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Measure solids and track dimensional scale

Compute volumes, surfaces and spatial distances with explicit units and boundaries.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

Model a three-dimensional measurement problem, distinguish its boundary from its interior and apply the correct scale power.

2. Build from plane geometry

Review rectangle and circle areas, circumference and the Pythagorean theorem. A prism with constant cross-section has volume equal to base area times perpendicular height. A right circular cylinder is an example with circular base. The perpendicular height is not a sloping edge unless the solid's geometry makes them equal. Multiplication of a square-unit area by a linear-unit height produces cubic units, which provides a quick dimensional check on the formula.

3. Spatial measurement terms

TermWhat it means
Cross-sectionA plane slice through a solid.
Lateral surfaceThe side boundary, excluding designated bases.
Space diagonalA segment connecting opposite vertices through the interior of a box.
CapacityThe interior volume available to hold material.

4. The requested dimension selects the model

A solid has several different measurements. Volume describes the space it occupies or holds. Surface area describes its boundary. A diagonal is a length through or along the figure. These quantities cannot be substituted for one another even when their numerical values happen to agree. A cube with side six has volume 216 cubic units and surface area 216 square units; the equal numerals do not make the measurements equivalent.

Begin with a labeled description of the solid and identify whether dimensions are inside, outside or idealized measurements. A container's capacity uses its inside dimensions. Material needed for an open box excludes the missing lid. A cylinder's lateral surface excludes both circular ends. The problem's words define which regions count, so reading the boundary condition is part of the mathematics.

Another way: steps

Identify the measured region; select lengths, areas or volumes; compute with consistent units; check boundaries and scale.

5. Rectangular prisms separate volume from surface

For perpendicular dimensions l, w and h, volume is lwh. Total surface area is 2lw + 2lh + 2wh because there are two faces of each pairwise dimension. An open-top box removes one lw face from that total. If only selected faces are painted, add exactly those faces rather than using the total surface formula automatically. Draw a list of face pairs when unsure. The units help diagnose errors: adding l, w and h gives a length, not a volume or an area.

6. Unroll a cylinder's curved side

The curved surface of a right circular cylinder can be unrolled into a rectangle. One dimension is the height h and the other is the circumference 2 pi r, giving lateral area 2 pi rh. A closed cylinder adds two circular bases, for total surface area 2 pi rh + 2 pi r squared. Its volume is pi r squared h. These formulas involve the same variables but represent different pieces. If a diameter is given, halve it before using r. Doubling the supplied diameter without correcting that interpretation causes a fourfold error in base area.

7. Use two right triangles for a space diagonal

A rectangular box's base diagonal has squared length l squared + w squared. That diagonal and the perpendicular height form another right triangle, so the space diagonal has squared length l squared + w squared + h squared. The segment runs through the box, not around its edges. A path constrained to the surface is a different problem and may require unfolding faces into a plane. Do not use the space diagonal as the shortest surface path unless the problem explicitly permits travel through the interior.

8. Similarity scales three independent directions

If all corresponding lengths of similar solids multiply by k, surface areas multiply by k squared and volumes by k cubed. A volume ratio of eight to twenty-seven corresponds to a positive length ratio of two to three, with area ratio four to nine. Equal volume alone does not imply similar solids. A flat box and a tall box can hold the same amount while having different surface areas. The scale rules depend on every length changing by the same factor, not just one dimension growing.

9. Convert units before multiplying dimensions

One meter is one hundred centimeters, so one square meter is ten thousand square centimeters and one cubic meter is one million cubic centimeters. The conversion factor is squared or cubed because each independent dimension changes units. If a tank's length is in meters and width and height are in centimeters, convert to a consistent length unit before multiplying. Converting a volume afterward is also valid if you apply the cubic conversion correctly. A liter is one thousand cubic centimeters, so a volume of twelve thousand cubic centimeters is twelve liters.

10. Composite solids need nonoverlapping pieces

A stepped solid can often be split into rectangular prisms whose interiors do not overlap. Add their volumes, or subtract a missing prism from an enclosing box. For surface area, shared interior faces are not exposed and must not be counted as exterior boundary. This makes surface-area addition more delicate than volume addition. Identify what is physically exposed before calculating. If two cubes are joined face to face, the total exterior area is the sum of their separate areas minus the two joined faces.

11. Changes in depth depend on container shape

In a vertical-sided tank with constant horizontal cross-section, added volume equals base area times the rise in liquid level. This linear relation may fail in a tapered container whose cross-section changes with depth. Use the shape specified in the question. For a rectangular tank, doubling depth doubles liquid volume while leaving the top surface area unchanged. A statement about a percent of full height therefore translates directly into the same percent of full volume only for a constant-cross-section container. The geometric assumption is what makes the proportion valid. Compare a box of dimensions one, two and twelve with a box of dimensions two, three and four. Both have volume twenty-four, but their face areas differ. A fixed volume therefore does not determine a unique surface area; changing proportions while preserving the product gives a concrete counterexample.

12. Check a storage limit with an independent bound

Suppose a rectangular container may be filled only to eighty percent of its inside height. Because the cross-section is constant, its usable volume is eighty percent of the full interior volume. You can calculate either the permitted depth first or the full volume first; both methods should agree. That agreement is an independent arithmetic check when the unit conversion is handled consistently.

Now imagine that the container's top is wider than its base. The same percentage of height need not mean the same percentage of volume. The original shortcut depends on the rectangular model, so it cannot be carried into the changed geometry without justification. This is a useful transfer check: identify which assumption makes a convenient proportional relationship work.

In practice, label the response unit before entering a number. A result in cubic centimeters may need conversion to liters, and an exterior measurement may need adjustment before it represents inside capacity. The automatic answer check cannot see your sketch, so use the explanation to compare your model, not merely the final digits.

13. Similar numerals do not imply similar quantities

Keep square and cubic units distinct. A radius is half a diameter. An open container has fewer boundary faces than a closed one. Surface paths differ from interior diagonals. Volume scaling uses the cube of a common length factor; if only height changes, volume changes only by that height factor. Capacity uses inside dimensions, including any stated allowance for wall thickness or required empty space above the contents.

14. Measure an open box's material

  1. Identify which faces exist.

    An open rectangular box is 6 by 4 by 3 units.

    The top 6-by-4 face is absent.

  2. Compute the base area.

    6 times 4 = 24.

    There is one base, not two.

  3. Add the four side faces.

    2(6 times 3) + 2(4 times 3) = 60.

    Opposite side faces come in equal pairs.

  4. Combine and check units.

    Total material area is 84 square units.

    This is a boundary area, not the interior volume of 72 cubic units.

15. Find a box's interior diagonal

  1. Read the perpendicular dimensions.

    A rectangular box measures 3 by 4 by 12.

    The target is the segment through its interior between opposite vertices.

  2. Calculate the base diagonal.

    The square root of 3 squared plus 4 squared is 5.

    The base is a rectangle with a right-angle corner.

  3. Form the second right triangle.

    Its legs are 5 and 12.

    The box height is perpendicular to the base plane.

  4. Calculate the space diagonal.

    The square root of 25 + 144 is 13.

    This uses the Pythagorean theorem a second time.

  5. Check the kind of path.

    Thirteen is an interior straight-line distance.

    The result does not describe a route constrained to the surface or edges.

16. Recover a linear factor from volume

  1. Read the similarity condition.

    Two similar containers have volumes 64 and 216.

    The larger is a uniformly scaled version of the smaller.

  2. Reduce the volume ratio.

    216/64 = 27/8.

    The ratio compares larger with smaller.

  3. Take the positive cube root.

    Linear factor k = 3/2.

    Three independent lengths contribute to volume.

  4. Convert to the requested area factor.

    k squared = 9/4.

    The target is surface area, not volume.

  5. Apply a supplied smaller area.

    If the smaller area is 80, the larger area is 180.

    Multiply eighty by nine-fourths.

  6. Check the roles of all ratios.

    Lengths scale by 1.5, areas by 2.25 and volumes by 3.375.

    Using the volume ratio directly on area would overstate the result.

17. Complete a water-level calculation

  1. Calculate the constant base area.

    A rectangular tank has base 20 by 35 centimeters: 700 square centimeters.

    The water level rises by six centimeters.

  2. Multiply by the perpendicular rise.

    Added volume = 700 times 6 = 4200 cubic centimeters.

    Convert this volume to liters.

  3. Your turn: work this step out. Its working is at the end of the packet.

    Convert and verify the scale.

18. Guided practice

A rectangular prism has dimensions 3, 4 and 7 centimeters. What is its volume in cubic centimeters?

Answer:

19. Guided practice

A cube has side length 5. What is its total surface area?

Answer:

20. Practice

A rectangular box is 8 by 5 by 3 centimeters. What is its total surface area in square centimeters?

Answer:

21. Practice

A right circular cylinder has radius 3 and height 10. What is its volume divided by pi?

Answer:

22. Practice

A right circular cylinder has radius 4 and height 6. What is its lateral surface area divided by pi?

Answer:

23. Practice

A rectangular prism has side lengths 2, 3 and 6. What is the length of a diagonal joining opposite vertices through its interior?

Answer:

24. Practice

Two similar solids have corresponding lengths in ratio 2:3. The smaller volume is 80. What is the larger volume?

Answer:

25. Practice

A cube has volume 216 cubic centimeters. What is its total surface area in square centimeters?

Answer:

26. Practice

A rectangular tank has base 50 centimeters by 30 centimeters. Water rises by 4 centimeters. How many liters of water were added? One liter equals 1000 cubic centimeters.

Answer:

27. Practice

A rectangular sheet 20 by 14 centimeters has a square of side 3 centimeters cut from each corner. The sides are folded up to form an open box. What is its volume in cubic centimeters?

Answer:

28. Somewhere new

A storage container has inside base dimensions 40 centimeters by 25 centimeters and inside height 30 centimeters. It may be filled to at most 80 percent of its height. What is the greatest permitted liquid volume in liters? One liter equals 1000 cubic centimeters.

Answer:

29. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

30. Test question

Every length of a model is multiplied by 4 to create a similar solid. By what factor does its volume increase?

Answer:

31. What you can do now

Tomorrow, explain how a cylinder's curved surface becomes a rectangle, then compare the length, area and volume factors when a model doubles in every dimension. Review plane geometry for base areas and the Pythagorean theorem.

Working for the steps left to you

17. Complete a water-level calculation, step 3

4200/1000 = 4.2 liters.

The conversion acts on cubic centimeters, not on a single length.