Solid Geometry: Chapter Test
20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.
Multiple choice
20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.
Core practice
10 problems from across the chapter. 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)
0 of 10 done · 0 skipped · 0 to revisit
Progress saved in this browser.
Progress can't be saved in this browser, so your choices last for this visit only.
-
Problem 1 Two cuts through a cylinder
A right circular cylinder is cut in two ways. Give the true section shape in each case and explain what parts of the cylinder bound it.
Cut A. The cutting plane is tilted relative to the circular bases. It passes through the interior and misses both end disks.
Cut B. The cutting plane contains the line through the two base centers.
- Hint 1
Imagine looking straight at each cutting plane so perspective cannot change the apparent shape.
- Hint 2
For Cut A, follow where the plane meets the curved side. For Cut B, trace the line through each end disk and the segments connecting their endpoints.
Answer
Cut A gives a noncircular ellipse. Cut B gives a rectangle.
Full solution
Cut A.
The tilt gives a noncircular ellipse because the plane cuts all the way around the curved side without being interrupted by either end disk.
The curved side supplies the entire boundary.
Cut B.
The plane cuts a diameter across each end disk.
The two diameters are parallel, and their corresponding endpoints are joined by two segments along the side of the cylinder.
These segments run perpendicular to the end disks, so the four segments bound a rectangle.
Answer
Cut A gives a noncircular ellipse. Cut B gives a rectangle.
Key idea
A section's true shape depends on which parts of the solid its plane intersects.
- Hint 1
-
Problem 2 Two colors for a roof
A model roof is a pyramid with a rectangular base by . Its apex is directly above the center of the rectangle.
The two faces on the edges use blue material, and the two faces on the edges use gray material. How much area of each material is needed? Give exact values and identify which color covers more area.
- Hint 1
Draw a section through the apex, the base center, and the midpoint of a blue face edge. Repeat for a gray face edge.
- Hint 2
The midpoint of a short edge is half the long dimension from the center. The midpoint of a long edge is half the short dimension from the center.
Answer
The blue faces need , and the gray faces need . The gray material covers more.
Full solution
Blue.
A short-edge midpoint is from the center.
With the perpendicular height, the slant height in centimeters is
Two such faces use square centimeters.
Gray.
A long-edge midpoint is from the center.
The slant height in centimeters is
Two such faces use square centimeters.
Both areas are positive.
Since , the gray area is larger.
Answer
The blue faces need , and the gray faces need . The gray material covers more.
Key idea
A rectangle can give two different face altitudes even when the apex is above its center.
- Hint 1
-
Problem 3 One container or many
A sealed sample container holds and has of exterior surface. One plan uses copies, coated separately. A second plan uses a single geometrically similar container with the same capacity as all copies together.
Find the total exterior area coated under each plan. What fraction of the first plan’s area does the second plan coat?
- Hint 1
First express the capacity of the single large container as a multiple of one small container.
- Hint 2
Recover the length factor from the capacity factor. Then compare its area with the sum of the small containers’ separate areas.
Answer
The small containers require . The single large container requires , one third as much.
Full solution
Coating the small copies separately adds all their areas.
In square centimeters,
The similar large container holds times as much as one copy.
Its length factor is , and its exterior area factor is .
Its area in square centimeters is
The second plan uses the fraction of the first plan’s coating area.
Answer
The small containers require . The single large container requires , one third as much.
Key idea
A collection of separate copies adds surface areas linearly, while one similar enlargement follows the square of its length factor.
- Hint 1
-
Problem 4 An open trough
An ideal straight trough is a right prism with two identical isosceles trapezoidal ends. At each end the bottom edge is , the open top edge is , the perpendicular depth is , and each sloped edge is .
The trough holds when filled to the rim. Find its length and the area of sheet material covering its bottom, two sloping sides, and two trapezoidal ends. Ignore thickness, seams, and waste.
Text description of this figure
Schematic trapezoidal end of the trough. The top edge spans an open 19-centimeter rim, the bottom is 9 centimeters wide, the perpendicular depth is 12 centimeters, and each sloped side is 13 centimeters.
- Hint 1
The shape at either end also describes the water at every position along a full trough.
- Hint 2
Find the trapezoid area first. The three long panels are rectangles whose widths are the three covered edges of an end.
Answer
The trough is long and requires of sheet material.
Full solution
The area of a trapezoidal end in square centimeters is
All cross-sections parallel to the ends have area .
The length in centimeters is
The bottom and two sloping side rectangles use a total of square centimeters.
The two ends use square centimeters.
There is no top panel.
The sheet area in square centimeters is
Answer
The trough is long and requires of sheet material.
Key idea
Capacity uses the filled section, while material area counts only the panels present.
- Hint 1
-
Problem 5 A footprint with one missing site
A solid is built from filled vertical columns of unit cubes on a grid two sites wide and three sites deep. The top footprint contains every site except the middle site in the left column. Every occupied site has positive integer height.
From left to right, the two front silhouette heights are . From front to back, the three right silhouette heights are . Only the outer silhouettes are specified.
Give two different height plans that fit all three views. Label the front, middle, and back rows, giving left and right heights in each row. Which occupied sites have a uniquely determined height? Explain.
Text description of this figure
The top footprint has two columns and three rows, with the front at the bottom. Five occupied sites are marked with crosses. Only the middle-left site is empty.
- Hint 1
Each front silhouette height is a maximum in one left-to-right position. Each right silhouette height is a maximum in one front-to-back row.
- Hint 2
Look for a row with just one occupied site. Then ask where the left column can reach height without exceeding a row's maximum.
- Hint 3
To test whether any other site is fixed, try varying the front row and back-right site while preserving all the maxima.
Answer
From front to back, one pair is
Only middle-right (height ) and back-left (height ) are fixed.
Full solution
The middle row has only its right site occupied, so the row maximum of fixes that height.
The left column must reach .
Its middle site is empty, and the front row has maximum .
The back-left site therefore has height .
Plan 1.
Set the front row to , the middle row to , and the back row to .
The column maxima are , the row maxima are , and only the middle-left site is empty.
Plan 2.
Set the front row to , the middle row to , and the back row to .
The maxima and occupied sites are unchanged.
The two plans disagree at front-left, front-right, and back-right.
Each of those heights can vary, so only the middle-right and back-left heights are fixed.
Answer
From front to back, one pair is
Only middle-right (height ) and back-left (height ) are fixed.
Key idea
Silhouette maxima can force some column heights while leaving others free.
- Hint 1
-
Problem 6 A pyramid from a corner measurement
A regular square pyramid has base side . The distance from the apex to a base corner is .
Find its exact volume and its lateral surface area.
- Hint 1
In the base, the distance from the center to a corner differs from the distance from the center to an edge midpoint.
- Hint 2
Use the center-to-corner distance with the measured edge to find perpendicular height. Use the center-to-midpoint distance with that height to find the face altitude.
Answer
The volume is , and the lateral surface area is .
Full solution
Half of a base diagonal is .
The apex, base center, and base corner form a right triangle.
The height in centimeters is
The square base has area .
The volume in cubic centimeters is
A base edge midpoint is from the center.
The face altitude in centimeters is
Four faces with base and altitude have combined area square centimeters.
The two horizontal legs start at the base center. Lengths are in centimeters. Answer
The volume is , and the lateral surface area is .
Key idea
A slanted edge goes to a corner. A face altitude goes to an edge midpoint. Their right triangles have different base legs.
- Hint 1
-
Problem 7 Two spins of the same rectangle
A filled rectangle has a marked side of length . The other side length is unknown. A full turn about the marked side sweeps a solid of volume .
Find the unknown side length. Then find the exact volume swept by a full turn of the original filled rectangle about a side of that unknown length.
Text description of this figure
Schematic rectangle. Its bottom side is the marked 8-centimeter side on the first rotation axis. The length of each perpendicular side is unknown.
- Hint 1
For each turn, distinguish the length that stays on the axis from the distance swept out away from the axis.
- Hint 2
For the first turn, the height is known and the square of the radius can be recovered from the volume. The roles of the two lengths switch in the second turn.
Answer
The unknown side is . The second turn sweeps a volume of .
Full solution
First turn.
The marked side gives height .
If the other side measures centimeters, it gives the radius of the resulting cylinder.
In cubic centimeters the volume condition is
Thus .
A side length is positive, so centimeters.
Second turn.
The axis now has length and the radius is .
The exact volume in cubic centimeters is
Answer
The unknown side is . The second turn sweeps a volume of .
Key idea
Switching a rectangle's rotation axis swaps the two lengths, and the off-axis length enters the volume squared.
- Hint 1
-
Problem 8 A model of a hollow casting
A casting's exterior shape encloses before a cavity is made. The cavity removes of material, and no other material is removed.
A model is made by multiplying all lengths of both the exterior and the cavity by .
Find the volume of material in the model and the fraction of the model's exterior-enclosed volume that is material. Explain how this fraction compares with that of the original casting.
- Hint 1
Think of the actual material as the part left after the cavity volume is removed.
- Hint 2
Both the full exterior-enclosed volume and the cavity volume are multiplied by the same cube of the length factor.
Answer
The model contains of material. The fraction filled by material is , the same as for the original.
Full solution
The original material volume in cubic centimeters is
Both exterior and cavity volumes change by factor
The difference changes by the same factor, so the model material volume in cubic centimeters is
The common factor cancels when material volume is divided by the full exterior-enclosed volume.
In either casting, the resulting fraction is
Answer
The model contains of material. The fraction filled by material is , the same as for the original.
Key idea
A uniform resize changes material volume by a cube factor while preserving the fraction occupied by material.
- Hint 1
-
Problem 9 Two hemispheres joined at their disks
Two solid hemispheres have radii and . Their flat disks lie in the same plane with the same center. The curved caps extend on opposite sides of the plane. The full smaller disk is glued to the larger disk, and the interiors do not overlap.
Find the exact volume and total exposed area of the assembly. Include every exposed portion of either flat disk.
Text description of this figure
Schematic larger hemispherical cap above the common disk plane and smaller cap below it. The disk outlines have the same center. A dashed semicircular arc marks the part of the smaller rim behind its cap.
- Hint 1
Decide which parts of the two flat disks are hidden by the join.
- Hint 2
Add the two hemisphere volumes. For exposed area, count the two curved caps and find the area of the flat ring left around the smaller disk.
Answer
The volume is and the exposed area is .
Full solution
The glued disks only meet at the common plane, so the enclosed volumes add.
In cubic centimeters,
The two curved caps are completely exposed.
Their combined area in square centimeters is
The smaller disk is fully hidden.
The larger disk has an exposed ring with area in square centimeters
The two caps and ring give an exposed area of square centimeters.
Answer
The volume is and the exposed area is .
Key idea
Unequal joining disks can leave a flat ring exposed even when both curved caps remain uncovered.
- Hint 1
-
Problem 10 Same area at one level
A right circular cylinder and a right circular cone each extend from a lower horizontal plane to an upper horizontal plane higher. The cone's apex is on the lower plane and its base is on the upper plane. Halfway between the planes, both horizontal sections have area .
Find the exact volume of each solid. Explain why the matching halfway sections are insufficient to establish equal volumes.
Schematic profiles. Use the stated section areas. Text description of this figure
A cylinder and a downward-pointing cone span the same two horizontal levels. A dashed line marks halfway. The two highlighted halfway cross-sections have the same radius.
- Hint 1
One solid has a constant section area. For the other, look at how a section radius changes with distance from its apex.
- Hint 2
The halfway section radius is half the cone's base radius. Think about the effect of doubling that length on the area before finding the cone's volume.
- Hint 3
Compare sections at a height below the halfway level and at a height above it. Cavalieri's comparison needs equal areas at every corresponding height.
Answer
The cylinder volume is . The cone volume is . The cone's sections are smaller below halfway and larger above halfway.
Full solution
Cylinder.
Every horizontal section has area square centimeters.
Its volume in cubic centimeters is
Cone.
Similar triangles through the axis show that the section radius grows in proportion to height measured from the apex.
The base radius is twice the halfway radius, so the base area is square centimeters.
The cone volume in cubic centimeters is
Below halfway, the cone section has radius smaller than at halfway and thus area below square centimeters.
Above halfway, the cone section is larger.
The cylinder area stays at square centimeters, so matching halfway fails the condition of matching at every height.
Answer
The cylinder volume is . The cone volume is . The cone's sections are smaller below halfway and larger above halfway.
Key idea
Equal area at one height cannot replace the full set of corresponding section areas in a volume comparison.
- Hint 1