2}\).

**Example 2**:

Find the length of canvas 1.25 \(m\) wide required to build a conical tent of radius 7 meters and height 24 meters.

**Solution**:

- Here, \(r\) = 7\(m\) and \(h\) = 24m.

So, \(l\) = \(\sqrt{r^{2} + h^{2}}\) = \(\sqrt{7^{2} + (24)^2}\) = \(\sqrt{625}\) = 25m.

Area of canvas = \(\pi rl\) = (\(\frac{22}{7}\) x 7 x 25)\(m^{2}\) = 550 \(m^{2}\).

∴ Length of canvas = (\(\frac{Area}{Width}\)) = (\(\frac{550}{1.25}\))m = 440 m.

A solid (3-dimensional) object with a circular flat base joined to a curved side that ends in an apex point is called cone. Let radius of the sphere be r. Then,

- Volume = (\(\frac{4}{3} \pi r^{3}\)) cubic units

Surface area = \(\pi r^{2}\) sq. units

**Example 1**:

Find the volume and surface area of a sphere of radius 10.5 cm.

**Solution**:

- Volume = \(\frac{4}{3} \pi r^{3}\) = (\(\frac{4}{3}\) x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\)) \(cm^{3}\) = 4851 \(cm^{3}\).

Surface area = \(4 \pi r^{2}\) = (4 x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\))\(cm^{2}\) = 1386 \(cm^{2}\).

**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 = \(\frac{150}{100}\) R = \(\frac{3R}{2}\).

Original volume = \(\frac{4}{3} \pi R^{3}\), New volume = \(\frac{4}{3} \pi (\frac{3R}{2})^{3}\) = \(\frac{9 \pi R^{3}}{2}\).

Increase % in volume = (\(\frac{19}{6} \pi R^{3}\) x \(\frac{3}{4 \pi R^{3}}\) x 100)% = 237.5%.

Original surface area = \(4 \pi R^{2}\). New surface area = \(4 \pi (\frac{3R}{2})^{2}\) = \(9 \pi R^{2}\)

Increase % in surface area = (\(\frac{5 \pi R^{2}}{4 \pi R^{2}}\) x 100)% = 125%.

In geometry, hemisphere is an exact half of sphere. Let the radius of a hemisphere be \(r\).

- Volume = \(\frac{2}{3} \pi r^3\) cubic units.

Curved surface area = 2\(\pi r^2\) sq. units.

Total surface area = 3\(\pi r^2 \) sq. units.

**Example 1**:

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

**Solution**:

- Volume = \(\frac{2}{3} \pi r^3\) = (\(\frac{2}{3}\) x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\)) \(cm^{3}\) = 2425.5 \(cm^{3}\).

Curved surface area = 2\(\pi r^2\) = (2 x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\)) \(cm^{2}\).

Total surface area = 3\(\pi r^2 \) = (3 x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\))\(cm^{2}\) = 693 \(cm^{2}\).

Total surface area = 3\(\pi r^2 \) = (3 x \(\frac{22}{7}\) x \(\frac{21}{2}\) x \(\frac{21}{2}\))\(cm^{2}\) = 1039.5 \(cm^{2}\).

**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 = (\(\frac{2}{3} \pi\) x 9 x 9 x 9) \(cm^{3}\).

Number of 1 bottle = (\(\pi \times \frac{3}{2} \times \frac{3}{2} \times 4\))\(cm^{3}\) = 9 \(\pi\) \(cm^{3}\).

Number of bottles = (\(\frac{486 \pi}{9 \pi}\)) = 54.

- Let each edge of a cube be of length \(a\). Then,

Volume = \(a^3\) cubic units

Surface area = 6\(a^2\) sq. units

Diagonal = \(\sqrt{3}a\)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 = \(\sqrt{l^2 + b^2 + h^2}\) units

**Cylinder**:

- Let radius of base = r and height( or length) = h. Then,

Volume = \(\pi r^2 h \) cubic units

Curved Surface area = 2\(\pi r h\) sq. units

Total surface area = 2(\(\pi rh + 2\pi r^2\)) = 2\(\pi r(h + r)\)sq. units

**Cone**:

- Let radius of base = r and height = h. Then,

Slant height, l = \(\sqrt{h^2 + r^2}\) units

Volume = \(\frac{1}{3}\pi r^2 h\) cubic units

Curved surface area = \(\pi r l\) sq. units

Total surface area = \(\pi r l + \pi r^2\) sq. units

**Sphere**:

- Let radius of the sphere be r. Then,

Volume = (\(\frac{4}{3} \pi r^3\)) cubic units

Surface area = \(\pi r^2\) sq. units

**Hemisphere**:

- Let the radius of a hemisphere be \(r\).

Volume = \(\frac{2}{3} \pi r^3\) cubic units.

Curved surface area = 2\(\pi r^2\) sq. units.

Total surface area = 3\(\pi r^2 \) 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 = \(\pi R^2 h – \pi r^2 h\) = \(\pi h(R^2 – r^2)\)

Curved surface area = 2\(\pi (R + r) h\)

**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 = \(\pi l(R + r)\) sq. units and \( l^2 \) = \( h^2 + (R – r)^2\)

volume of frustum of a cone = \(\frac{\pi h}{3}(R^2 + r^2 + Rr)\) cubic units

Total surface area of frustum of right circular cone = Area of base + area of top + lateral surface area

= \(\pi R^2 + \pi r^2 + \pi l(R + r)\)

= \(\pi [R^2 + r^2 + l(R + r)]\) 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 = \(\frac{1}{3}(area of the base) * height \)cu. units

Surface area of aright pyramid = \(\frac{1}{2}\) (perimeter of the base) x slant height sq. units

Total surface area of a right pyramid = surface area + area of the base sq. units.

- 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\(\pi R^2\) = 2\(\pi 3^2\) = 18\(\pi\)

Similarly, surface area of cone = \(\pi r l\)

= \(\pi r\sqrt{h^2 + r^2}\)

= \(\pi * 3\sqrt{3^2 + 4^2}\)

= 15\(\pi\)

Hence, Surface area of toy = surface area of hemispherical base + surface area of cone

= 18\(\pi\) + 15\(\pi\) = 33\(\pi\) = 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, \(r_{2}\) = 5 cm, \(r_{1}\) = 15 cm, \(h\) = 24 cm

Therefore, volume of bucket (i.e. capacity) = \(\frac{\pi h}{3}[(r_{1})^2 + (r_{2})^2 + r_{1}r_{2}]\)

= \(\frac{22}{7 * 3}[(15)^2 + (5)^2 + 15 * 5](cm)^2\)

= \(\frac{22 * 8}{7}[225 + 25 + 75](cm)^3\)

= \(\frac{22 * 8 * 325}{7}(cm)^3\)

= 8171.43\((cm)^3\)

Therefore, the capacity of bucket = 8171.43\((cm)^3\)

**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 = (\(\frac{4}{3} \pi r^3\)) cubic units

â‡’ Volume of sphere = (\(\frac{4}{3} \pi 9^3\)) cubic units

â‡’ Volume of sphere = 972 \(\pi (cm)^3\)

Volume of wire = \(\pi\) x 0..2 x 0.2 x 0.2 x h\((cm)^3\)

Therefore,

972 \(\pi (cm)^3\) = \(\pi * \frac{2}{10} * \frac{2}{10} * h \)

â‡’ h = (972 x 5 x 5)cm

â‡’ h = \(\frac{972 x 5 x 5}{100}\) m

â‡’ 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 = \(\frac{1}{3} \pi R^2 * R\) : \(\frac{2}{3} \pi R^3\) : \(\pi R^2 * R\) = 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 = \(\frac{2}{3} \pi r^3\) cubic units = \(\frac{2}{3} * \frac{22}{7} * \frac{21}{2} * \frac{21}{2} * \frac{21}{2}\)\((cm)^3\) = 2425.5 \((cm)^3\)

Curved surface area = 2\(\pi r^2\) sq. units = 2 * \(\frac{22}{7} * \frac{21}{2} * \frac{21}{2} \) sq. units = 693 sq. units

Total surface area = 3\(\pi r^2\) sq. units = 3 * \(\frac{22}{7} * \frac{21}{2} * \frac{21}{2} \) sq. units = 1039.5 sq. units