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Surface Area is the sum of the areas of all faces of a solid. A three-dimensional object is considered a solid. A pyramid is a solid with a polygonal base and triangular faces that meet at a point called the apex (or vertex). Here, we will focus on finding the surface area of pyramids....Read MoreRead Less

The surface area of a pyramid calculates the total area occupied by all of a pyramid’s faces. Triangular pyramids, square pyramids, hexagonal pyramids and so on are examples of different types of pyramids based on the shape of their bases.

A pyramid’s surface area is calculated by adding the areas of all its faces. The altitude or height of the pyramid is the perpendicular distance from the apex to the center of the base. The slant height is the length of the perpendicular drawn from the apex to the edge of the base of the triangular faces (side faces).

The area of the base and lateral triangles can be found separately using the respective formulas for finding the area of polygons. Then, the results are added to get the surface area of the pyramid. The sum of the areas of the pyramid’s side faces (triangles) alone, excluding the area of the base, is the Lateral Surface Area (LSA).

Surface Area = Area of base + Area of lateral faces

Lateral Surface Area = Area of lateral faces

Alternatively, we can also use the following formulas.

Consider a regular pyramid with a base perimeter of ‘P’, a base area of ‘B’, and a slant height of ‘H’. Then,

- The Lateral Surface Area \(=~(\frac{1}{2})~PH\)
- The Surface Area \(=~(\frac{1}{2})~PH+B\)

**1) What is the surface area of a pyramid with a square base whose side length is 3 cm and has a slant height of 4cm?**

**Solution: **We know that a square pyramid has a square base and 4 triangular faces. All the triangles are isosceles and congruent to each other.

So, the surface area of a square pyramid is the sum of the area of four of its triangular side faces and the base area.

Area of the base = \( (\text{side length})^2\)

= \( 3^2\)

= 9 sq.cm

Area of a triangle = \( \frac{1}{2}\times~\text{base}\times~\text{height}\)

Area of the triangular face = \( \frac{1}{2}\times~3\times~4\)

= \( ~3\times~2\)

= 6 sq.cm

Area of four triangular faces = 4 \( \times\) 6

= 24 sq.cm

Therefore, surface area = Area of the base + Area of four triangular faces

= 9 + 24

= 33 sq.cm

**2) The image is of a toy whose shape is that of a pyramid mounted on the top of a square prism as shown. Now, the pyramid alone is covered with wrapping paper and the rest is left uncovered. What will be the area of the wrapping material needed to cover the top of the toy?**

**Solution:**

The base length of the pyramid = 8 inches.

The slant height of the pyramid = 6 inches.

In the figure, the base of the pyramid will not be covered with wrapping material, hence we need to find the lateral surface area of the pyramid.

Lateral surface area = surface area of the 4 triangular faces of the pyramid

= \( 4\times~(\frac{1}{2}\times~8\times~6)\)

= 96 sq.in

The area of the wrapping material needed to cover the top of the toy is 96 sq. inches.

**3) Find the surface area of a pyramid with height \( 3\sqrt{3}\)**** cm and a square base with sides of 6 cm in length.**

**Solution:**

As we know, surface area = area of square + area of 4 triangles

We need to find the slant height AC from triangle ABC.

\(BC=\frac{6}{2}=3\) cm

\(AC^2=3^2+(3\sqrt{3})^2\) (Using pythagoras theorem)

\(AC^2=36\)

\(AC=\sqrt{36}=6\)

Therefore, surface area = area of square + area of 4 triangles

= \( 6\times~6+4~ \left ( \frac{1}{2}\times6\times6\right )\)

= 108 sq.cm

**4) Find the surface area of a pyramid whose base edge is 14 cm long and the slant height is 9 cm.**

**Solution:**

To understand the question better, let’s draw a rough diagram.

Base edge of the pyramid = 14 cm.

Lateral edge of the pyramid = 9 cm.

Surface Area of the pyramid \(= \text{area of base}+4 \times\left (\frac{1}{2}\times\text{base}\times\text{slant height} \right )\)

By Pythagoras’ Theorem,

Slant height, \(h=\sqrt{9^2-7^2}\)

\(=\sqrt{81-49}\)

\(=\sqrt{32}\)

\(=4\sqrt{2}\) cm

Surface area of the pyramid \(=(14\times14)+4\times\left ( \frac{1}{2}\times14\times4\sqrt{2} \right )\)

\(=196+4\times~7\times~4\sqrt{2}\)

\(=354.39\) sq.cm

Frequently Asked Questions

A pyramid’s surface area is calculated by adding the areas of all its faces (both the base and the side faces).

Surface Area \(=\left ( \frac{1}{2} \right )PH+B\) can be used to calculate the surface area of a pyramid with a base perimeter of ‘P’, a base area of ‘B’ and a slant height of ‘H’.

A pyramid’s lateral surface area is the sum of the areas of all its side faces (which are triangles).

Lateral Surface Area \(=\left ( \frac{1}{2} \right )PH\) can be used to calculate the lateral surface area of a pyramid, where ‘P’ is the base perimeter and ‘H’ is the slant height.

If the altitude is known, the surface area of a pyramid can be calculated. Consider a pyramid with a regular polygon at its base with a side length of ‘a’, a slant height of ‘H’ and an altitude of ‘h’. If only ‘a’ and ‘h’ are provided to us, we can calculate the slant height. Let’s understand this using the steps below.

- Step 1: Using Pythagoras’ Theorem, \(H^2=\left ( \frac{a}{2} \right )^2+h^2\) we can find ‘H’.

- Step 2: Determine the perimeter ‘P’ of the base.
- Step 3: Calculate the base area ‘B’.
- Step 4: Using the formula\(\left ( \frac{1}{2} \right )PH+B\), calculate the pyramid’s total surface area.

Volume is basically defined as the amount of space occupied by a three dimensional object, whereas surface area is the sum of areas of all its faces.

the same as that of the improper fraction. Thus, the required fraction is obtained.

Pyramids are three dimensional objects that have only 1 base, whereas prisms are solids that have 2 bases (top and bottom).