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Surface Areas of Pyramids and Cones: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The five pieces

A regular square pyramid has base side $8$ cm and slant height $6$ cm.

Find its lateral area and its total surface area when closed.

Text description of this figure

Complete net of a regular square pyramid. The central square has side 8 cm. Four congruent triangles attach, one to each square edge. A dotted perpendicular within the top triangle labels face altitude 6 cm. Dashed square edges are folds.
1. Hint 1
  Identify which piece is the base and which pieces are lateral faces.
2. Hint 2
  Each of the four triangular faces has base $8$ cm and altitude $6$ cm.
Answer

$\begin{gathered}L=96\text{ cm}^2\\S=160\text{ cm}^2\end{gathered}$

Full solution

Each triangle has area $24$ square centimeters.
There are four side triangles, so
$\begin{aligned}L&=4(24)\\L&=96\text{ cm}^2.\end{aligned}$

The square base has area $64$ square centimeters.
A closed pyramid includes it once.
$\begin{aligned}S&=96+64\\S&=160\text{ cm}^2.\end{aligned}$

Answer

$\begin{gathered}L=96\text{ cm}^2\\S=160\text{ cm}^2\end{gathered}$

Key idea

A closed pyramid includes its triangular sides and one base.
Problem 2 Two roles for height

A regular square pyramid has base side $18$ cm and perpendicular height $12$ cm.

Find its slant height and lateral area.

Text description of this figure

True right triangle VOM in a regular pyramid. VO is perpendicular height 12 cm; OM is base apothem 9 cm; VM is slant height ℓ. The right angle is at O.
1. Hint 1
  The perpendicular height is not the altitude of a side face.
2. Hint 2
  The base center is $9$ cm from a side midpoint.
3. Hint 3
  Use the right triangle from apex to center to midpoint.
Answer

$\begin{gathered}\ell=15\text{ cm}\\L=540\text{ cm}^2\end{gathered}$

Full solution

The center-to-midpoint distance is half the square side, or $9$ cm.
In triangle VOM, the height and this distance are perpendicular legs.

The slant height is the positive hypotenuse.
$\begin{aligned}\ell^2&=12^2+9^2\\\ell&=15\text{ cm}.\end{aligned}$

The base perimeter is $72$ cm.
Hence
$\begin{aligned}L&=\tfrac12(72)(15)\\L&=540\text{ cm}^2.\end{aligned}$

Answer

$\begin{gathered}\ell=15\text{ cm}\\L=540\text{ cm}^2\end{gathered}$

Key idea

The base apothem and perpendicular height determine the face altitude.
Problem 3 The semicircular pattern

A semicircular sheet has radius $18$ cm. Its two straight radii are joined without overlap to form a right cone, with the whole semicircular arc becoming the base rim.

Find the cone's radius, perpendicular height, and lateral area exactly.

Text description of this figure

A circular sector with radius 18 cm and selected angle 180 degrees. The straight radii are cut edges.
1. Hint 1
  Match the curved edge of the sheet to the entire cone rim.
2. Hint 2
  The sheet radius becomes the cone slant height.
3. Hint 3
  Find the rim circumference first, then use the cone's right triangle.
Answer

$\begin{gathered}r=9\text{ cm}\\h=9\sqrt3\text{ cm}\\L=162\pi\text{ cm}^2\end{gathered}$

Full solution

The semicircle arc is $18\pi$ cm long.
That is the cone base circumference, so
$r=9\text{ cm}.$

The slant height is $18$ cm.
The perpendicular height follows from
$\begin{aligned}h^2&=18^2-9^2\\h&=9\sqrt3\text{ cm}.\end{aligned}$

The curved area equals the area of the original semicircle.
$\begin{aligned}L&=\tfrac12\pi(18)^2\\L&=162\pi\text{ cm}^2.\end{aligned}$

Answer

$\begin{gathered}r=9\text{ cm}\\h=9\sqrt3\text{ cm}\\L=162\pi\text{ cm}^2\end{gathered}$

Key idea

A pattern preserves area while its arc closes into the base circumference.
Problem 4 A hexagonal boundary

A regular hexagonal pyramid has base side $10$ cm and slant height $14$ cm. Give exact areas for both cases.

Case A: find the area of the boundary consisting of all six triangular side faces, excluding the hexagonal base.

Case B: find the closed total surface area, including the hexagonal base.

Text description of this figure

A regular hexagonal base has side 10 cm. One congruent triangular face is attached to each of its six edges. The altitude of the top triangular face is 14 cm. Dashed base edges are folds. The base polygon is excluded in Case A and included in Case B.
1. Hint 1
  List the surfaces included in each case.
2. Hint 2
  The six triangle bases total $60$ cm.
3. Hint 3
  The hexagon separates into six equilateral triangles of side $10$ cm.
Answer

$\begin{gathered}\text{A: }420\text{ cm}^2\\\text{B: }(420+150\sqrt3)\text{ cm}^2\end{gathered}$

Full solution

Case A: all six side triangles share altitude $14$ cm.
$\begin{aligned}L&=\tfrac12(60)(14)\\L&=420\text{ cm}^2.\end{aligned}$

Case B: each equilateral base triangle has area $25\sqrt3$ square centimeters.
The full hexagonal base has area $150\sqrt3$ square centimeters.

Case B includes the same sides and this one base, for total area $(420+150\sqrt3)$ square centimeters.

Answer

$\begin{gathered}\text{A: }420\text{ cm}^2\\\text{B: }(420+150\sqrt3)\text{ cm}^2\end{gathered}$

Key idea

Adding a base changes the included area by precisely that base area.
Problem 5 The shade blank

An ideal right-conical shade has base radius $7$ cm and perpendicular height $\sqrt{95}$ cm. It includes its curved side and no base disk.

Find the sheet area and the exact sector angle of a pattern that covers this shade without overlap. Ignore seams and thickness.

Text description of this figure

True right triangle VOM in a right cone. VO is perpendicular height √95 cm; OM is radius 7 cm; VM is slant height ℓ. The right angle is at O.
1. Hint 1
  The sheet is the curved side, so its radius is the slant height.
2. Hint 2
  Add the squared radius to the squared perpendicular height.
3. Hint 3
  The sector occupies the fraction $r/\ell$ of a full circle.
Answer

$\begin{gathered}L=84\pi\text{ cm}^2\\\theta=210^\circ\end{gathered}$

Full solution

Half of the axial section is right, with legs $7$ cm and $\sqrt{95}$ cm.
Thus
$\begin{aligned}\ell^2&=49+95\\\ell&=12\text{ cm}.\end{aligned}$

The included area is only the curved side.
$\begin{aligned}L&=\pi(7)(12)\\L&=84\pi\text{ cm}^2.\end{aligned}$

The sector radius is $12$ cm, and its arc must be $14\pi$ cm.
Its angle is
$\begin{aligned}\theta&=360^\circ\cdot\frac7{12}\\\theta&=210^\circ.\end{aligned}$

Answer

$\begin{gathered}L=84\pi\text{ cm}^2\\\theta=210^\circ\end{gathered}$

Key idea

An open conical covering needs the sector but no base disk.
Problem 6 A side-face edge

A regular square pyramid has apex-to-base-vertex edges $26$ cm long. Each triangular side face has altitude $24$ cm.

Find the square base side and the pyramid's closed total surface area.
1. Hint 1
  Work within one isosceles triangular face.
2. Hint 2
  The face altitude bisects the base edge.
3. Hint 3
  The half-edge and face altitude are legs with hypotenuse $26$ cm.
Answer

$\begin{gathered}s=20\text{ cm}\\S=1360\text{ cm}^2\end{gathered}$

Full solution

In half of a triangular face, the base half-edge has square
$(s/2)^2=26^2-24^2.$
Thus the half-edge is $10$ cm and the full base side is $20$ cm.

The base perimeter is $80$ cm.
The four triangles have total area
$\begin{aligned}L&=\tfrac12(80)(24)\\L&=960\text{ cm}^2.\end{aligned}$

The square base adds $400$ square centimeters.
Therefore
$S=1360\text{ cm}^2.$

Answer

$\begin{gathered}s=20\text{ cm}\\S=1360\text{ cm}^2\end{gathered}$

Key idea

A face edge is a hypotenuse in a different triangle from the solid height.
Problem 7 The two altitude pairs

A pyramid has a square base of side $12$ cm. Two of its side-face altitudes are $\sqrt{52}$ cm and the other two are $\sqrt{28}$ cm.

Find its lateral area exactly. Explain why half the base perimeter times one of the listed altitudes is not justified.

Text description of this figure

Two triangular face types. Each type occurs twice. The first has base 12 cm and altitude √52 cm. The second has base 12 cm and altitude √28 cm. Dotted altitudes are perpendicular to the face bases.
1. Hint 1
  Compute each pair of triangular face areas separately.
2. Hint 2
  Each face has base $12$ cm, but the face altitudes differ.
Answer

$12(\sqrt{52}+\sqrt{28})\text{ cm}^2$ ; unequal face altitudes prevent using either as a common altitude.

Full solution

Two triangles have altitude $\sqrt{52}$ cm.
Their combined area is $12\sqrt{52}$ square centimeters.

The other two have altitude $\sqrt{28}$ cm.
Their combined area is $12\sqrt{28}$ square centimeters.

Adding the pairs gives
$L=12(\sqrt{52}+\sqrt{28})\text{ cm}^2.$

The regular-pyramid shortcut factors out one shared face altitude.
These face altitudes are unequal, so that factorization does not apply.

Answer

$12(\sqrt{52}+\sqrt{28})\text{ cm}^2$ ; unequal face altitudes prevent using either as a common altitude.

Key idea

The sum of actual face areas remains valid when a common-altitude shortcut does not.
Problem 8 Every allowed sector

A circular sector has positive radius $R$ cm and angle $\theta$ degrees, strictly between $0$ and $360$ . Its two straight radii are joined to form a right cone without overlap, using the whole arc as the base rim.

Let $r$ denote the cone base radius and $L$ its lateral area. Derive expressions for both in terms of $R$ and $\theta$ . Explain why the cone has positive perpendicular height.
1. Hint 1
  The sheet radius becomes the slant height.
2. Hint 2
  Match the sector arc to the base circumference.
3. Hint 3
  Use the strict angle bounds to compare the cone radius with its slant height.
Answer

$\begin{gathered}r=\frac{\theta R}{360}\\L=\frac{\theta}{360}\pi R^2\end{gathered}$
The perpendicular height is positive.

Full solution

For an instance, a radius- $16$ cm sector of angle $90$ degrees has arc length $8\pi$ cm.
It makes a cone of radius $4$ cm and slant height $16$ cm, with lateral area $64\pi$ square centimeters.

For general R and theta, match the whole selected arc to the base circumference.
$\begin{aligned}2\pi r&=\frac{\theta}{360}\cdot2\pi R\\r&=\frac{\theta R}{360}.\end{aligned}$

The curved area is exactly the sector area.
$L=\frac{\theta}{360}\pi R^2.$

The angle fraction is strictly between zero and one, so the base radius is strictly between zero and R.
The slant height is R, giving positive square difference $R^2-r^2$ and therefore positive perpendicular height.

Answer

$\begin{gathered}r=\frac{\theta R}{360}\\L=\frac{\theta}{360}\pi R^2\end{gathered}$
The perpendicular height is positive.

Key idea

Matching the arc length determines the radius and also checks that the pattern makes a genuine cone.
Problem 9 Two linked surface records

A closed right cone has total surface area $252\pi\text{ cm}^2$ . Its lateral area is $108\pi\text{ cm}^2$ greater than its base area.

Find its base radius, slant height, and perpendicular height exactly. Check both surface records.
1. Hint 1
  Use the total and the difference to separate base area from lateral area.
2. Hint 2
  Write the lateral area as the base area plus $108\pi$ .
3. Hint 3
  After finding the radius, reverse the lateral-area formula.
Answer

$\begin{gathered}r=6\sqrt2\text{ cm}\\\ell=15\sqrt2\text{ cm}\\h=3\sqrt{42}\text{ cm}\end{gathered}$

Full solution

Let B be base area and L be lateral area.
The two records give
$\begin{aligned}L+B&=252\pi\\L-B&=108\pi.\end{aligned}$
Adding gives lateral area $180\pi$ square centimeters, leaving base area $72\pi$ square centimeters.

The base radius is the positive solution of
$r^2=72.$
Hence the radius is $6\sqrt2$ cm.

Reverse the lateral-area product.
$\begin{aligned}\ell&=\frac{180}{6\sqrt2}\\\ell&=15\sqrt2\text{ cm}.\end{aligned}$

The slant height exceeds the radius.
Subtract their squares.
$\begin{aligned}h^2&=450-72\\h&=3\sqrt{42}\text{ cm}.\end{aligned}$

The recovered base area is $72\pi$ and the lateral area is $180\pi$ , in square centimeters.
Their sum is $252\pi$ and their difference is $108\pi$ , as required.

Answer

$\begin{gathered}r=6\sqrt2\text{ cm}\\\ell=15\sqrt2\text{ cm}\\h=3\sqrt{42}\text{ cm}\end{gathered}$

Key idea

Separate the included areas before recovering lengths from them.
Problem 10 Rebuild the pyramid

A closed regular square pyramid has total area $800\text{ cm}^2$ . Its lateral area is three times its base area.

Find the base side, slant height, and perpendicular height exactly.

Check that the recovered side faces have the required lateral area and that the solid has positive height.
1. Hint 1
  First divide the total into one base share and three lateral shares.
2. Hint 2
  Once the square side is known, its perimeter determines the slant height.
3. Hint 3
  The square apothem is half its side.
Answer

$\begin{gathered}s=10\sqrt2\text{ cm}\\\ell=15\sqrt2\text{ cm}\\h=20\text{ cm}\end{gathered}$
Lateral area $600\text{ cm}^2$ ; height is positive.

Full solution

The total consists of four equal area shares, one for the base and three for the sides.
The base area is $200$ square centimeters and the lateral area is $600$ square centimeters.

The positive square side is
$s=10\sqrt2\text{ cm}.$
The perimeter is $40\sqrt2$ cm.

Use the lateral-area relation to recover slant height.
$\begin{aligned}600&=\tfrac12(40\sqrt2)\ell\\\ell&=15\sqrt2\text{ cm}.\end{aligned}$

The apothem is $5\sqrt2$ cm.
The perpendicular height satisfies
$\begin{aligned}h^2&=(15\sqrt2)^2-(5\sqrt2)^2\\h^2&=400\\h&=20\text{ cm}.\end{aligned}$

Each face has area $150$ square centimeters, because half of $(10\sqrt2)(15\sqrt2)$ is $150$ .
Four faces give the required $600$ square centimeters, and adding the $200$ square-centimeter base returns $800$ .
The height is positive, so the dimensions describe a nondegenerate regular pyramid.

Answer

$\begin{gathered}s=10\sqrt2\text{ cm}\\\ell=15\sqrt2\text{ cm}\\h=20\text{ cm}\end{gathered}$
Lateral area $600\text{ cm}^2$ ; height is positive.

Key idea

Reconstruction needs both the area relations and a compatible perpendicular height.

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