Geometry · 6.G.A.4 · G.MG.A.1

Surface area of pyramids and cones

A pyramid or a cone narrows to a point. How much material does it take to cover one?

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The Pythagorean theoremMath 8 Nets of solidsMath 6

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Modeling with solids: density and designGeometry

Where you are 3 in Surface area

Geometry11

Solids8

Surface area3

Surface area of a sphere Surface area of prisms and cylinders

Surface area of pyramids and cones

When you reach for it

Canvas for a tent, tiles for a pointed roof, card for a party hat, paint for a spire: each asks for the area of the outside of a solid that rises from a flat base to a single point. The surface area of a solid with flat faces is the sum of the areas of its faces, and a net lays those faces out flat so that none is missed. For a prism the faces round the side are rectangles whose height is the height of the prism, which is how prisms and cylinders work.

A pyramid is different in one way that matters. Its side faces are triangles that lean inward to meet at the top, so the height of each triangle is measured along the sloping face, not straight up. That length is the slant height, and it is longer than the height of the solid. When a problem gives you a different length, the real work is finding the slant height, and that is a job for the Pythagorean theorem. A cone is the round version of the same idea: its curved side unrolls into a piece of a circle, and the slant height is the radius of that piece.

Reach for this when a solid has a flat base and one point, the apex, joined to every point of the base’s edge: a pyramid on any polygon, or a cone on a circle.

The procedure

The parts of a pyramid

A pyramid has a polygon for its base and one point off the base, the apex, joined by a straight edge to every vertex of the base. Its faces are the base and one triangle for each edge of the base, the lateral faces. The total area of the lateral faces is the lateral area, and

\text{surface area} = \text{base area} + \text{lateral area}

A pyramid is regular when its base is a regular polygon (all sides equal, all angles equal) and its apex sits directly above the center of the base. Its height h is the distance from the apex V straight down to the base, and in a regular pyramid that perpendicular meets the base at its center O.

In a regular pyramid every lateral face is the same isosceles triangle. Each edge from the apex runs to a vertex of the base, and every vertex of a regular polygon is the same distance R from the center. The line VO is perpendicular to the base, so the triangle formed by V, O and any base vertex has a right angle at O, and every one of these edges has length \displaystyle \sqrt{h^2 + R^2}. So every lateral face has two equal sides, and its third side is an edge of the base, also all equal. Three equal sides in each face make all the faces congruent.

The slant height \ell of a regular pyramid is the height of one lateral face: the distance from the apex, down the middle of the face, to the midpoint M of a base edge. It ends at the midpoint because the altitude from the apex of an isosceles triangle bisects the base. Since the faces are congruent, one slant height serves them all.

Figure 1The height h runs straight down inside the pyramid. The slant height \ell runs down the face. They are two sides of the right triangle VOM.

Three lengths and two right triangles

Three different lengths run from the apex down to the base: the height h, straight down inside the solid; the slant height \ell, down the middle of a face; and the lateral edge e, down the crease between two faces. A problem can hand you any one of them. Two right triangles connect them.

Inside the pyramid, take V, O and M. The line VO is perpendicular to the base, so it is perpendicular to every line in the base through O, including OM. The distance OM from the center of a regular polygon to the midpoint of a side is its The apothem of a regular polygon, written a. For a square of side s it is simply \displaystyle a = \frac{s}{2}. By the Pythagorean theorem,

\ell^2 = h^2 + a^2

On a face, take V, M and a vertex of the base. The slant height is the altitude of the face, so the angle at M is right, and M is half an edge from the vertex. So

e^2 = \ell^2 + \left(\frac{s}{2}\right)^2

Figure 2On a face, the slant height and half an edge are the legs of a right triangle whose hypotenuse is the lateral edge e.

In the first triangle \ell is the hypotenuse, so \ell > h. In the second it is a leg, so \ell < e. The slant height always lies strictly between the height and the lateral edge.

Adding up the faces

Figure 3Unfolded, a square pyramid is one square and four triangles, each of base s and height \ell.

Unfold a regular pyramid whose base has n sides of length s. The net is the base and n triangles, each with base s and height \ell, so each has area \displaystyle \frac{1}{2}s\ell. All of them together have area \displaystyle n \cdot \frac{1}{2}s\ell = \frac{1}{2}(ns)\ell, and ns is the perimeter P of the base. Writing B for the area of the base:

\text{lateral area} = \frac{1}{2}P\ell \qquad\qquad \text{surface area} = B + \frac{1}{2}P\ell

The base area B comes from whatever area rule fits the base: a square, a triangle, or a regular polygon. The formula is nothing more than the sum of the faces written short. When the faces are not all alike, as in Example 5, add them one at a time instead.

The cone

A right circular cone has a circle of radius r for its base and its apex directly above the center, at height h. Every segment from the apex to the rim of the base is the hypotenuse of a right triangle with legs h and r, so they all have the same length, the slant height of the cone:

\ell^2 = r^2 + h^2

Slit the curved surface along one of those segments and flatten it. Every point of the rim was \ell from the apex, so the flattened piece is a sector of a circle of radius \ell, and its curved edge is the rim of the base, which has length 2\pi r, the Circumference of the base. A whole circle of radius \ell has circumference 2\pi\ell and area \pi\ell^2. The sector’s arc is the fraction \displaystyle \frac{2\pi r}{2\pi\ell} = \frac{r}{\ell} of the whole circumference, and a sector’s area is the same fraction of the whole circle’s area. So

\text{lateral area} = \frac{r}{\ell}\cdot\pi\ell^2 = \pi r\ell

Figure 4The cone with r = 3 and h = 4 (so \ell = 5) and its net. The curved surface unrolls into a sector of radius \ell whose arc is the rim, 2\pi r, long.

Adding the circular base, \pi r^2,

\text{surface area} = \pi r^2 + \pi r\ell = \pi r(r + \ell)

The pyramid formula says the same thing. Put P = 2\pi r in \displaystyle \frac{1}{2}P\ell and you get \displaystyle \frac{1}{2}(2\pi r)\ell = \pi r\ell. That is no accident: regular pyramids drawn inside the cone with more and more sides have bases closing in on the circle and slant heights closing in on \ell, so their lateral areas close in on the cone’s.

The same fraction gives the angle of the sector in the net. It is \displaystyle \frac{r}{\ell} of a full turn:

\theta = 360^\circ \cdot \frac{r}{\ell}

Worked examples

Example 1

A square pyramid from its slant height

A square pyramid has base edges of 10 cm and slant height 13 cm. Find its surface area.

Solution

The base is a square, so B = 10^2 = 100 cm². Each of the four lateral faces is a triangle with base 10 cm and height 13 cm, since the slant height is exactly the height of a face. One face has area \displaystyle \frac{1}{2} \cdot 10 \cdot 13 = 65 cm², and four have 260 cm². The formula gives the same thing in one line: P = 4 \cdot 10 = 40, and \displaystyle \frac{1}{2} \cdot 40 \cdot 13 = 260.

\text{surface area} = 100 + 260 = 360 \text{ cm}^2

Example 2

A square pyramid from its height

A square pyramid has base edges of 12 m and height 8 m. Find its surface area.

Solution

The 8 m is the height of the solid, straight down from the apex, and it is not the height of a face. Find the slant height from the right triangle VOM inside the pyramid. The apothem of a square is half its side, a = 6, so

\ell^2 = h^2 + a^2 = 8^2 + 6^2 = 100, \qquad \ell = 10

Now B = 12^2 = 144, and with P = 48 the lateral area is \displaystyle \frac{1}{2} \cdot 48 \cdot 10 = 240. The surface area is 144 + 240 = 384 m².

Using 8 in place of 10 would give a lateral area of 192, too small, because the faces lean: each one is longer from apex to base than the pyramid is tall.

Example 3

A square pyramid from its lateral edge

A square pyramid has base edges of 10 cm and lateral edges of 13 cm. Find its surface area.

Solution

These are the same two numbers as Example 1, but the 13 is now a different length: the edge where two faces meet, not the middle of a face. Use the right triangle on a face. Its hypotenuse is the lateral edge, 13, and one leg is half a base edge, 5:

\ell^2 = e^2 - \left(\frac{s}{2}\right)^2 = 13^2 - 5^2 = 144, \qquad \ell = 12

The lateral area is \displaystyle \frac{1}{2} \cdot 40 \cdot 12 = 240 and the base is 100, so the surface area is 340 cm². Example 1 gave 360 cm². Same numbers, different lengths, different solids: always ask which length you have been given.

Example 4

A hexagonal pyramid

A regular hexagonal pyramid has base edges of 6 cm and slant height 10 cm. Find its surface area exactly and to two decimal places.

Solution

The lateral faces need only the perimeter: P = 6 \cdot 6 = 36, so the lateral area is \displaystyle \frac{1}{2} \cdot 36 \cdot 10 = 180 cm².

For the base, draw lines from the center of the hexagon to its six vertices. They cut it into six equilateral triangles of side 6. Each triangle’s height splits it into two right triangles with hypotenuse 6 and one leg 3, so the height is \displaystyle \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}. Each triangle has area \displaystyle \frac{1}{2} \cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3}, and the hexagon has \displaystyle B = 54\sqrt{3} cm².

\text{surface area} = 180 + 54\sqrt{3} \approx 273.53 \text{ cm}^2

That height \displaystyle 3\sqrt{3} is also the apothem of the hexagon, so the pyramid’s own height would be \displaystyle \sqrt{10^2 - 27} = \sqrt{73} cm. The surface area did not need it.

Example 5

A rectangular base, two slant heights

A pyramid has a rectangular base 10 m by 18 m, and its apex is 12 m directly above the center of the rectangle. Find its surface area.

Solution

The base is not a regular polygon, so the faces are not all alike. They come in two pairs. The apex is above the center, which is equally far from the two ends of any edge, so every face is still isosceles and its altitude still lands at the midpoint of its edge. What differs is how far that midpoint is from the center.

The faces standing on the 18 m edges: the midpoint of an 18 m edge is half of 10, so 5 m, from the center. Their height is \displaystyle \sqrt{12^2 + 5^2} = 13, and each has area \displaystyle \frac{1}{2} \cdot 18 \cdot 13 = 117. Two of them make 234.

The faces standing on the 10 m edges: that midpoint is 9 m from the center. Their height is \displaystyle \sqrt{12^2 + 9^2} = \sqrt{225} = 15, and each has area \displaystyle \frac{1}{2} \cdot 10 \cdot 15 = 75. Two of them make 150.

The base is 10 \cdot 18 = 180, so

\text{surface area} = 180 + 234 + 150 = 564 \text{ m}^2

There is no single slant height here, so \displaystyle \frac{1}{2}P\ell has nothing to put in for \ell. Adding the faces always works.

Example 6

A cone

A cone has radius 3 cm and height 4 cm. Find its surface area, exactly and to two decimal places, and the angle of the sector in its net.

Solution

The slant height comes from the right triangle of height, radius and slant: \displaystyle \ell = \sqrt{3^2 + 4^2} = 5 cm. Then

\begin{aligned} \text{lateral area} &= \pi r\ell = \pi \cdot 3 \cdot 5 = 15\pi \\ \text{base area} &= \pi r^2 = 9\pi \\ \text{surface area} &= 24\pi \approx 75.40 \text{ cm}^2 \end{aligned}

The sector in the net is \displaystyle \frac{r}{\ell} = \frac{3}{5} of a full turn, \displaystyle 360^\circ \cdot \frac{3}{5} = 216^\circ. This is the cone in the figure above.

Example 7

An open cone

A paper cup is a cone with no lid. Its rim has diameter 7 cm and the cup is 12 cm deep. How much paper does it take, ignoring any overlap?

Solution

The cup is open at the rim, so only the curved surface is made of paper; there is no base to add. The radius is half the diameter, r = 3.5, and the depth is the height, h = 12:

\ell = \sqrt{3.5^2 + 12^2} = \sqrt{156.25} = 12.5

\text{lateral area} = \pi \cdot 3.5 \cdot 12.5 = 43.75\pi \approx 137.44 \text{ cm}^2

Example 8

Working backwards from the surface area

A cone has radius 5 cm and surface area 90\pi cm². Find its slant height and its height.

Solution

The base accounts for \pi \cdot 5^2 = 25\pi, so the curved surface is 90\pi - 25\pi = 65\pi. That is \pi r\ell with r = 5:

5\pi\ell = 65\pi, \qquad \ell = 13

The height is the other leg of the right triangle with hypotenuse \ell and leg r: \displaystyle h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 cm.

Example 9

Rolling a sector into a cone

A sector with radius 20 cm and angle 144^\circ is cut from card, and its two straight edges are joined, without overlap, to make an open cone. Find the radius and height of the cone, and compare \pi r\ell with the area of the card.

Solution

Run the net backwards. Every point of the arc ends up on the rim, 20 cm from the tip, so the slant height is the radius of the sector, \ell = 20. The arc becomes the rim of the base, so its length is the circumference 2\pi r:

\frac{144}{360} \cdot 2\pi \cdot 20 = 16\pi = 2\pi r, \qquad r = 8

The height is \displaystyle h = \sqrt{20^2 - 8^2} = \sqrt{336} = 4\sqrt{21} \approx 18.33 cm.

The formula gives a lateral area of \pi \cdot 8 \cdot 20 = 160\pi cm². The card itself is \displaystyle \frac{144}{360} of a circle of radius 20, which is \displaystyle \frac{144}{360} \cdot \pi \cdot 20^2 = 160\pi cm². They agree, as they must: the curved surface is the card.

Boundaries

\displaystyle \frac{1}{2}P\ell needs every lateral face to have the same height. That is true for a regular pyramid. A rectangular base that is not a square gives two different face heights (Example 5), and a pyramid whose apex is not above the center of its base has faces of several heights, each found from its own right triangle. In every case the surface area is still the sum of the faces; only the short formula fails.

\pi r\ell is for a right circular cone, apex directly above the center of a circular base. If the apex leans to one side, the segments from apex to rim have different lengths, there is no single slant height, and there is no formula of this simple kind for the curved surface.

Slant height, height and lateral edge are three different lengths, with h < \ell < e for a regular pyramid and h < \ell for a cone. Surface area uses \ell, because it is the height of a face. The volume of a pyramid or cone uses h.

Lateral area or total surface area depends on the object. A cup, a hat or a roof has no base to cover, and a solid built from pieces does not count the faces where the pieces touch.

All lengths in one unit, and the answer in square units. A base in meters and a slant height in centimeters must be put in the same unit before multiplying.

The sector in a cone’s net is always less than a full circle. Its angle is \displaystyle 360^\circ \cdot \frac{r}{\ell}, and r < \ell because r is a leg and \ell the hypotenuse of the same right triangle.

Where this goes next

Pyramids and cones finish the solids with flat or unrollable sides. Set this lesson beside prisms and cylinders and the pattern is the same: base area plus the area of the pieces round the side, with the prism’s Ph replaced by the pyramid’s \displaystyle \frac{1}{2}P\ell, and the cylinder’s 2\pi rh by the cone’s \pi r\ell. A sphere breaks the pattern, since no part of it can be flattened without stretching, and its surface area needs a new argument.

The two heights part company in the next lesson along: the volume of pyramids and cones is \displaystyle \frac{1}{3}Bh, with the vertical height, where surface area used the slant height. Working with both on the same solid is the core of modeling with solids, where a design must hold a given amount while using as little material as possible.

The cone’s net also leads on. Its arc is an arc length and its area a sector’s, so any question about cutting a cone from flat stock, or about a cone with its tip cut off (a lampshade, a bucket), is a question about sectors: the cut-off cone’s curved surface is the large cone’s minus the small one’s.

Exercises

Regular pyramids

A square pyramid has base edges of 8 m and slant height 5 m. Find its surface area.
A square pyramid has base edges of 14 cm and height 24 cm. Find its slant height and its surface area.
A square pyramid has base edges of 16 inches and lateral edges of 17 inches. Find its slant height and its surface area.
A regular triangular pyramid has an equilateral triangle for its base, with edges of 6 cm, and slant height 7 cm. Find its surface area exactly and to one decimal place.
A regular pentagonal pyramid has base edges of 4 m and slant height 9 m, and its base has an area of about 27.53 m². Find its surface area.
A regular hexagonal pyramid has base edges of 4 cm and height 2 cm. Find its slant height and its lateral area.

A rectangular base

A pyramid has a rectangular base 16 m by 40 m, and its apex is 15 m directly above the center of the base. Find its surface area.

Cones

A cone has radius 6 cm and slant height 10 cm. Find its surface area exactly and to two decimal places.
A cone has radius 8 m and height 15 m. Find its slant height and its exact surface area.
A cone has a base of diameter 14 cm and height 24 cm. Find its surface area exactly and to two decimal places.
A party hat is an open cone with radius 9 cm and height 12 cm. How much card does it take, to two decimal places, ignoring any overlap?
A cone has radius 4 cm and slant height 10 cm. What is the angle of the sector in its net?

Working backwards

A square pyramid has base edges of 6 cm and surface area 96 cm². Find its slant height and its height.
A cone has radius 3 cm and lateral area 24\pi cm². Find its slant height and its height.
A cone has radius 4 cm and surface area 36\pi cm². Find its slant height and its height.
A sector of radius 12 cm and angle 120^\circ is rolled into an open cone without overlap. Find the radius, the height and the lateral area of the cone.

Putting solids together

A tower is a cube with edges of 6 m, topped by a square pyramid whose base is the top of the cube and whose slant height is 5 m. The whole outside is painted, except the floor. What area is painted?
A spinning top is two cones of radius 6 cm glued base to base, one of height 8 cm and one of height 4.5 cm. Find its surface area exactly and to two decimal places.
A square building 16 m wide has a roof shaped as a square pyramid, with its peak 6 m above the top of the walls. What area of roofing covers it?

Reasoning

Explain why, in any regular pyramid, the slant height is longer than the height and shorter than the lateral edge.
A pyramid has a rectangular base that is not a square, with its apex above the center. Explain why the formula \displaystyle \frac{1}{2}P\ell cannot be used for its lateral area, and what to do instead.
Explain why the lateral area of a cone, \pi r\ell, is what \displaystyle \frac{1}{2}P\ell gives when P is the circumference of the base, and why it is reasonable that the pyramid formula carries over to cones.
Every length of a cone (radius, height and slant height) is doubled. By what factor does its surface area change? Show why.
Explain why the angle of the sector in a cone’s net is \displaystyle 360^\circ \cdot \frac{r}{\ell}, and why that angle is always less than 360^\circ.

Answers

144 m². The base is 8^2 = 64 and the lateral area is \displaystyle \frac{1}{2} \cdot 32 \cdot 5 = 80.
\ell = 25 cm and surface area 896 cm². The apothem is 7, so \ell^2 = 24^2 + 7^2 = 625; the base is 196 and the lateral area is \displaystyle \frac{1}{2} \cdot 56 \cdot 25 = 700.
\ell = 15 inches and surface area 736 square inches. On a face, \ell^2 = 17^2 - 8^2 = 225; the base is 256 and the lateral area is \displaystyle \frac{1}{2} \cdot 64 \cdot 15 = 480.
\displaystyle 63 + 9\sqrt{3} \approx 78.6 cm². The lateral area is \displaystyle \frac{1}{2} \cdot 18 \cdot 7 = 63. The base triangle has height \displaystyle \sqrt{6^2 - 3^2} = 3\sqrt{3}, so its area is \displaystyle \frac{1}{2} \cdot 6 \cdot 3\sqrt{3} = 9\sqrt{3}.
About 117.53 m². The lateral area is \displaystyle \frac{1}{2} \cdot 20 \cdot 9 = 90, and the base adds 27.53.
\ell = 4 cm and lateral area 48 cm². The apothem of the hexagon is the height of an equilateral triangle of side 4, which is \displaystyle \sqrt{4^2 - 2^2} = 2\sqrt{3}, so \displaystyle \ell^2 = 2^2 + (2\sqrt{3})^2 = 16. The lateral area is \displaystyle \frac{1}{2} \cdot 24 \cdot 4 = 48.
1720 m². Faces on the 40 m edges have height \displaystyle \sqrt{15^2 + 8^2} = 17 and area \displaystyle \frac{1}{2} \cdot 40 \cdot 17 = 340 each. Faces on the 16 m edges have height \displaystyle \sqrt{15^2 + 20^2} = 25 and area \displaystyle \frac{1}{2} \cdot 16 \cdot 25 = 200 each. The base is 640, and 640 + 680 + 400 = 1720.
96\pi \approx 301.59 cm². The base is 36\pi and the lateral area is \pi \cdot 6 \cdot 10 = 60\pi.
\ell = 17 m and surface area 200\pi m². \displaystyle \ell = \sqrt{8^2 + 15^2}; the base is 64\pi and the lateral area is \pi \cdot 8 \cdot 17 = 136\pi.
224\pi \approx 703.72 cm². r = 7 and \displaystyle \ell = \sqrt{7^2 + 24^2} = 25; the base is 49\pi and the lateral area is 175\pi.
135\pi \approx 424.12 cm². Only the curved surface: \displaystyle \ell = \sqrt{9^2 + 12^2} = 15, and \pi \cdot 9 \cdot 15 = 135\pi.
144^\circ. The sector is \displaystyle \frac{4}{10} of a full turn, and \displaystyle 360^\circ \cdot \frac{4}{10} = 144^\circ.
\ell = 5 cm and h = 4 cm. \displaystyle 36 + \frac{1}{2} \cdot 24 \cdot \ell = 96 gives 12\ell = 60; then h^2 = 5^2 - 3^2 = 16.
\ell = 8 cm and \displaystyle h = \sqrt{55} \approx 7.42 cm. 3\pi\ell = 24\pi, and h^2 = 8^2 - 3^2 = 55.
\ell = 5 cm and h = 3 cm. 16\pi + 4\pi\ell = 36\pi gives 4\pi\ell = 20\pi; then h^2 = 5^2 - 4^2 = 9.
r = 4 cm, \displaystyle h = 8\sqrt{2} \approx 11.31 cm, lateral area 48\pi cm². The arc, \displaystyle \frac{120}{360} \cdot 2\pi \cdot 12 = 8\pi, becomes the rim 2\pi r. The slant height is 12, so h^2 = 12^2 - 4^2 = 128, and \pi \cdot 4 \cdot 12 = 48\pi, which is also the area of the sector, \displaystyle \frac{120}{360} \cdot \pi \cdot 12^2.
204 m². The cube’s top is covered by the roof and its bottom is the floor, so only the four walls are painted, 4 \cdot 36 = 144, plus the roof, \displaystyle \frac{1}{2} \cdot 24 \cdot 5 = 60.
105\pi \approx 329.87 cm². The two bases are glued together and hidden. The slant heights are \displaystyle \sqrt{6^2 + 8^2} = 10 and \displaystyle \sqrt{6^2 + 4.5^2} = 7.5, so the area is \pi \cdot 6 \cdot 10 + \pi \cdot 6 \cdot 7.5 = 60\pi + 45\pi.
320 m². The roof has no base to cover. The apothem of the square is 8, so \displaystyle \ell = \sqrt{6^2 + 8^2} = 10, and the lateral area is \displaystyle \frac{1}{2} \cdot 64 \cdot 10 = 320.
The height, the apothem and the slant height form a right triangle with the right angle at the center of the base, so \ell is its hypotenuse and \ell^2 = h^2 + a^2 > h^2. On a face, the slant height and half an edge are the legs of a right triangle whose hypotenuse is the lateral edge, so \displaystyle e^2 = \ell^2 + \left(\frac{s}{2}\right)^2 > \ell^2. Both inequalities are strict because a and \displaystyle \frac{s}{2} are not zero.
The formula adds n triangles that all have height \ell. Here the center of the base is nearer the long edges than the short ones, so the faces on the long edges have a smaller height than the faces on the short edges, and there is no single \ell. Find each pair’s height from its own right triangle with the pyramid’s height, and add the four faces and the base, as in Example 5.
With P = 2\pi r, \displaystyle \frac{1}{2}P\ell = \frac{1}{2} \cdot 2\pi r \cdot \ell = \pi r\ell, exactly the cone’s lateral area. Regular pyramids drawn inside a cone with more and more sides have perimeters closing in on the circumference and slant heights closing in on the cone’s, while their lateral surfaces close in on the cone’s curved surface. So their lateral areas, each \displaystyle \frac{1}{2}P\ell, close in on \pi r\ell.
It is multiplied by 4. The new surface area is \pi(2r)(2r + 2\ell) = 4\pi r(r + \ell), four times the old one. Each area is a product of two lengths, and doubling both multiplies the product by 2 \cdot 2.
The rim of the base, of length 2\pi r, becomes the arc of a sector of radius \ell, and a full circle of that radius would have circumference 2\pi\ell. So the arc is the fraction \displaystyle \frac{2\pi r}{2\pi\ell} = \frac{r}{\ell} of the full circle, and the angle is that fraction of 360^\circ. Since r is a leg and \ell the hypotenuse of the right triangle with legs r and h, r < \ell, so the fraction is less than 1.