Check: A box with the same 6-by-6 base and 10 in height would hold 36 \times 10 = 360\text{ in}^3. A pyramid holds exactly one-third of that box, and 360 \div 3 = 120 ✓.
Tip:B is the area of the base, not the length of one side. For every pyramid, find the matching prism and divide by 3.
Question 2 of 36
Fill in
A cone has a radius of 3 cm and a height of 7 cm. What is the volume of the cone, rounded to the nearest tenth of a cubic centimeter?
cm³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The cone formula is on the formula sheet: V = \tfrac{1}{3}\pi r^2 h. Square the radius before you do anything else.
Step-by-step solution
Write the formula (on the formula sheet): V = \tfrac{1}{3}\pi r^2 h.
Square the radius: r^2 = 3^2 = 9.
Substitute: V = \tfrac{1}{3}\pi(9)(7) = \tfrac{1}{3}\pi(63) = 21\pi.
Use the \pi key: 21\pi \approx 65.97.
Round to the nearest tenth: 65.97 \approx 66.0.
Answer: about 66.0 cm³
Check: A cylinder with the same radius and height holds \pi(9)(7) = 63\pi \approx 197.9. One-third of 197.9 is about 66.0 ✓.
Tip: Leave \pi until the last step. Writing the exact answer 21\pi first keeps rounding errors out. (With \pi \approx 3.14 you get 65.94 \approx 65.9, which is also accepted.)
Question 3 of 36
The square pyramid below has base edges of 10 ft, a height of 12 ft, and a slant height of 13 ft.
What is the total surface area of the pyramid, including the base?
Hint
Surface area uses the slant height, not the height. The formula sheet gives SA = \tfrac{1}{2}ps + B.
Step-by-step solution
Find the perimeter of the base: p = 4 \times 10 = 40\text{ ft}.
Find the base area: B = 10 \times 10 = 100\text{ ft}^2.
Use the formula sheet: SA = \tfrac{1}{2}ps + B, with slant height s = 13.
Lateral area (the 4 triangles): \tfrac{1}{2}(40)(13) = 20 \times 13 = 260\text{ ft}^2.
Add the base: 260 + 100 = 360.
Answer: 360 ft²
Check: Each triangular face has area \tfrac{1}{2}(10)(13) = 65. Four faces: 4 \times 65 = 260. Plus the base 100 gives 360 ✓.
Tip: The height (12 ft) is a distractor here. Height goes in volume formulas; slant height goes in surface area formulas.
Question 4 of 36
A museum gift shop sells a stone model of a pyramid. The model is a square pyramid with base edges of 9 cm and a height of 8 cm.
What is the volume of the model?
Hint
Find the area of the square base first. Then use V = \tfrac{1}{3}Bh from the formula sheet.
Step-by-step solution
The formula for any pyramid is V = \tfrac{1}{3}Bh, where B is the area of the base (on the formula sheet).
The base is a square with side 9 cm, so B = 9^2 = 81\text{ cm}^2.
Multiply by the height: Bh = 81 \times 8 = 648.
Take one-third: V = \tfrac{1}{3}(648) = 216.
Answer: 216 cm³
Check: A box with the same base and height holds 81 \times 8 = 648\text{ cm}^3. A pyramid holds exactly one-third of that box, and 648 \div 3 = 216 ✓.
Tip:B is an area, not a side length. For a square base, always square the edge before you multiply by the height.
Question 5 of 36
Fill in
A candle maker pours wax into a cone-shaped mold. The mold has a radius of 5 cm and a height of 12 cm.
How many cubic centimeters of wax fill the mold? Round to the nearest whole number.
cm³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Use V = \tfrac{1}{3}\pi r^2 h. Square the radius first, and press the \pi key only at the end.
Step-by-step solution
Formula (on the formula sheet): V = \tfrac{1}{3}\pi r^2 h.
Square the radius: r^2 = 5^2 = 25.
Multiply by the height: 25 \times 12 = 300.
Take one-third: \tfrac{1}{3}(300) = 100, so V = 100\pi.
Use the \pi key: 100\pi \approx 314.16, which rounds to 314.
Answer: about 314 cm³
Check: A cylinder with the same radius and height holds \pi(25)(12) = 300\pi \approx 942.5\text{ cm}^3. One-third of 942.5 is about 314 ✓.
Tip: Keep the exact form (100\pi) until the last step. With \pi \approx 3.14 you get 314 as well, so either way of using \pi gives the same rounded answer here.
Question 6 of 36
A glass skylight is shaped like a pyramid with a rectangular base. The base is 12 ft long and 5 ft wide, and the skylight is 4 ft tall.
What is the volume of the space inside the skylight?
Hint
The base is a rectangle, so its area is length × width. Then use the pyramid formula.
Step-by-step solution
Pyramid formula (on the formula sheet): V = \tfrac{1}{3}Bh.
The base is a rectangle: B = 12 \times 5 = 60\text{ ft}^2.
Multiply by the height: 60 \times 4 = 240.
Take one-third: V = \tfrac{1}{3}(240) = 80.
Answer: 80 ft³
Check: Work backwards: h = \dfrac{3V}{B} = \dfrac{3(80)}{60} = 4 ft, which matches the height in the problem ✓.
Tip: The pyramid formula V = \tfrac{1}{3}Bh works for any base shape. Only the way you find B changes: square, rectangle, triangle, and so on.
Question 7 of 36
A café serves drinks in two kinds of glasses. One glass is a cylinder, and the other is a cone. Both glasses have the same radius and the same height. The cylinder glass holds 18 fluid ounces.
How many fluid ounces does the cone glass hold?
Hint
Compare the two formulas: V = \pi r^2 h and V = \tfrac{1}{3}\pi r^2 h. What is the only difference?
Step-by-step solution
Cylinder: V = \pi r^2 h.
Cone with the same r and h: V = \tfrac{1}{3}\pi r^2 h.
The only difference is the factor \tfrac{1}{3}. So the cone holds one-third of the cylinder.
\tfrac{1}{3} \times 18 = 6.
Answer: 6 fluid ounces
Check: Three full cone glasses would fill the cylinder glass: 3 \times 6 = 18 fluid ounces ✓.
Tip: Remember the pair "cone = \tfrac{1}{3} cylinder" and "pyramid = \tfrac{1}{3} prism" (same base, same height). Many GED questions can be solved with this fact and no calculator at all.
Question 8 of 36
A solid cone has a radius of 6 cm, a height of 8 cm, and a slant height of 10 cm, as shown.
What is the total surface area of the cone, in terms of \pi?
Hint
Use SA = \pi r s + \pi r^2. The s is the slant height, measured along the side of the cone.
Step-by-step solution
Formula (on the formula sheet): SA = \pi r s + \pi r^2.
Curved (lateral) part: \pi r s = \pi(6)(10) = 60\pi.
Circular base: \pi r^2 = \pi(6^2) = 36\pi.
Add: 60\pi + 36\pi = 96\pi.
Answer: 96\pi cm² (about 301.6\text{ cm}^2)
Check: The three lengths fit the Pythagorean theorem, 6^2 + 8^2 = 36 + 64 = 100 = 10^2, so 10 really is the slant height, and the answer uses 10 (not 8) in the \pi r s term ✓.
Tip: In a cone, r, h, and s form a right triangle with s as the hypotenuse. The slant height is always the longest of the three.
Question 9 of 36
Fill in
A paper snow-cone cup is shaped like a cone. The opening has a diameter of 3 inches, and the cup is 4 inches deep. How many cubic inches of shaved ice does the cup hold when filled level with the top? Round to the nearest tenth.
in³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The formula needs the radius. What is half of the diameter?
Step-by-step solution
Find the radius: r = 3 \div 2 = 1.5\text{ in}.
Formula (on the formula sheet): V = \tfrac{1}{3}\pi r^2 h.
Square the radius: 1.5^2 = 2.25.
Substitute: V = \tfrac{1}{3}\pi(2.25)(4) = \tfrac{1}{3}\pi(9) = 3\pi.
Evaluate: 3\pi \approx 9.42, which rounds to 9.4.
Answer: about 9.4 in³
Check: A can with the same size opening and depth would hold \pi(2.25)(4) = 9\pi \approx 28.3. The cone holds one-third of that: 28.3 \div 3 \approx 9.4 ✓.
Tip: Word problems often give the diameter (the width across the top). Always cut it in half before squaring. Using 3 as the radius would make the answer four times too big.
Question 10 of 36
Drop-down
The rectangular pyramid below has a base that is 8 m long and 6 m wide. Its volume is 192 m³.
Complete the statements.
The area of the base is m², so the height of the pyramid is m.
Choose an answer from each menu.
Hint
Put the known values into V = \tfrac{1}{3}Bh and solve for h. To undo \tfrac{1}{3}, multiply by 3.
Step-by-step solution
Base area of the rectangle: B = 8 \times 6 = 48\text{ m}^2.
Formula: V = \tfrac{1}{3}Bh, so 192 = \tfrac{1}{3}(48)h.
Tip: When you solve for a radius, the last step is almost always a square root. If your answer is a perfect square like 64, ask yourself whether you forgot that step.
Question 12 of 36
Select all
Select all of the cones that have a volume of exactly 12\pi cubic units.
Select all that apply.
Hint
For each cone, compute \tfrac{1}{3}r^2 h (the number in front of \pi). Watch for any cone given by its diameter.
Step-by-step solution
For each cone, V = \tfrac{1}{3}\pi r^2 h. We need \tfrac{1}{3}r^2 h = 12, which means r^2 h = 36.
r = 3, h = 4: r^2 h = 9 \times 4 = 36 → V = 12\pi ✓
r = 4, h = 3: r^2 h = 16 \times 3 = 48 → V = 16\pi ✗
diameter 4, so r = 2, h = 9: r^2 h = 4 \times 9 = 36 → V = 12\pi ✓
r = 2, h = 3: r^2 h = 4 \times 3 = 12 → V = 4\pi ✗
r = 6, h = 1: r^2 h = 36 \times 1 = 36 → V = 12\pi ✓
Answer: A, C, and E
Check: Each correct cone has r^2 h = 36, and \tfrac{36}{3} = 12 ✓.
Tip: Radius and height are not interchangeable — only the radius is squared. Also, very different-looking cones can hold the same amount.
Question 13 of 36
A square pyramid has base edges of 16 in and a height of 15 in.
What is the slant height s of the pyramid?
Hint
Look for the right triangle inside the pyramid: the height, half of a base edge, and the slant height.
Step-by-step solution
The height meets the base at its center. From the center to the middle of a base edge is half of the edge: 16 \div 2 = 8\text{ in}.
The height (15), that half-edge (8), and the slant height s form a right triangle with s as the hypotenuse.
Check: 8-15-17 is a known Pythagorean triple: 64 + 225 = 289 = 17^2 ✓. Also, s (17) is longer than h (15), as it must be.
Tip: For a square pyramid, s^2 = h^2 + \left(\tfrac{b}{2}\right)^2. For a cone, s^2 = h^2 + r^2. Always use half the base edge for a pyramid.
Question 14 of 36
A gift box is shaped like a closed square pyramid. Each edge of the square base is 6 in, and the slant height of each triangular face is 5 in.
How much cardboard covers the whole box, including the bottom?
Hint
Count every face: one square base and four triangles. Each triangle has base 6 in and height 5 in (the slant height).
Step-by-step solution
Area of the square base: 6 \times 6 = 36\text{ in}^2.
Area of one triangular face: \tfrac{1}{2}bh = \tfrac{1}{2}(6)(5) = 15\text{ in}^2. The slant height is the height of the triangle.
Four triangles: 4 \times 15 = 60\text{ in}^2 (the lateral area).
Total surface area: 36 + 60 = 96.
Answer: 96 in²
Check: The formula sheet gives SA = \tfrac{1}{2}ps + B, where p is the base perimeter. p = 4 \times 6 = 24, so \tfrac{1}{2}(24)(5) + 36 = 60 + 36 = 96 ✓.
Tip: Surface area is a sum of flat faces. Make a quick list (1 square + 4 triangles) so no face is forgotten, and leave out the bottom only when the problem says the figure has no floor.
Question 15 of 36
A paper water cup is shaped like a cone with no lid. Its radius is 4 cm, and its slant height is 9 cm.
How much paper is needed to make the cup, in terms of \pi?
Hint
An open cup has only the curved side. The curved (lateral) area of a cone is \pi r s.
Step-by-step solution
The cup is open at the top, so there is no circular base to cover. We need only the lateral (curved) area.
The formula sheet gives the cone's surface area as SA = \pi r s + \pi r^2. The part \pi r s is the curved side; \pi r^2 is the circle.
Substitute r = 4 and s = 9: \pi r s = \pi(4)(9) = 36\pi.
Answer: 36\pi cm² (about 113.1 cm²)
Check: If the cup were closed, you would add \pi(4)^2 = 16\pi to get 52\pi. Since the cup has no lid, 36\pi is the right amount ✓.
Tip: Read the situation: cups, funnels, party hats and ice-cream cones have no base, so use only \pi r s. A solid cone uses \pi r s + \pi r^2.
Question 16 of 36
Fill in
A cone-shaped tank in a science lab holds 96\pi cubic inches of liquid when full. The radius of the tank's base is 6 inches.
What is the height of the tank?
in
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Put V = 96\pi and r = 6 into V = \tfrac{1}{3}\pi r^2 h. The \pi on both sides cancels.
Step-by-step solution
Formula: V = \tfrac{1}{3}\pi r^2 h, so 96\pi = \tfrac{1}{3}\pi(6^2)h.
Square the radius: 6^2 = 36, so 96\pi = \tfrac{1}{3}\pi(36)h = 12\pi h.
Tip: When the volume is given "in terms of \pi," the \pi cancels right away, and no calculator rounding is needed. The shortcut h = \dfrac{3V}{\pi r^2} gives \dfrac{3(96)}{36} = 8.
Question 17 of 36
A crystal paperweight is a square pyramid with a volume of 100 cm³ and a height of 12 cm.
What is the length of each edge of its square base?
Hint
Solve 100 = \tfrac{1}{3}s^2(12) for s^2 first. What final step turns s^2 into s?
Step-by-step solution
For a square base, B = s^2, so V = \tfrac{1}{3}s^2 h.
Tip: Solving for a side of a square base always ends with a square root. Ask yourself whether your answer is the edge s or the area s^2 before you choose.
Question 18 of 36
Fill in
A chocolate shop makes candies in molds shaped like square pyramids. Each mold has base edges of 3 cm and a height of 5 cm.
How many cubic centimeters of melted chocolate fill one mold?
cm³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Square the base edge to get the base area B, then use V = \tfrac{1}{3}Bh.
Step-by-step solution
Formula (on the formula sheet): V = \tfrac{1}{3}Bh.
Base area of the square: B = 3^2 = 9\text{ cm}^2.
Multiply by the height: Bh = 9 \times 5 = 45.
Take one-third: V = 45 \div 3 = 15.
Answer: 15 cm³
Check: A square box 3 cm by 3 cm by 5 cm holds 45\text{ cm}^3, and three pyramid molds would fill it: 3 \times 15 = 45 ✓.
Tip: Multiply in whatever order is easiest. Here \tfrac{1}{3} of 9 is 3, and 3 \times 5 = 15 — fast enough to do in your head.
Question 19 of 36
Fill in
A paperweight is a pyramid with a triangular base, as shown. The base is a right triangle with legs of 9 in and 4 in. The height of the pyramid is 10 in.
What is the volume of the paperweight?
in³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First find the area of the triangle base with \tfrac{1}{2}bh. Then use that area as B in V = \tfrac{1}{3}Bh.
Step-by-step solution
The base is a right triangle, so its two legs are its base and height: B = \tfrac{1}{2}(9)(4) = 18\text{ in}^2.
Pyramid formula: V = \tfrac{1}{3}Bh with h = 10 (the height of the pyramid, not of the triangle).
V = \tfrac{1}{3}(18)(10) = \tfrac{1}{3}(180) = 60.
Answer: 60 in³
Check: A triangular prism with the same base and height would hold 18 \times 10 = 180\text{ in}^3; the pyramid is one-third of that, 60 ✓.
Tip: A triangular pyramid uses two fractions: \tfrac{1}{2} for the triangle's area and \tfrac{1}{3} for the pyramid. Forgetting the \tfrac{1}{2} gives 120, a common wrong answer.
Question 20 of 36
A landscaping company dumps gravel in a cone-shaped pile. The pile has a radius of 6 ft and a height of 4 ft. The gravel costs $1.50 per cubic foot.
To the nearest dollar, what is the cost of the gravel in the pile?
Hint
Find the volume of the cone first. Then think: if each cubic foot costs $1.50, do you multiply or divide?
Step-by-step solution
Volume of the cone: V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(6^2)(4) = \tfrac{1}{3}\pi(144) = 48\pi.
Use the \pi key: 48\pi \approx 150.80\text{ ft}^3.
Each cubic foot costs $1.50, so multiply: 150.80 \times 1.50 \approx 226.19.
Round to the nearest dollar: $226.
Answer: $226
Check: Estimate: the pile is about 150 cubic feet, and 150 \times 1.5 = 225. That is very close to $226 ✓.
Tip: "Price per cubic foot" means cost = volume × price. Find the volume first, keep a couple of decimals, and round only the final money amount.
Question 21 of 36
Jordan found the volume of a cone with a diameter of 10 in and a height of 6 in. Here is the work.
V = \tfrac{1}{3}\pi r^2 h
r = 10
V = \tfrac{1}{3}\pi(100)(6)
V = 200\pi \text{ in}^3
In which step did Jordan make the first mistake?
Hint
Check each step against the facts in the problem. What is the radius when the diameter is 10?
Step-by-step solution
Step 1 is the correct cone formula from the formula sheet.
Step 2 is wrong: 10 in is the diameter. The radius is half of it, r = 10 \div 2 = 5.
Steps 3 and 4 do the arithmetic correctly for r = 10, but they start from the wrong radius.
Check: Jordan's answer 200\pi is 4 times the correct 50\pi. Doubling the radius multiplies the volume by 2^2 = 4, which is exactly what using the diameter does ✓.
Tip: Before you substitute, circle the word "radius" or "diameter" in the problem. If it says diameter, halve it first.
Question 22 of 36
Fill in
A children's play tent is shaped like a square pyramid with no floor. The base is 8 ft on each side, and the center pole is 3 ft tall.
How many square feet of fabric are needed to make the four triangular sides?
ft²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The fabric covers only the triangles, so you need the lateral area \tfrac{1}{2}ps. First find the slant height.
Step-by-step solution
Find the slant height. Half of a base edge is 8 \div 2 = 4 ft, and the height is 3 ft.
Pythagorean theorem: s^2 = 3^2 + 4^2 = 9 + 16 = 25, so s = 5 ft.
Perimeter of the base: p = 4 \times 8 = 32 ft.
Lateral area (no floor, so no B): \tfrac{1}{2}ps = \tfrac{1}{2}(32)(5) = 80.
Answer: 80 ft²
Check: Each triangle has base 8 and height 5, so its area is \tfrac{1}{2}(8)(5) = 20. Four triangles: 4 \times 20 = 80 ✓.
Tip: Read what the object covers. "No floor", "open top", or "just the sides" means you drop the base term B from the formula. The 3-4-5 triangle shows up a lot.
Question 23 of 36
Drag and drop
A cone has volume V = \tfrac{1}{3}\pi r^2 h. Drag the correct tile to show how the volume changes in each case. One tile is not used.
Radius doubled, height unchanged
Height doubled, radius unchanged
Radius and height both doubled
Tap a tile, then tap a box to place it.
Hint
Try it with numbers: start with r = 1, h = 1, then change one value at a time and compare r^2 h.
Step-by-step solution
Compare r^2 h before and after (the \tfrac{1}{3}\pi is the same every time). Start with r = 1, h = 1: r^2 h = 1.
Radius doubled: r = 2, h = 1 → r^2 h = 4 \times 1 = 4. The volume is multiplied by 4, because the radius is squared (2^2 = 4).
Height doubled: r = 1, h = 2 → r^2 h = 1 \times 2 = 2. The volume is multiplied by 2.
Both doubled: r = 2, h = 2 → r^2 h = 4 \times 2 = 8. The volume is multiplied by 8 (4 \times 2, not 4 + 2).
Answer: ×4, ×2, ×8 (the ×6 tile is not used)
Check: With real numbers, r = 3, h = 5: V = 15\pi. Doubling both gives r = 6, h = 10: V = \tfrac{1}{3}\pi(36)(10) = 120\pi, and 120\pi \div 15\pi = 8 ✓.
Tip: Effects of changes multiply. When every length of a 3-D figure is doubled, the volume is multiplied by 2^3 = 8.
Question 24 of 36
Drop-down
A cone-shaped pile of salt is stored for winter roads. Complete each statement.
If the radius of the pile is tripled and the height stays the same, the volume is multiplied by .
If the radius and the height are both cut in half, the volume is multiplied by .
Choose an answer from each menu.
Hint
In V = \tfrac{1}{3}\pi r^2 h, the radius is squared but the height is not. Try r = 1, h = 1 and change the values.
Step-by-step solution
Only the part r^2 h changes; \tfrac{1}{3}\pi stays the same.
Radius tripled: r^2 becomes 3^2 = 9 times as large. The height is unchanged, so the volume is multiplied by 9.
Both halved: r^2 is multiplied by \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}, and h is multiplied by \tfrac{1}{2}.
Check: Use numbers. r = 2, h = 6: V = \tfrac{1}{3}\pi(4)(6) = 8\pi. Tripled radius, r = 6: \tfrac{1}{3}\pi(36)(6) = 72\pi = 9 \times 8\pi ✓. Halved, r = 1, h = 3: \tfrac{1}{3}\pi(1)(3) = \pi = \tfrac{1}{8} \times 8\pi ✓.
Tip: Square the factor for the radius, use the factor once for the height, and multiply the results. Halving every length of any solid gives \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8} of the volume.
Question 25 of 36
Fill in
A solid wooden cone for a toy set has a radius of 7 cm and a slant height of 15 cm. The whole cone, including its circular base, will be painted.
What is the total surface area to be painted? Round to the nearest square centimeter.
cm²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
A solid cone has two parts: the curved side (\pi r s) and the circular base (\pi r^2).
Step-by-step solution
Formula (on the formula sheet): SA = \pi r s + \pi r^2.
Curved side: \pi r s = \pi(7)(15) = 105\pi.
Circular base: \pi r^2 = \pi(7^2) = 49\pi.
Total: 105\pi + 49\pi = 154\pi.
Use the \pi key: 154\pi \approx 483.81, which rounds to 484.
Answer: about 484 cm²
Check: The curved part (105\pi \approx 329.9) should be bigger than the base (49\pi \approx 153.9) because the slant height is longer than the radius; 329.9 + 153.9 = 483.8 ✓.
Tip: Add the "\pi parts" first (105 + 49 = 154) and multiply by \pi once. Fewer calculator steps means fewer rounding mistakes.
Question 26 of 36
A park sculpture is a solid concrete square pyramid. Its base is 6 yd on each side, and it contains 48 cubic yards of concrete.
How tall is the sculpture?
Hint
Use V = \tfrac{1}{3}Bh with V = 48 and B = 6^2. To undo the \tfrac{1}{3}, multiply both sides by 3.
Tip: The shortcut h = \dfrac{3V}{B} works for every pyramid and cone. Here \dfrac{3 \times 48}{36} = \dfrac{144}{36} = 4. Remember that B is the base area (36 yd²), not the edge (6 yd) or the perimeter (24 yd).
Question 27 of 36
A cone-shaped paper cup holds 75\pi cubic centimeters of water when it is full. The cup is 9 cm deep.
What is the diameter of the top of the cup?
Hint
Solve 75\pi = \tfrac{1}{3}\pi r^2(9) for r. Then read the question again: does it ask for the radius?
Step-by-step solution
Formula: V = \tfrac{1}{3}\pi r^2 h, so 75\pi = \tfrac{1}{3}\pi r^2(9).
Simplify: \tfrac{1}{3}(9) = 3, so 75\pi = 3\pi r^2.
Divide both sides by 3\pi: r^2 = 25.
Square root: r = 5 cm.
The question asks for the diameter: d = 2r = 10 cm.
Answer: 10 cm
Check: With r = 5: \tfrac{1}{3}\pi(25)(9) = \tfrac{1}{3}\pi(225) = 75\pi ✓, and the diameter is twice the radius, 10 cm ✓.
Tip: Work-backwards cone problems usually take three moves: undo the \tfrac{1}{3} and h, take a square root, and then check whether the question wants the radius or the diameter.
Question 28 of 36
Fill in
A town stores road salt in a cone-shaped pile with a diameter of 18 ft and a height of 8 ft. Salt is ordered in cubic yards (1\text{ yd}^3 = 27\text{ ft}^3).
How many cubic yards of salt are in the pile? Round to the nearest tenth.
yd³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Halve the diameter, find the volume in cubic feet, and then convert. How many cubic feet are in one cubic yard?
Step-by-step solution
Radius: r = 18 \div 2 = 9 ft.
Volume in cubic feet: V = \tfrac{1}{3}\pi(9^2)(8) = \tfrac{1}{3}\pi(648) = 216\pi \approx 678.58\text{ ft}^3.
One cubic yard is 3 \times 3 \times 3 = 27 cubic feet, so divide by 27: 678.58 \div 27 \approx 25.13.
Round to the nearest tenth: 25.1.
Answer: about 25.1 yd³
Check: Convert the lengths first instead: r = 3 yd and h = \tfrac{8}{3} yd. \tfrac{1}{3}\pi(9)\left(\tfrac{8}{3}\right) = 8\pi \approx 25.13 ✓.
Tip: 1 yard = 3 feet, but 1 cubic yard = 3^3 = 27 cubic feet. Dividing by 3 (or 9) is the most common unit trap on volume problems.
Question 29 of 36
Maria wants the total surface area of a square pyramid with base edges of 8 cm and a height of 3 cm. Here is her work.
One triangle: \tfrac{1}{2}(8)(8.5) = 34\text{ cm}^2
Total: 64 + 4(34) = 200\text{ cm}^2
In which step did Maria make the first mistake?
Hint
Picture the right triangle inside the pyramid that has the slant height as its hypotenuse. How long is its horizontal leg?
Step-by-step solution
Step 1 is correct: the base is 8 \times 8 = 64\text{ cm}^2.
Step 2 is wrong. The right triangle inside the pyramid runs from the center of the base to the middle of one edge, so its horizontal leg is half the base edge: 8 \div 2 = 4 cm, not 8 cm.
Correct slant height: s = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 cm.
Correct finish: one triangle = \tfrac{1}{2}(8)(5) = 20; total = 64 + 4(20) = 144\text{ cm}^2.
Answer: Step 2
Check: 3-4-5 is a well-known right triangle, so s = 5 makes sense; 64 + 80 = 144\text{ cm}^2 ✓.
Tip: In a square pyramid, height, slant height and half the base edge form a right triangle. Draw it before using the Pythagorean theorem.
Question 30 of 36
Select all
A display stand is a square prism, and a decoration next to it is a square pyramid. Both have a square base 6 ft on each side and a height of 9 ft.
Select all of the statements that are true.
Select all that apply.
Hint
Find both volumes: prism V = Bh, pyramid V = \tfrac{1}{3}Bh. Then test each statement.
Step-by-step solution
Base area for both: B = 6^2 = 36\text{ ft}^2.
Prism: V = Bh = 36 \times 9 = 324\text{ ft}^3 → statement B is true.
Pyramid: V = \tfrac{1}{3}(36)(9) = 108\text{ ft}^3 → statement A is true.
324 \div 108 = 3, so the prism holds 3 times as much → statement C is true, and D (one-half) is false.
A cone with radius 6 ft and height 9 ft: \tfrac{1}{3}\pi(36)(9) = 108\pi \approx 339.3\text{ ft}^3, not 108 → E is false.
Answer: A, B, and C
Check:108 + 108 + 108 = 324, so three pyramids exactly match one prism ✓.
Tip: A pyramid is one-third of the prism with the same base and height. A cone whose radius equals the square's side is a different solid — its base is a circle of area \pi r^2, not a square.
Question 31 of 36
Drag and drop
Drag the correct volume under each pyramid. One tile is not used.
Square base 6 in by 6 in, height 5 in
Rectangular base 10 in by 4 in, height 6 in
Triangular base with base 6 in and height 10 in; pyramid height 12 in
Tap a tile, then tap a box to place it.
Hint
For each pyramid, find the base area B first (square, rectangle, or triangle), then take \tfrac{1}{3}Bh.
Step-by-step solution
Use V = \tfrac{1}{3}Bh for every pyramid; only B changes.
Square base: B = 6 \times 6 = 36. V = \tfrac{1}{3}(36)(5) = 60\text{ in}^3.
Rectangular base: B = 10 \times 4 = 40. V = \tfrac{1}{3}(40)(6) = 80\text{ in}^3.
Triangular base: B = \tfrac{1}{2}(6)(10) = 30. V = \tfrac{1}{3}(30)(12) = 120\text{ in}^3.
Answer: 60 in³, 80 in³, 120 in³ (the 240 in³ tile is not used)
Check: Work backwards for the triangle pyramid: h = \dfrac{3V}{B} = \dfrac{3(120)}{30} = 12 ✓.
Tip: Keep the two heights apart in a triangular pyramid: the triangle's height (10 in) is used for B, and the pyramid's height (12 in) is used for V.
Question 32 of 36
A cone-shaped grain hopper (point down) has a radius of 3 ft and a height of 7 ft. A conveyor belt pours grain into the empty hopper at 2 cubic feet per minute.
To the nearest minute, how long does it take to fill the hopper?
Hint
Find the volume in cubic feet. If 2 cubic feet go in each minute, how do you find the number of minutes?
Check: In 33 minutes the belt pours 33 \times 2 = 66\text{ ft}^3, which matches the hopper's volume of about 66 ft³ ✓.
Tip: Look at the units: \dfrac{\text{ft}^3}{\text{ft}^3/\text{min}} = \text{min}. If the units of your answer do not come out to minutes, you multiplied when you should have divided.
Question 33 of 36
A kitchen funnel is shaped like a cone (ignore the spout). The top opening has a radius of 9 cm, and the slant height along the side is 15 cm.
To the nearest cubic centimeter, how much liquid can the funnel hold?
Hint
The volume formula needs the height, but you were given the slant height. Use the right triangle formed by r, h, and s.
Step-by-step solution
Find the height. In a cone, h^2 + r^2 = s^2, so h^2 = 15^2 - 9^2 = 225 - 81 = 144.
Square root: h = \sqrt{144} = 12 cm.
Volume formula: V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(81)(12).
Simplify: 81 \times 12 = 972 and 972 \div 3 = 324, so V = 324\pi.
Evaluate: 324\pi \approx 1017.9, which rounds to 1{,}018.
Answer: about 1,018 cm³
Check: 9-12-15 is the 3-4-5 triple times 3, so h = 12 is right ✓. A cylinder of radius 9 and height 12 would hold 972\pi \approx 3{,}054; one-third of that is about 1,018 ✓.
Tip: Two-step cone and pyramid problems are common on the GED: get the missing length with the Pythagorean theorem first, then use the formula.
Question 34 of 36
A tent at a summer camp is shaped like a square pyramid. Each edge of the square floor is 14 ft, and the slant height of each triangular side is 25 ft.
What is the volume of the space inside the tent?
Hint
The volume needs the true height, not the slant height. Use a right triangle with the slant height as its hypotenuse and half the base edge as one leg.
Step-by-step solution
Find the height. The right triangle has hypotenuse s = 25 and one leg equal to half the base edge, 14 \div 2 = 7.
h = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 ft.
Base area: B = 14^2 = 196\text{ ft}^2.
Volume: V = \tfrac{1}{3}(196)(24) = \tfrac{1}{3}(4{,}704) = 1{,}568.
Answer: 1,568 ft³
Check:7^2 + 24^2 = 49 + 576 = 625 = 25^2 ✓, so the height is right. Then 196 \times 24 = 4{,}704 and 4{,}704 \div 3 = 1{,}568 ✓.
Tip: Volume always uses the height (straight up), and surface area uses the slant height (along a face). When only one is given, the Pythagorean theorem gives the other.
Question 35 of 36
A cone-shaped feeding funnel on a farm holds 40 quarts of grain. The farm buys a larger funnel with the same height, but its radius is 1.5 times as long.
How many quarts does the larger funnel hold?
Hint
Only the radius changes, and the radius is squared in V = \tfrac{1}{3}\pi r^2 h. What is 1.5^2?
Step-by-step solution
In V = \tfrac{1}{3}\pi r^2 h, the height and \tfrac{1}{3}\pi do not change.
The radius is multiplied by 1.5, so r^2 is multiplied by 1.5^2 = 1.5 \times 1.5 = 2.25.
New volume: 40 \times 2.25 = 90 quarts.
Answer: 90 quarts
Check: Try numbers. Old funnel: r = 2, h = 3 → \tfrac{1}{3}\pi(4)(3) = 4\pi. New funnel: r = 3 → \tfrac{1}{3}\pi(9)(3) = 9\pi. 9\pi \div 4\pi = 2.25 ✓.
Tip: You do not need the actual radius or height. Square the radius factor, multiply by the height factor (here 1), and apply that to the old volume.
Question 36 of 36
Three students found the volume of a square pyramid with base edges of 18 cm and a slant height of 15 cm.
Is 15 cm the height of the pyramid? Find the true height with the Pythagorean theorem, using half the base edge.
Step-by-step solution
Base area: 18^2 = 324\text{ cm}^2 — all three students used this correctly.
The 15 cm is the slant height, not the height. Find the height with a right triangle: hypotenuse 15, leg 18 \div 2 = 9.
h = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 cm.
Volume: V = \tfrac{1}{3}(324)(12) = \tfrac{1}{3}(3{,}888) = 1{,}296\text{ cm}^3. That is Ben's work.
Answer: Ben
Check:9^2 + 12^2 = 81 + 144 = 225 = 15^2 ✓, so the height is 12 cm, and 324 \times 12 \div 3 = 1{,}296 ✓.
Tip: When you judge someone's work, check three things: the right formula, the right measurements (height vs. slant height, radius vs. diameter), and the arithmetic.
