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The surface area of a triangular pyramid calculator is a free online tool that can be used to calculate the surface area of the pyramid as well as its base length, base height and slant height. Let us familiarize ourselves with the calculator....Read MoreRead Less

The steps that follow helps in using the surface area of a triangular pyramid calculator:

**Step 1:** Enter any two known measures into the respective input boxes and the unknown measures will be calculated.

**Step 2:** Select the appropriate units for the input and output.

**Step 3: **Click on the ‘Solve’ button to obtain the result.

**Step 4: **Click on the ‘Show Steps’ button to view the stepwise solution applied to find the missing measures.

**Step 5: **Click on the button to enter new inputs and start again.

**Step 6: **Click on the ‘Example’ button to play with different input values.

**Step 7: **Click on the ‘Explore’ button to visualize a triangular pyramid.

**Step 8**: When on the ‘Explore’ page, click the ‘Calculate’ button, if you want to go back to the calculator.

A triangular pyramid is a three dimensional shape that has a triangular base and three lateral faces that are also triangles. These triangles meet at a vertex called the apex. The perpendicular dropped from the apex to the center of the base is the height of the pyramid. The slant height is the perpendicular distance from from the apex to the midpoint of the side of the base. The height of the base is the altitude of the base triangle.

The surface area of a solid is the sum of the areas of all its faces. Hence, the surface area of a triangular pyramid is the sum of the area of its triangular base and its three lateral triangular faces.

In this calculator, the triangle considered as the base is an **equilateral** **triangle**.

Consider a triangular pyramid of side of base b, height of base a and slant height h.

Surface area, S = Area of the base + Area of the three lateral faces

Area of base = \(\frac{1}{2}\times b \times a\)

Area of one lateral face = \(\frac{1}{2}\times b \times h\)

Therefore, S = \(\frac{1}{2}\times b \times a+3\times\frac{1}{2}\times b \times h\)

– When the surface area S and side of base b is known, height of base a and slant height h can be calculated as,

The height of an equilateral triangle can be found by multiplying its side by \(\frac{\sqrt{3}}{2}\).

Therefore, the height of the base, a = \(\frac{\sqrt{3}}{2}b\)

Therefore, slant height, h = \(\frac{S-\text{Area of base}}{3\times \frac{1}{2}\times b}\)

– When the surface area S and slant height h is known, side of base b and height of base a can be calculated as,

We know that, a = \(\frac{\sqrt{3}}{2}b\)

Therefore, the area of base = \(\frac{1}{2}\times b \times \frac{\sqrt{3}}{2}\times b\)

= \(\frac{\sqrt{3}}{4}\times b^2\)

So, surface area, S = \(\frac{\sqrt{3}}{4}\times b^2+3 \times \frac{1}{2}\times b \times h\)

\(\frac{\sqrt{3}}{4}b^2+\frac{3}{2}hb-S\) = 0 (1)

Therefore, this is a quadratic equation, where b is the variable.

A quadratic equation \(ax^2+bx+c=0\), can be solved for x as:

x = \(\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

On applying the above expression to equation (1):

a = \(\frac{\sqrt{3}}{4}\), b = \(\frac{3}{2}h\) and c = – S

The side of the base is a measure of length, hence, the value will always be positive. Therefore, we consider only the positive root of the equation.

Therefore, side of the base, b = \(\frac{-\frac{3}{2}h+\sqrt{\left(\frac{3}{2}g\right)^2-4\times \frac{\sqrt{3}}{4}\times(-S)}}{2\times \frac{\sqrt{3}}{4}}\)

**Example 1: ****Find the surface area and the height of the base of a triangular pyramid with a base side length of 8 centimeters and slant height of 6 centimeters**.

**Solution:**

b = 8 cm

h = 6 cm

S = Area of the base + Area of the three lateral faces

Since the base of the pyramid is an equilateral triangle,

So, a = \(\frac{\sqrt{3}}{2}b\)

a = \(\frac{\sqrt{3}}{2}\times 8\)

a = 6.928 cm

Area of base = \(\frac{1}{2}\times b \times a\)

= \(\frac{1}{2}\times 8 \times 6.928\)

= 27.712 \(cm^2\)

Area of one lateral face = \(\frac{1}{2}\times b \times h\)

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

= 24 \(cm^2\)

S = Area of base + Area of three lateral faces

= 27.712 + 3 \(\times\) 24

= 99.713 \(cm^2\)

**The surface area of the triangular pyramid is 99.713 square centimeters and the height of its base is 6.928 centimeters.**

**Example 2: ****Find the side of the base and height of the base of a triangular pyramid whose surface area is 50 square meters and slant height is 5 meters.**

**Solution: **

S = 50 \(m^2\)

h = 5 m

Surface area, S = \(\frac{\sqrt{3}}{4}\times b^2+3\times \frac{1}{2}\times b \times h\)

\(\frac{\sqrt{3}}{4}b^2+\frac{3}{2}hb-S\) = 0

This is a quadratic equation, where b is the variable.

Therefore, side of base, b = \(\frac{-\frac{3}{2}h+\sqrt{\left(\frac{3}{2}h\right)^2-4\times \frac{\sqrt{3}}{4}\times(-S)}}{2\times \frac{\sqrt{3}}{4}}\)

b = \(\frac{-\frac{3}{2}\times 5+\sqrt{\left(\frac{3}{2}\times 5\right)^2-4\times \frac{\sqrt{3}}{4}\times(-50)}}{2\times \frac{\sqrt{3}}{4}}\)

b = 5.141 m

a = \(\frac{\sqrt{3}}{2}\times 5.141\)

= 4.452 m

**So, the side of the base of the triangular pyramid is 5.141 meters and the height of its base is 4.452 meters.**

**Example 3: ****Find the surface area and side of the base of a triangular pyramid whose height of its base is \(3\sqrt{3}\) meters and the slant height is 10 meters.**

**Solution:**

a = \(3\sqrt{3}\) m

h = 10 m

We know that, a = \(\frac{\sqrt{3}}{2}b\)

So, b = \(a\div \frac{\sqrt{3}}{2}\)

= \(3\sqrt{3}\div \frac{\sqrt{3}}{2}\)

= 6 m

Area of the base = \(\frac{1}{2}\times b \times a\)

= \(\frac{1}{2}\times 6 \times 3\sqrt{3}\)

= 15.588 \(m^2\)

Area of one lateral face = \(\frac{1}{2}\times b \times h\)

= \(\frac{1}{2}\times 6 \times 10\)

= 30 \(m^2\)

S = Area of the base + Area of the three lateral faces

= 15.588 \(m^2\) + 330 \(m^2\)

= 105.588 \(m^2\)

**Therefore, the surface area of the triangular pyramid is 105.588 square meters and the side of its base is 6 meters.**

**Example 4: ****Find the slant height and side of the base of a triangular pyramid with a base of height \(8\sqrt{3}\) inches, and whose surface area is 398.85 square inches.**

**Solution:**

a = \(8\sqrt{3}\) in

S = 398.85 \(in^2\)

We know that, a = \(\frac{\sqrt{3}}{2}b\)

So, b = \(a\div\frac{\sqrt{3}}{2}\)

= \(8\sqrt{3}\div\frac{\sqrt{3}}{2}\)

= 16 in

Area of the base = \(\frac{1}{2}\times b \times a\)

= \(\frac{1}{2}\times 16 \times 8\sqrt{3}\)

= 110.851 \(in^2\)

Slant height, h = \(\frac{S-\text{Area of base}}{3\times \frac{1}{2}\times b}\)

h = \(\frac{398.85-110.851}{3\times \frac{1}{2}\times 16}\)

h = 12 in

**Therefore, the slant height of the pyramid is ****12**** inches and its base length is 16 inches.**

**Example 5: ****John was playing with a triangular pyramid shaped puzzle of surface area 42.93 \(cm^2\). The base length of the puzzle is 4 centimeters. Help John in finding its slant height and the height of its base. **

**Solution:**

S = 42.93 \(cm^2\)

b = 4 cm

Area of the triangular base = \(\frac{1}{2}\times b \times a\)

a = \(\frac{\sqrt{3}}{2}b\)

= \(a\frac{\sqrt{3}}{2}\times 4\)

= 3.464 cm

Area of base = \(\frac{1}{2}\times b \times a\)

= \(\frac{1}{2}\times 4 \times 3.466\)

= 6.928 \(cm^2\)

Slant height, h = \(\frac{S-\text{Area of base}}{3\times \frac{1}{2}\times b}\)

= \(\frac{42.93-6.928}{3\times \frac{1}{2}\times 4}\)

= 6 cm

**The slant height of the triangular pyramid is 6 centimeters and the height of its base is 3.464 centimeters.**

Frequently Asked Questions

A triangular pyramid is a three dimensional solid that has a triangular base and three lateral faces that are triangles.

The faces of a solid can be laid flat such that when they are folded the original solid can be obtained. The flat pattern thus formed by its faces is termed as the **net** of a solid. In other words, a net is a two-dimensional representation of a three-dimensional object. It is used to calculate the surface area of the three dimensional object.

A pyramid that has a triangular base is termed as a tetrahedron. A tetrahedron is regular when all its faces are equilateral triangles.

There are 6 edges, 4 faces and 4 vertices in a triangular pyramid.