Surface Area of Pyramid

A pyramid is a solid whose base is a polygon and whose other faces are triangles meeting at a single point at the top. When the base is a square, the solid is a square-based pyramid. The four triangles standing on the four sides of the square are its lateral faces, and the point where they meet is the apex, also called the vertex.

In the pyramid below, ABCD is the square base and P is the apex. O is the centre of the base and lies directly below P. M is the midpoint of the side BC. Four lengths describe a pyramid like this, and each one has its own letter.

Two right-angled triangles sit inside the pyramid, and they tie these four lengths together. The first is triangle POM, with its right angle at O. Its sides are h, OM and l. Since O is the centre of the square, OM is half a side, so \( OM = \frac{a}{2} \). Here l is the hypotenuse, so Pythagoras' theorem gives:

\[ l^2 = h^2 + \left(\frac{a}{2}\right)^2 \]

The second is triangle PMC, which lies in the face PBC. Its right angle is at M, because in the isosceles triangle PBC the line from the apex to the midpoint of the base side is perpendicular to that side. Its sides are l, \( MC = \frac{a}{2} \) and the edge e, which is the hypotenuse:

\[ e^2 = l^2 + \left(\frac{a}{2}\right)^2 \]

Take a paper model of a square-based pyramid, cut it along its four slanting edges and press it flat on the table. The flat shape you get is the net of the pyramid. The net lays every face out side by side, so the whole surface can be counted one piece at a time.

Looking at the net tells us three things about the surface:

• (a) The pyramid has four triangular faces, one on each side of the square.
• (b) All four triangles have the same area. Each has base a, and because the apex is directly above the centre of the base, each has the same height l. So the four faces are congruent isosceles triangles.
• (c) The surface is made of two kinds of shape: one square of area \( a^2 \) and four triangles, each of area \( \frac{1}{2} \times a \times l \).

The base is a square of side a. The area of a square is its side multiplied by itself, so the area of the base, A, is:

\[ A = a^2 \]

The formula comes straight out of the net.

Step 1: The area of any triangle is:

\[ \frac{1}{2} \times \text{base} \times \text{height} \]

Step 2: Each face stands on a side of the base, so its base is a, and its height is the slant height l. One triangular face therefore has area:

\[ \frac{1}{2} \times a \times l \]

Step 3: Adding the four equal faces together:

\[ \text{LSA} = 4 \times \frac{1}{2} \times a \times l \]

Step 4: Simplifying, since \( 4 \times \frac{1}{2} = 2 \):

\[ \text{LSA} = 2al \]

Counting the square and all four triangles of the net gives the whole surface:

\[ \text{TSA} = A + \text{LSA} \]

\[ \text{TSA} = a^2 + 2al \]

Both terms share the factor a. Taking it outside gives a shorter form:

\[ \text{TSA} = a(a + 2l) \]

Both forms give the same number, and the factorised one is quicker for checking an answer. An area is a length multiplied by a length, so it is measured in square units: when the lengths are in cm, every area is in \( \text{cm}^2 \).

Question: A square-based pyramid has base side 6 cm and slant height 5 cm. Find its lateral surface area and its total surface area.

Here \( a = 6 \text{ cm} \) and \( l = 5 \text{ cm} \). Both lengths the formulas need are given, so they go straight in.

Step 1: Finding the area of the square base:

\[ A = a^2 = 6^2 = 36 \text{ cm}^2 \]

Step 2: Putting a and l into the formula for the lateral surface area:

\[ \text{LSA} = 2al = 2 \times 6 \times 5 = 60 \text{ cm}^2 \]

Step 3: Adding the base to the four faces to get the total surface area:

\[ \text{TSA} = 36 + 60 = 96 \text{ cm}^2 \]

Check: \( a(a + 2l) = 6 \times (6 + 10) = 96 \), which agrees. So the lateral surface area is 60 cm² and the total surface area is 96 cm².

Question: A square-based pyramid has base side 10 cm and edge 13 cm. Find its total surface area.

Here \( a = 10 \text{ cm} \) and \( e = 13 \text{ cm} \). The formula \( 2al \) needs the slant height, but the question gives the edge instead, so l must first be found from the right-angled triangle GMR.

Step 1: Writing the relation between the edge and the slant height:

\[ e^2 = l^2 + \left(\frac{a}{2}\right)^2 \]

Step 2: Substituting \( e = 13 \) and \( a = 10 \):

\[ 13^2 = l^2 + \left(\frac{10}{2}\right)^2 \]

Step 3: Squaring the numbers:

\[ 169 = l^2 + 25 \]

Step 4: Subtracting 25 from both sides:

\[ l^2 = 144 \]

Step 5: Taking the positive square root, since a length is positive:

\[ l = 12 \text{ cm} \]

Step 6: Finding the area of the base:

\[ A = 10^2 = 100 \text{ cm}^2 \]

Step 7: With the slant height known, finding the lateral surface area:

\[ \text{LSA} = 2 \times 10 \times 12 = 240 \text{ cm}^2 \]

Step 8: Adding the two areas:

\[ \text{TSA} = 100 + 240 = 340 \text{ cm}^2 \]

So the total surface area of the pyramid is 340 cm².

Question: A square-based pyramid has total surface area 144 cm² and slant height 5 cm. Find its vertical height and the length of its edge.

Here \( \text{TSA} = 144 \text{ cm}^2 \) and \( l = 5 \text{ cm} \). Both h and e need the base side a, so a is found first.

Step 1: Putting the known values into the total surface area formula:

\[ 144 = a^2 + 2 \times a \times 5 \]

Step 2: Simplifying and moving every term to one side, which gives a quadratic equation:

\[ a^2 + 10a - 144 = 0 \]

Step 3: Splitting the middle term, because \( 18 \times 8 = 144 \) and \( 18 - 8 = 10 \):

\[ a^2 + 18a - 8a - 144 = 0 \]

Step 4: Taking the common factor out of each pair:

\[ a(a + 18) - 8(a + 18) = 0 \]

Step 5: Taking \( (a + 18) \) out as a common factor:

\[ (a + 18)(a - 8) = 0 \]

Step 6: Either \( a + 18 = 0 \), giving \( a = -18 \), or \( a - 8 = 0 \), giving \( a = 8 \). A length cannot be negative, so \( a = -18 \) is rejected and:

\[ a = 8 \text{ cm} \]

Step 7: For the vertical height, writing the relation between the slant height and the vertical height:

\[ l^2 = h^2 + \left(\frac{a}{2}\right)^2 \]

Step 8: Substituting \( l = 5 \) and \( \frac{a}{2} = 4 \), and squaring:

\[ 25 = h^2 + 16 \]

Step 9: Subtracting 16 from both sides and taking the positive square root:

\[ h^2 = 9 \]

\[ h = 3 \text{ cm} \]

Step 10: For the edge, substituting into the relation between the edge and the slant height:

\[ e^2 = 5^2 + 4^2 = 25 + 16 = 41 \]

Step 11: Taking the positive square root:

\[ e = \sqrt{41} \text{ cm} \]

So the vertical height is 3 cm and the edge is \( \sqrt{41} \text{ cm} \). Since 41 is not a perfect square, the answer is left as a square root.

Question: A square-based pyramid has base side 12 cm and lateral surface area 240 cm². Find its slant height and its vertical height.

Here \( a = 12 \text{ cm} \) and \( \text{LSA} = 240 \text{ cm}^2 \). The only unknown in the lateral surface area formula is l, so l is found first.

Step 1: Putting the known values into \( \text{LSA} = 2al \):

\[ 240 = 2 \times 12 \times l \]

Step 2: Dividing both sides by \( 2 \times 12 = 24 \):

\[ l = \frac{240}{24} = 10 \text{ cm} \]

Step 3: For the vertical height, substituting \( l = 10 \) and \( \frac{a}{2} = 6 \) into the relation for triangle POQ:

\[ 10^2 = h^2 + 6^2 \]

Step 4: Squaring the numbers:

\[ 100 = h^2 + 36 \]

Step 5: Subtracting 36 from both sides:

\[ h^2 = 64 \]

Step 6: Taking the positive square root:

\[ h = 8 \text{ cm} \]

So the slant height is 10 cm and the vertical height is 8 cm. The slant height came out longer than the vertical height, exactly as it must.

Which formula to start from depends on what the question gives you:

• Given a and l: use \( 2al \) and \( a^2 + 2al \) directly.
• Given a and e: find l first from \( e^2 = l^2 + \left(\frac{a}{2}\right)^2 \).
• Given a and h: find l first from \( l^2 = h^2 + \left(\frac{a}{2}\right)^2 \).
• Given an area and one length: put the values into the formula and solve for the unknown. If a quadratic appears, factorise it and reject the negative root.