# Volume – Surface Area Problems

#### Chapter 26

5 Steps - 3 Clicks

# Volume – Surface Area Problems

### Introduction

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.

### Methods

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 $$cm^{3}$$.

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.

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.

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.

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%.

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.

### Formulae

Cube:

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.

### Samples

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$$\pi R^2$$ = 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