Quantitative Aptitude - SPLessons

Volume – Surface Area Problems

Chapter 26

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

Volume – Surface Area Problems

shape 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.


shape 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.

shape 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.

shape 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

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

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

    ⇒ h = (972 x 5 x 5)cm

    ⇒ h = \(\frac{972 x 5 x 5}{100}\) 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 = \(\frac{1}{3} \pi R^2 * R\) : \(\frac{2}{3} \pi R^3\) : \(\pi R^2 * R\) = 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