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Cross-Sections and Solids of Revolution: 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 A prism at a new level

A prism has hexagonal bases. A plane cuts its interior parallel to those bases. Describe the section and explain how its size compares with a base. Sketch one possible section on paper. The bases are not assumed to be regular.
1. Hint 1
  Use the orientation of the plane relative to the bases.
2. Hint 2
  Follow the prism edges joining corresponding base vertices.
Answer

A hexagon congruent to a base; a sketch must have six sides and match a possible base shape.

Full solution

Parallel sections of a prism repeat the base.
The joining edges carry corresponding base vertices to the section, preserving the base shape and dimensions.

Thus the section is a congruent hexagon.
A regular hexagon is one possible sketch, but the conditions also allow nonregular hexagons.

Answer

A hexagon congruent to a base; a sketch must have six sides and match a possible base shape.

Key idea

A base-parallel prism section repeats the base, including its size.
Problem 2 A section viewed directly

A right circular cylinder has radius 3.5 cm and height 8 cm. A plane cuts its interior parallel to the bases. Name the true section shape and find its exact area.
1. Hint 1
  Distinguish the true shape from an oval perspective drawing.
2. Hint 2
  A base-parallel section has the cylinder radius. Square that radius for disk area.
Answer

A circular disk of area $\frac{49\pi}{4}$ square cm, or $12.25\pi$ square cm.

Full solution

The section is a circular disk congruent to a base.
Its radius is 3.5 cm, regardless of the plane's level inside the cylinder.

Its exact area is
$\begin{aligned}A&=\pi(3.5)^2\\A&=12.25\pi\text{ cm}^2.\end{aligned}$
The fraction form is $49\pi/4$ square cm.

The right panel faces the cutting plane directly; no section diameter is inferred from the projected oval.

Answer

A circular disk of area $\frac{49\pi}{4}$ square cm, or $12.25\pi$ square cm.

Key idea

Use the actual section radius, not the apparent width of its perspective drawing.
Problem 3 A plane above a base diagonal

A right rectangular box has a 5 cm by 12 cm base and perpendicular height 9 cm. A plane perpendicular to the base contains a base diagonal. Find the shape, both dimensions, and area of the true section.
1. Hint 1
  Trace the plane along the base and the upright edges.
2. Hint 2
  The bottom edge of the section is a base diagonal. Its length comes from a right triangle in the base.
3. Hint 3
  The other section dimension is the box height.
Answer

A rectangle 13 cm by 9 cm, with area 117 square cm.

Full solution

The plane rises perpendicularly above the diagonal.
The diagonal and the upright direction are perpendicular, so the true section is a rectangle.

Find its diagonal width d.
$\begin{aligned}d^2&=5^2+12^2\\d^2&=169\\d&=13\text{ cm}.\end{aligned}$

The section height is 9 cm.
Its area is 13 times 9, or 117 square cm.

The true rectangle is shown beside the perspective projection. All lengths are in centimeters.

Answer

A rectangle 13 cm by 9 cm, with area 117 square cm.

Key idea

A diagonal on the base can become a side of a rectangular plane section.
Problem 4 Two cuts of one pyramid

A right square pyramid has apex directly above the center of its base. Describe the true section in each case. Explain what parts of the pyramid form its boundary.

Case A: An interior plane is parallel to the base.

Case B: A plane contains the apex and a diagonal of the base.
1. Hint 1
  Use the flat faces crossed by each plane.
2. Hint 2
  Case B includes the whole base diagonal and the apex.
Answer

Case A: a smaller square, bounded by the four side faces. Case B: an isosceles triangle, bounded by the base diagonal and two apex-to-vertex edges.

Full solution

Case A: The plane crosses the four side faces parallel to the base edges.
It makes a square smaller than the base because the pyramid narrows toward the apex.

Case B: The diagonal joins two opposite base vertices, and the plane joins both to the apex.
These three segments form a triangle.

The apex lies above the square center, so its distances to those opposite vertices are equal.
Thus Case B is an isosceles triangle.

Case B is a triangle through the apex, not a square parallel to the base.

Answer

Case A: a smaller square, bounded by the four side faces. Case B: an isosceles triangle, bounded by the base diagonal and two apex-to-vertex edges.

Key idea

The same pyramid gives different sections when the plane orientation changes.
Problem 5 Choose a plane for each boundary

For a right circular cylinder of radius 6 cm and height 13 cm, describe one cutting plane producing each requested true section. Every cut must pass through the interior.

Case A: A circle of radius 6 cm.

Case B: A rectangle 12 cm wide and 13 cm high.

Case C: A noncircular ellipse whose whole boundary lies on the curved side.
1. Hint 1
  Relate each plane to the bases or the axis.
2. Hint 2
  For Case C, the plane must tilt without meeting either end disk.
Answer

Case A: parallel to the bases. Case B: through the cylinder axis. Case C: oblique to the bases, sufficiently gently tilted to miss both end disks.

Full solution

Case A: Use any interior plane parallel to the bases.
Its disk has the same radius as a base, 6 cm.

Case B: Use a plane containing the axis.
It cuts each end in a 12 cm diameter, with 13 cm upright edges.

Case C: Start near the middle and tilt a base-parallel plane a small nonzero amount.
Choose a tilt small enough that the whole closed boundary still lies on the curved side, missing the end disks.
Its true section is a noncircular ellipse.

Case C needs a genuine oblique plane, with room above and below the entire cut.

Answer

Case A: parallel to the bases. Case B: through the cylinder axis. Case C: oblique to the bases, sufficiently gently tilted to miss both end disks.

Key idea

Section shape depends on the plane orientation and which surfaces it intersects.
Problem 6 A turning template

A maker uses the filled right triangle ABC as a rotating template. The right angle is A, leg AB is 12 cm, and leg AC is 9 cm. The template turns a full revolution about AB.

Name the generated solid and give its height, base radius, and base diameter. State what boundary the hypotenuse sweeps.

The dashed line through AB names the axis. The picture shows only the given template.

Text description of this figure

Triangle ABC is right at A, with vertical axis AB of 12 centimeters and perpendicular leg AC of 9 centimeters.
1. Hint 1
  Identify which leg stays fixed and which reaches away from the axis.
2. Hint 2
  The distance from the axis to C is a radius, not a diameter.
Answer

A right circular cone of height 12 cm, radius 9 cm, and diameter 18 cm. The hypotenuse sweeps the curved side.

Full solution

AB stays fixed and becomes the 12 cm height-axis.
AC reaches 9 cm perpendicularly from that axis and sweeps the circular base disk.

The base radius is 9 cm, so its diameter is 18 cm.
The hypotenuse sweeps the cone's curved side, while B stays at the apex.

The fixed leg supplies height; the perpendicular leg supplies radius. The perspective cone is schematic.

Answer

A right circular cone of height 12 cm, radius 9 cm, and diameter 18 cm. The hypotenuse sweeps the curved side.

Key idea

A rotating right triangle gives a cone when the axis is one of its legs.
Problem 7 Exchange the fixed side

Rectangle ABCD has AB of 4 cm and perpendicular AD of 11 cm. Give the cylinder height, radius, and axial-section dimensions for each full rotation.

Case A: Turn the filled rectangle about AB.

Case B: Turn the filled rectangle about AD.

The two possible axes are named in the text; no generated solid is drawn.

Text description of this figure

Rectangle ABCD has vertical side AB of 4 centimeters and perpendicular side AD of 11 centimeters. The lines through AB and AD are the two stated axis choices.
1. Hint 1
  The fixed side gives the height in each case.
2. Hint 2
  The axial section is twice the radius wide and one height tall.
Answer

Case A: height 4 cm, radius 11 cm, axial section 22 cm by 4 cm. Case B: height 11 cm, radius 4 cm, axial section 8 cm by 11 cm.

Full solution

Case A: AB is the axis, so height is 4 cm.
The 11 cm perpendicular side sweeps radius 11 cm.
The axial section spans a 22 cm diameter and the 4 cm height.

Case B: AD is the axis, so height is 11 cm.
The 4 cm perpendicular side sweeps radius 4 cm.
The axial section spans an 8 cm diameter and the 11 cm height.

Answer

Case A: height 4 cm, radius 11 cm, axial section 22 cm by 4 cm. Case B: height 11 cm, radius 4 cm, axial section 8 cm by 11 cm.

Key idea

Changing the named axis exchanges the height and radius roles, not height and diameter.
Problem 8 A slice away from the center

A sphere has radius 17 cm. A plane cuts its interior at a perpendicular distance of 8 cm from the center. Find the radius of the circle where the plane meets the sphere and the exact area of the corresponding disk in the ball.

The center view is a flat plane containing O, M, and P. The true circular section is shown beside it.

Text description of this figure

The sphere center O, perpendicular foot M, and boundary point P form a right triangle with OP of 17 centimeters, OM of 8 centimeters, and MP unknown.
1. Hint 1
  Join the sphere center to the section center and to a boundary point.
2. Hint 2
  The sphere radius is the hypotenuse; the perpendicular center distance is one leg.
3. Hint 3
  Use the positive section radius, then the disk-area rule.
Answer

Section radius 15 cm; disk area $225\pi$ square cm.

Full solution

Let O be the sphere center, M the perpendicular foot in the cutting plane, and P a point on the section circle.
M is the circle center, and triangle OMP is right at M.

Its hypotenuse OP is 17 cm, and leg OM is 8 cm.
The section radius r satisfies
$\begin{aligned}r^2&=17^2-8^2\\r^2&=225\\r&=15\text{ cm}.\end{aligned}$

The corresponding ball section fills that circle.
Its disk area is
$A=225\pi\text{ cm}^2.$

Answer

Section radius 15 cm; disk area $225\pi$ square cm.

Key idea

The radius of an off-center sphere section is a leg of a right triangle.
Problem 9 Identify a solid from several cuts

An unknown solid is one of a right circular cylinder, a right circular cone, or a ball. It has a stated symmetry axis. Every interior plane perpendicular to that axis cuts a disk of the same positive radius.

Which solid is it? Describe a plane making a rectangular section, and explain why the other two candidates fail the disk condition.
1. Hint 1
  Think about what happens as a perpendicular plane moves along each candidate's axis.
2. Hint 2
  A cone narrows toward an apex; ball sections change size toward the ends.
Answer

A right circular cylinder. A plane through its axis makes a rectangle. Cone sections shrink toward the apex; ball sections shrink away from the central plane.

Full solution

A cylinder repeats the same circular section throughout its interior height, so it meets the condition.

A cone's perpendicular disks shrink toward its apex.
A ball's disks are largest at its center and shrink toward either end.
Neither candidate has a constant positive section radius.

For the cylinder, choose a plane containing its axis.
Its boundary uses two end diameters and two upright side segments, making a rectangle.

Answer

A right circular cylinder. A plane through its axis makes a rectangle. Cone sections shrink toward the apex; ball sections shrink away from the central plane.

Key idea

A family of specified sections can distinguish solids that share one circular slice.
Problem 10 Recover three generating regions

For each solid, give a filled region of the allowed kind and name its axis of rotation. Check that the resulting dimensions match.

Case A: A right cylinder has an axial rectangle 18 cm wide and 7 cm high. Use a rectangle turning about a side.

Case B: A right cone has an axial triangle with base 18 cm and perpendicular height 7 cm. Use a right triangle turning about a leg.

Case C: A ball has diameter 18 cm. Use a semicircular region turning about its diameter.
1. Hint 1
  The full axial width is twice the radius for both the cylinder and the cone.
2. Hint 2
  In Cases A and B, the axis length is the stated height.
3. Hint 3
  For Case C, use half a disk with the ball radius.
Answer

A: rectangle 9 cm by 7 cm, about a 7 cm side. B: right triangle with legs 9 cm and 7 cm, about its 7 cm leg. C: semicircular region of radius 9 cm, about its diameter.

Full solution

Case A: Half the 18 cm axial width is radius 9 cm.
A rectangle reaching 9 cm from a fixed 7 cm side produces height 7 cm and diameter 18 cm, as required.

Case B: The cone also has radius 9 cm and height 7 cm.
Use perpendicular legs of those lengths and fix the 7 cm leg.
The whole axial triangle has a base twice the radius, 18 cm, and height 7 cm.

Case C: The ball radius is half its diameter, or 9 cm.
A filled semicircular region of radius 9 cm turns about its 18 cm diameter and sweeps the entire ball.

Answer

A: rectangle 9 cm by 7 cm, about a 7 cm side. B: right triangle with legs 9 cm and 7 cm, about its 7 cm leg. C: semicircular region of radius 9 cm, about its diameter.

Key idea

Work from full axial width to radius before choosing the generating half-region.

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