2.    Find where there is a need of finding surface area of a solid figure.

3.    Find the surface areas of cuboids, cubes, cylinders, cones, spheres and hemispheres using their respective formulae.

4.    Solve some problems related to daily life situations involving surface areas of above solid figures.

Measuring the surface (or boundary) constituting the solid. It is called the surface area of the solid figure.

Just as area is measured in square units, surface area is also measured in square units.

In daily life, there are many situations, where we have to find the surface area. For example, if we are interested in white washing the walls and ceiling of a room, we shall have to find the surface areas of the walls and ceiling.

Fig. 1.1 represents a cuboid. It can be easily seen from the figure that a cuboid has six rectangular regions as its faces.

The two adjacent faces meet in a line segment called an edge of the cuboid. For example, faces ABCD and ABFE meet in the edge AB. There are in all 12 edges of a cuboid.

Points A,B,C,D,E,F,G and H are called the corners or vertices of the cuboid. So, there are 8 corners or vertices of a cuboid.

It can also be seen that at each vertex, three edges meet. One of these three edges is taken as the length, the second as the breadth and third is taken as the height of the cuboid. These are usually denoted by l, b and h respectively.

Thus, we may say that AB (= EF = CD = GH) is the length, AE (=BF = CG = DH) is the breadth and AD (= EH = BC = FG) is the height of the cuboid.

= l×b + b×h + h×l + l×b + b×h + h×l

= 2(lb + bh + hl)

Example 1: Length, breadth and height of cuboid are 4 cm, 3 cm and 12 cm respectively. Find its surface area.

Solution:

Surface area of the cuboid = 2 (lb + bh +hl)

= 2 (4×3+3×12+12×4) cm²

= 2 (12 + 36 + 48) cm²

= 192 cm²

Example 2: Find the surface area of a cuboid of length 6m, breadth 3m and height 2m.

Solution:

Surface area of the cuboid = 2 (lb + bh +hl)

= 2 (6×3+3×2+2×6) m²

= 2 (18 + 6 + 12) m²

= 72 m²

= 2 (a×a + a×a + a×a)

= 6a²

Example 1: Find the surface area of a cube of edge 6 cm.

Solution:

Surface area of a cube = 6a²

= 6(6)²

= 216 cm²

Example 2: Find the surface area of a cube of edge 10 m.

Solution:

Surface area of a cube = 6a²

= 6(10)²

= 600 m²

It can be seen that the two ends (or bases) of a right circular cylinder are congruent circles. In Fig. 3.1, A and B are the centres of these two circles of radii AD (= BC). Further, AB is perpendicular to each of these circles.

Here, AD (or BC) is called the base radius and AB is called the height of the cylinder. It can also be seen that the surface formed by two circular ends are flat and the remaining surface is curved.

So, curved surface area of the cylinder

= area of the rectangle

= 2π r × h

= 2πrh

In case, the cylinder is closed at both the ends, then the total surface area of the cylinder

= 2π

= 2πrh + 2πr²

= 2πr (r + h)

Example 1: The radius and height of a right circular cylinder are 7cm and 10cm respectively. Find its

(i) curved surface area

(ii) total surface area

Solution:

     (i) curved surface area = 2πrh

= 2×22/7 ×7×10

= 440 cm²

    (ii) total surface area = 2πrh + 2πr²

          = (2× 22/7×7×10) + (2× × (10)²)

= 440 cm² + 308 cm²

             = 748 cm²

Example 2: Find the curved surface area, total surface area and volume of a right circular cylinder of radius 7 m and height 14 m.

Solution:

(i) curved surface area = 2πrh

= 2×22/7 ×7×14

= 616 cm²

(ii) total surface area = 2πrh + 2πr²

= (2×22/7×7×14) + (2×22/7× (7)²)

= 616 cm² + 308 cm²

     = 924 cm²

It can be seen that end (or base) of a right circular cone is a circle. In Fig. 4.1, BC is the radius of the base with centre B and AB is the height of the cone and it is perpendicular to the base. Further, A is called the vertex of the cone and AC is called its slant height. from the Pythagoras Theorem, we have

slant height l = sqrt (r²+h²), where r, h and l are respectively the base radius, height and slant height of the cone. You can also observe that surface formed by the base of the cone is flat and the remaining surface of the cone is curved.

Area of this sector = {Arc length of the sector / Circumference of the circle with radius l} × Area of circle with radius l

= [2πr/2πl] × πrl ² = πrl

Clearly, curved surface of the cone = Area of the sector

= πrl

If the area of the base is added to the above, then it becomes the total surface area.

So, total surface area of the cone

= πrl + πr² = πr (l + r)

Let us rotate a semicircle about its diameter. The solid so generated with this rotation is called a sphere. It can also be defined as follows:

The locus of a point which moves in space in such a way that its distance from a fixed point remains the same is called a sphere. The fixed point is called the centre of the sphere and the same distance is called the radius of the sphere (Fig. 5.1). A football, cricket ball, a marble etc. are examples of spheres that we come across in daily life.

HEMISPHERE

If a sphere is cut into two equal parts by a plane passing through its centre, then each part is called a hemisphere.

Now draw a circle of radius r (i.e. the same radius as that of the ball and cover it with a similar string starting from the centre of the circle [See Fig. 5.1(iii)]. Measure the length of the string used to cover the circle. What do you observe? You will observe that length of the string used to cover the hemisphere is twice the length of the string used to cover the circle.

Since the width of the two strings is the same, therefore

surface area of the hemisphere = 2 × area of the circle = 2πr²

So, surface area of the sphere = 2 × 2πr²

= 4πr²

Thus, we have:

Surface area of a sphere = 4πr²

Surface area of a solid hemisphere = 2πr² + πr² = 3πr²

Where r is the radius of the sphere (hemisphere)

The measure of the boundary constituting the solid figure itself is called its surface.

Some solid figures have only flat surfaces, some have only curved surfaces and some have both flat as well as curved surfaces.

Surface area of a cuboid = 2(lb + bh + hl) where l, b and h are respectively length, breadth and height of the cuboid.

Cube is a special cuboid whose each edge is of same length.

Surface area of a cube of edge a is 6a².

Curved surface area of a right circular cylinder = 2πrh; its total surface area = 2πrh+2πr², where r and h are respectively the base radius and height of the cylinder.

Curved surface area of a right circular cone is πrl, its total surface area = πrl + πr², where r, h and l are respectively the base radius, height and slant height of the cone.

Surface area of sphere = 4πr², where r is the radius of the sphere.

Curved surface area of a hemisphere of radius r = 2πr²; its total surface area = 3πr².

(i) Surface area of a cube of side a is 6a².

(ii) Total surface area of a cone is πrl, where r and l are resepctively the base radius and slant height of the cone.

(iii) Surface area of a hemisphere of radius r is 2πr².

(iv) Curved surface area of a right circular cylinder is 2πrh.

2. Find the surface area of a cuboid of length 3m, breadth 2.5 m and height 1.5 m.

   
   

3. Find the surface area of a cube of edge 1.6 cm.

   
   

   
   

4. Find the total surface area of a hollow cylindrical pipe open at the ends if its height is 10 cm, external diameter 10 cm and thickness 12 cm (use π = 3.14).

   
   

5. Slant height and radius of the base of a right circular cone are 25 cm and 7 cm respectively. Find its (i) curved surface area (ii) total surface area.

6. Diameter of a hemispherical toy is 56 cm. Find its (i) curved surface area (ii) total surface area

= l×b + b×h + h×l + l×b + b×h + h×l

= 2(lb + bh + hl)

Therefore,

TSA of cone =πrl+πr^2

= πr(l+r)

where, r is the radius and l is the slant height of the cone.

Surface area of sphere= 4πr^2

Therefore,

TSA of cone =πrs+πr^2

= πr(r+s)

where, r is the radius and s is the slant height of the cone.

Therefore, CSA of hemisphere= 2πr^2. Total surface area of hemisphere is the sum of the curved surface area and the flat circular base.

Therefore, TSA of hemisphere

=2πr^2+πr^2

= 3πr^2

The right-angled triangle is rotated around one of its two short sides to generate cone.

Each solid figure has a unique formula for calculating its surface area, depending on its shape and structure. Understanding these formulas helps in problem-solving and practical applications, such as determining the amount of paint needed to cover an object or the material required to create a container. Mastering these calculations enhances mathematical proficiency and aids in real-life problem-solving across various fields, including engineering, architecture, and manufacturing.

The surface area of solids is a critical factor in many aspects of daily life and industrial applications. From construction and manufacturing to medical advancements and energy efficiency, understanding surface area helps optimize resources, improve designs, and enhance functionality. Mastering surface area calculations is essential for problem-solving in science, engineering, and practical applications, making it a valuable mathematical and real-world concept.