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A child's wooden building block is a cube with edges 4\text{ in} long. A parent wants to paint every face of the block.
How many square inches will be painted?
Hint
A cube has 6 faces, and every face is a square with the same side length.
Step-by-step solution
Each face is a square 4\text{ in} by 4\text{ in}: area = 4^2 = 16\text{ in}^2.
A cube has 6 identical faces.
Surface area: SA = 6 \times 16 = 96.
The parent will paint 96\text{ in}^2.
Check: using the formula sheet, SA = ph + 2B with p = 16, h = 4, B = 16: 64 + 32 = 96 ✓.
Tip: For a cube, memorize two quick facts: volume = s^3 and surface area = 6s^2. Paint, wrapping and fabric questions want the surface area; "holds" or "fills" questions want the volume.
Question 2 of 10 · Rectangular prisms
Select all
Use the rectangular prism shown.
Select all statements that are true.
Select all that apply.
Hint
Work out B, p, V = Bh, the lateral area ph, and SA = ph + 2B, then compare each statement.
Step-by-step solution
Take the 10 \times 4 face as the base and 3\text{ ft} as the height.
B = 10 \times 4 = 40\text{ ft}^2; p = 2(10) + 2(4) = 28\text{ ft}.
Volume: V = Bh = 40 \times 3 = 120\text{ ft}^3. A is true.
Lateral area (4 sides): ph = 28 \times 3 = 84\text{ ft}^2. C is true.
Surface area: SA = ph + 2B = 84 + 80 = 164\text{ ft}^2. B is true.
D (82) is half of the surface area, and E (17) adds the edges — both false.
Tip: In a select-all question, judge each statement on its own. Computing all the key measures first (V, ph, SA) makes every choice quick to check.
Question 3 of 10 · Cylinders
Drop-down
A company prints paper labels that wrap exactly once around the curved side of a can, with no overlap. Each can has a radius of 1.5\text{ in} and a height of 4.5\text{ in}. The labels do not cover the top or bottom.
Select the correct choice for each blank.
The size of one label is found with , so each label has an area of about (to the nearest tenth).
Choose an answer from each menu.
Hint
A label is the curved side only. Unrolled, it is a rectangle: its width is the distance around the can and its height is the height of the can.
Step-by-step solution
The label covers only the curved side, so use the lateral area 2\pi r h.
2\pi r h = 2\pi(1.5)(4.5) = 13.5\pi.
With the π key: 13.5\pi \approx 42.41, so about 42.4\text{ in}^2.
Check: the label is a rectangle 2\pi(1.5) \approx 9.42\text{ in} wide and 4.5\text{ in} tall: 9.42 \times 4.5 \approx 42.4 ✓.
Tip: Decide which parts are covered before choosing a formula: label = 2\pi r h; closed can = 2\pi r h + 2\pi r^2; contents = \pi r^2 h.
Question 4 of 10 · Right prisms
Fill in
A kids' play fort is shaped like a triangular prism with no floor. Each triangular end has a base of 10\text{ ft}, a height of 12\text{ ft}, and two slanted sides of 13\text{ ft}. The fort is 15\text{ ft} long.
How many square feet of fabric cover the two triangular ends and the two slanted sides?
\text{ft}^2
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Picture the fort: two triangles and two slanted rectangles. The bottom rectangle (the floor) is missing.
Step-by-step solution
One triangular end: \tfrac{1}{2} \times 10 \times 12 = 60\text{ ft}^2. Two ends: 120\text{ ft}^2.
One slanted side: 13 \times 15 = 195\text{ ft}^2. Two sides: 390\text{ ft}^2.
Total: 120 + 390 = 510.
The fort needs 510\text{ ft}^2 of fabric.
Check: with a floor it would also need 10 \times 15 = 150\text{ ft}^2, giving 660. Removing the floor: 660 - 150 = 510 ✓.
Tip: The slanted rectangles use the slanted side (13), not the height (12). The height is used only inside the triangle's area.
Question 5 of 10 · Pyramids and cones
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.
Evaluate: 256\pi \approx 804.25, which rounds to 804.
Answer: about 804 in²
Check: Estimate with \pi \approx 3: 4 \times 3 \times 64 = 768. The answer should be a bit more than 768, and 804 is ✓.
Tip: Words like cover, paint, wrap, coat, material signal surface area (square units). Words like hold, fill, contain, capacity signal volume (cubic units).
Question 7 of 10 · Composite 3-D figures
A bowling trophy has a solid cylinder base with a diameter of 6 in and a height of 2 in. On top sits a solid ball with a diameter of 4 in.
Which expression gives the total volume of the trophy?
Hint
Use the cylinder and sphere volume formulas, and change each diameter to a radius first.
Step-by-step solution
Cylinder base: radius = 6 \div 2 = 3, so V = \pi(3)^2(2).
Ball: radius = 4 \div 2 = 2, so V = \tfrac{4}{3}\pi(2)^3.
The parts are joined, so add: \pi(3)^2(2) + \tfrac{4}{3}\pi(2)^3.
Check: Each term matches one part: the first has r = 3 and a height (cylinder), the second has r = 2 cubed (sphere) ✓.
Tip: For "which expression" items, check three things in order: the right formula for each piece, radii (not diameters), and + vs. - (joined vs. removed).
Question 8 of 10 · Cylinders
Drag and drop
Drag each cylinder into the column that shows its volume, rounded to the nearest cubic inch. One cylinder will not be used.
about 113\text{ in}^3
about 157\text{ in}^3
about 226\text{ in}^3
Tap a tile, then tap a box to place it.
Hint
Find each volume in terms of \pi first, then multiply by \pi. Check whether each cylinder gives a radius or a diameter.
Step-by-step solution
Use V = \pi r^2 h:
r = 3, h = 4: \pi(9)(4) = 36\pi \approx 113.1 → about 113.
r = 5, h = 2: \pi(25)(2) = 50\pi \approx 157.1 → about 157.
r = 3, h = 8: \pi(9)(8) = 72\pi \approx 226.2 → about 226.
d = 6, h = 2: r = 3, so \pi(9)(2) = 18\pi \approx 56.5 → not used.
Check: doubling the height of cylinder 1 gives cylinder 3, and 113 \times 2 \approx 226 ✓.
Tip: Read every label: "diameter 6" and "radius 6" are very different cylinders. Halve a diameter before squaring.
Question 9 of 10 · Right prisms
A foam wedge is a triangular prism, as shown. The slanted side of the triangle is not labeled.
What is the total surface area of the wedge?
Hint
You need all three sides of the triangle for the perimeter. Find the slanted side with a^2 + b^2 = c^2 first.
Step-by-step solution
Missing side (hypotenuse): c^2 = 6^2 + 8^2 = 36 + 64 = 100, so c = 10\text{ in}.
Perimeter of the triangle: p = 6 + 8 + 10 = 24\text{ in}.
Tip: When a figure leaves a side unlabeled, the question usually expects the Pythagorean theorem. Watch for the 6-8-10 (a 3-4-5 doubled) right triangle.
Question 10 of 10 · Pyramids and cones
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.