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Secondary 4 G3 Mathematics Tutor Bukit Timah | Composite Solids: Water Filter Cartridge Volume and Surface Area

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by

eduKate Asia
Three learners review open books together at a classroom table, with stacks of textbooks, stationery and a whiteboard in the bright room.
Three secondary pupils studying geometry and Mathematics at a small-group tuition table.

Did you know? When a cone is joined to a cylinder, the circular face where they meet does not count as an exterior surface. A student may calculate every separate shape correctly and still overstate the composite surface area by counting an internal face that cannot be seen from outside.

Quick answer: divide the composite solid into familiar shapes, record radius and perpendicular heights, calculate their volumes separately and add them. For exterior surface area, count only the exposed faces: the curved cylinder, curved cone and bottom circular base in this particular model. Use Pythagoras to find the cone’s slant height before calculating its curved area.

The official 2027 SEC G3 Mathematics K310 syllabus includes volume and surface area of cylinders, cones, spheres and composite solids. This imaginary water-filter cartridge is a purely geometrical classroom model, not a practical filter design, manufacturing specification or claim about any real product.

Five answers at a glance

  • Cylinder: volume 90πcm³, curved area60πcm².
  • Cone: volume12πcm³, slant height5cm, curved area15πcm².
  • Whole solid volume 102πcm³ and total height14cm.
  • Outside area is 84πcm²; the shared joining discs remain internal.
  • A similar model with all lengths multiplied by 1.5 has area factor2.25 and volume factor3.375.

Find the exact worked question

Start with an invented composite solid that has exactly two parts

Imagine a solid cylinder of radius 3 cm and perpendicular height 10 cm. Attach a solid right circular cone on top, with the same radius 3 cm and perpendicular height 4 cm.

The cone sits directly on the cylinder’s upper circular face. Their surfaces meet with no gap, and the joined circular interface becomes an internal boundary. We assume both components are solid, uniformly shaped and measured by their ideal geometric dimensions.

These are fictional teaching numbers. A real water filter might be hollow or made of several porous layers, but those practical properties are deliberately absent. The model is chosen to test mensuration and careful interpretation.

Which measurements are radii and which are heights?

The radius r=3 cm is measured from the centre of a circular cross-section to its boundary. The diameter is twice the radius, so it equals 6 cm.

The cylindrical perpendicular height is 10 cm. The conical perpendicular height is 4 cm. They are measured along the common central axis, not around the sloping outside of the cone.

Write the two height labels separately before calculating. A common mistake is to put the cone’s slant height into the volume formula, which requires its vertical perpendicular height.

A simple word diagram before any formula

From bottom to top, picture a flat circular base, a ten-centimetre-tall cylinder, then a four-centimetre-tall cone coming to a point. The radius of both is three centimetres.

Draw a central axis through the cylinder and cone. Label the cylinder height 10, the cone perpendicular height 4, and the shared circular interface r=3. The total outside height is 14 cm.

Now mark the circular interface as hidden inside the composite. A visible-face inventory before surface-area calculations is a much better safeguard than trying to remember whether to subtract ‘one circle’ or ‘two circles’ later.

Cylinder volume: base area times height

The cylinder’s base is a circle with radius three, so its area is πr²=9π cm².

Multiply by the cylinder’s perpendicular height of ten centimetres: V_cylinder=π(3²)(10)=90π cm³.

Check dimensions: square centimetres multiplied by centimetres gives cubic centimetres. A result in cm² would be an area and could not describe the amount of space occupied by the solid cylinder.

Cone volume: why the one-third matters

The cone has the same radius three and vertical height four. Its ideal solid volume is V_cone=(1/3)πr²h.

Substituting gives (1/3)π(3²)(4)=(1/3)×36π=12π cm³.

The one-third coefficient is not optional. A cone occupies one-third of the volume of a cylinder sharing its base area and perpendicular height. Leaving it out would produce 36π instead of 12π.

Total composite volume

The two solid components meet at a face but do not overlap in their interiors. Therefore their volumes add directly.

V_total=90π+12π=102π cm³, approximately 320 cm³ to three significant figures. The exact form 102π is useful while further calculations are pending.

Do not subtract the shared circular face from volume. A face has area, not positive three-dimensional volume, so it does not create a volume overlap in this ideal joined-solid model.

What percentage of volume comes from the cone?

The cone’s volume is 12π and the entire composite volume is 102π. Its share is 12π/(102π)=2/17.

Expressed as a percentage, the cone contributes approximately 11.8% of total solid volume. The remaining cylinder contributes about 88.2%.

This is not the percentage of the object’s height taken by the cone. Height share is 4/14≈28.6%, a different measure with a different denominator.

Total height is simpler than total volume

The cylinder stands ten centimetres high and the cone rises another four centimetres directly above it. The outside height is therefore 10+4=14 cm.

That total is a one-dimensional length, unlike the 102π cm³ occupied volume or the exterior area we will calculate next.

An exam answer should not confuse 14 cm with an area or a volume. The units often reveal whether the correct physical quantity has been found.

A side view of the cone reveals a right triangle

Slice the cone through its central axis. In the cross-section, the perpendicular height is four centimetres and the base radius is three centimetres.

The slant height l is the hypotenuse of a right triangle with legs four and three. Pythagoras gives l²=4²+3²=16+9=25, so l=5 cm.

The cone’s full triangular cross-section has base diameter six centimetres, but half of that triangle has the three-centimetre radius. It is the half-cross-section that gives the 3–4–5 relationship.

Slant height is not cone height

The cone’s perpendicular height h=4 cm controls volume. Its slant height l=5 cm controls the length along the conical surface.

Using l=5 in the cone-volume formula would give a fictitious volume 15π rather than the correct 12π. Using h=4 in the curved-surface formula would understate the outside surface.

Write h and l beside a small right-triangle sketch. Different formulas use different dimensions; the same physical object can have several related but non-interchangeable lengths.

Curved surface area of the cylinder

The cylinder’s curved lateral area is 2πrh. Using r=3 and h=10 gives 2π×3×10=60π cm².

One explanation is that the side unwraps to a rectangle whose width is the circumference 2πr=6π cm and whose height is ten centimetres. Its rectangular area is 6π×10=60π.

The two circular faces are not included in this curved-area expression. Keeping curved area separate from flat ends helps prevent double-counting in the composite solid.

Curved surface area of the cone

The curved lateral area of a cone is πrl, where l is the slant height, not the vertical height.

With r=3 and l=5, the conical curved surface has area π×3×5=15π cm².

Do not add the cone’s own circular base when describing the exposed composite surface: its base is attached to the cylinder’s upper face, so that circular disc is hidden inside.

The bottom base remains exposed

The outside bottom of the cylinder is a circle of radius three. Its area is πr²=9π cm².

This bottom face is visible or externally accessible under the invented solid model. It therefore belongs in the total external surface-area count.

The upper cylinder disc and cone base do not belong in the visible-face count because they meet each other inside the joined object. A student must reason about which faces are exposed, not add all the faces from two separate solids.

The correct exterior surface-area total

The three exposed components are cylinder curved area 60π, cone curved area 15π and cylinder bottom area 9π, all in cm².

Add them: A_exterior=60π+15π+9π=84π cm², approximately 264 cm² to three significant figures.

This is the outside geometric area of our ideal solid model. It is not an estimate of material thickness, filter capacity or water-contact area within a hollow cartridge.

Why the common double-counting method fails

Suppose a pupil calculates total surface area of the separate cylinder as 2πrh+2πr²=78π cm², and total surface area of the separate cone as πrl+πr²=24π cm².

Adding them gives 102π cm², but the two circular faces at the interface—one on each separate component—are internal after assembly. Subtract 2×9π=18π cm² to obtain 84π cm².

This is an alternative correct route. It also explains why the hidden circular interface must effectively be removed twice from the separate total surface areas, while the same interface is irrelevant to volume addition.

Two routes to the same exposed area

Direct route: add the exposed cylinder curve, cone curve and bottom base, obtaining 60π+15π+9π=84π.

Separate-solids route: 78π+24π−18π=84π. Both approaches independently verify the external surface-area answer.

On a longer exam question, use the route that makes the diagram’s exposed faces clearest. A list of visible faces is often less error-prone than a formula followed by a hidden-face correction.

A full calculation in exact and decimal forms

The volume is 102π cm³, which is about 320.442 cm³ before final rounding. The external area is 84π cm², about 263.894 cm².

If the question asks for three significant figures, the reports become 320 cm³ and 264 cm². If it asks for answers in terms of π, retain the exact expressions.

Do not round π to 3 midway through the calculation. The published K310 assessment notes explain how to handle π and final accuracy; use the question’s explicit instruction when it differs from a default.

Estimate to catch a tenfold mistake

The cylinder volume is roughly 9×3.14×10, around 283 cm³. The cone adds about 38 cm³, so a total near 320 cm³ makes sense.

Curved cylinder area is roughly 6×3.14×10≈188 cm²; the cone and bottom add roughly 47 and28 cm², giving about263 cm².

A final value of 3200 cm³ or 26 cm² would be suspicious under these dimensions. A short estimate is an independent check on calculator entries and missing powers.

How a rectangular packaging box compares

Imagine a fictional rectangular display box with internal dimensions 8 cm by 8 cm by 16 cm. The composite object has maximum diameter six centimetres and height fourteen centimetres.

Those maximum bounding dimensions fit within the stated dimensions when aligned centrally: 6≤8 in both horizontal directions and14≤16 vertically. This shows dimension-wise clearance for the ideal object under a chosen orientation.

It does not establish real packaging suitability, shock resistance or manufacturability. The box is simply a mathematical cuboid used to practise dimension comparisons.

Volume of the fictional display box

The ideal cuboid’s internal volume is 8×8×16=1024 cm³.

The composite object occupies 102π≈320.44 cm³. The difference in the geometrical volumes is 1024−102π≈704 cm³.

That is an unused-volume calculation for two ideal shapes, not a claim about whether multiple solid cartridges can pack into the empty regions. Volume subtraction alone does not solve a packing arrangement.

The object’s fraction of box volume

Divide the object volume by the box volume: 102π/1024≈0.313.

The object occupies about 31.3% of the ideal cuboid’s internal volume. The rest is 68.7% under pure volume subtraction.

This does not mean the other 68.7% is all available as one continuous usable region. Its shape matters for physical packing and is not determined by the total-volume difference.

A fictional material density calculation

Suppose, separately, the solid display model is made of a homogeneous imaginary material with density 0.8 g/cm³. This is a classroom figure, not the property of a real filter.

Mass equals density times volume: 0.8×102π=81.6π g, about 256 g to three significant figures.

Check unit cancellation: g/cm³ times cm³ leaves grams. A negative or extremely large mass would suggest a unit or decimal error.

Why density and surface area answer different questions

Mass at fixed density depends on volume, measured in cm³. Surface coating required at fixed coverage depends on exposed area, measured in cm².

For our model, volume is 102π cm³, while exposed area is 84π cm². Their numerical coefficients do not make the two quantities interchangeable.

A student should identify whether a price is given per cubic centimetre of solid material or per square centimetre of exterior coating before multiplying anything.

A coating-rate exercise

Assume one fictional coating tube covers up to 150 cm² of exposed area, under an ideal uniform coating assumption. The composite’s exposed surface area is 84π≈263.89 cm².

Divide by coverage: 263.89/150≈1.76 tubes. Since the fictional supplier sells only whole tubes, at least 2 tubes are required.

This is a mathematical coverage calculation, not advice about coating real filtration equipment. A real coating’s suitability, waste and safety properties are not specified.

A ten-percent coating allowance

Suppose the question requires ten percent extra coverage beyond the calculated surface area. Required nominal coverage becomes 1.10×84π=92.4π≈290.28 cm².

At 150 cm² per tube, the ratio is about1.94, so 2 whole tubes still suffice for the stated allowance.

The choice of two remains valid because the supplied nominal coverage 300 cm² exceeds the new requirement. A student should test against the actual rounded-up product capacity, not assume every percentage increase adds one whole tube.

A twenty-five-percent coating allowance

Increase the allowance to 25% instead. Required coverage is 1.25×84π=105π≈329.87 cm².

Two tubes provide only 300 cm², below 329.87. Therefore the model now requires 3 whole tubes, with nominal capacity450 cm².

The relationship between required area and purchase count changes in steps. A relatively small increase beyond a pack threshold can cause the required number of whole packs to jump.

Fictional tube cost and whole-number budgets

Suppose each imaginary coating tube costs S$9.50 as a final quoted price. Two tubes cost S$19, while three cost S$28.50.

If the fictional coating budget is S$25, the ten-percent allowance fits but the twenty-five-percent allowance does not. The shortfall for the three-tube case is S$3.50.

Do not conclude that the coating is actually safe or appropriate for a functional water filter. The calculation only compares stated nominal coverage and invented prices.

How surface area changes when the model is enlarged

Imagine making a geometrically similar copy with every length scaled by 1.5. The cylinder radius becomes4.5cm, its height15cm, and the cone perpendicular height6cm.

Every exposed surface area scales by the square of the linear factor: 1.5²=2.25. The new external area is 2.25×84π=189π cm², about594cm².

This applies because all linear dimensions are changed by the same factor. If the radius alone changed while heights stayed fixed, the shapes would not form a similar enlargement and the simple area factor would not apply.

How volume scales under the same enlargement

For similar solids, volume scales by the cube of the length factor. Here 1.5³=3.375.

The new volume is 3.375×102π=344.25π cm³, about 1081 cm³. The overall height increases from fourteen to twenty-one centimetres.

An area factor of 2.25 and volume factor of3.375 are different. A child who applies the area factor to volume is confusing two-dimensional and three-dimensional measurements.

An inverse volume-ratio calculation

Suppose two similar cartridge models have a volume ratio small:large of 8:27. The corresponding positive length ratio is the cube root, 2:3.

If the smaller model has an overall height14cm, the larger has height 14×3/2=21cm. The corresponding external area ratio is 4:9.

This matches the earlier scale factor1.5. Taking a square root of a volume ratio instead of a cube root would produce a different, incorrect length relationship.

A hollow cartridge would need a different model

Real filter cartridges can have internal cavities, but our primary object is deliberately solid. If a cylindrical hole were introduced, its volume would need to be subtracted, and its exposed internal walls might affect any surface-area request.

Without an explicit inner radius and hole depth, no numerical hollow-volume calculation can be justified. A student should not invent a hollow core simply because the title mentions a filter.

This illustrates the importance of reading the stated geometry rather than relying on what a real product usually looks like.

A separate hollow-cylinder illustration

Consider an independent hollow cylinder of outer radius3cm, inner radius1cm and height10cm, open at both ends. Its material volume is π(3²−1²)×10=80π cm³.

This is smaller than a completely solid cylinder of the same outer dimensions, which has volume90πcm³. The difference 10π cm³ is the removed inner cylindrical space.

This is a separate example, not a modification secretly applied to the original composite with a solid cone. Keeping examples independent prevents accidental mixing of assumptions.

An open hollow cylinder has several exposed surfaces

For the separate open-ended hollow cylinder, the outer curved surface area is2π×3×10=60πcm², while the inner curved surface is2π×1×10=20πcm².

Each ring-shaped open end has areaπ(3²−1²)=8πcm², and there are two such end faces. The full material surface area, including inner walls and annular ends, is 60π+20π+16π=96πcm².

The phrase ‘surface area’ needs interpretation: an exterior-only coating may exclude the inner wall, while total exposed material surface includes it. State what is counted before summing.

Why a right cone’s cross-section is helpful

A central cross-section through a right circular cone is an isosceles triangle with base diameter6cm, equal sloping sides5cm and perpendicular altitude4cm.

The half-triangle has horizontal leg3cm and vertical leg4cm, so the five-centimetre slant height follows directly from Pythagoras.

Do not calculate the slant height using the full six-centimetre diameter together with the four-centimetre height. That would describe a different triangle and produce √52 instead of five.

A slant-height change without equal scaling

Imagine a new cone with the same base radius3cm but perpendicular height8cm rather than4cm. Its slant height is √(3²+8²)=√73≈8.54cm.

Its volume is (1/3)π×9×8=24πcm³, twice the original cone volume because height doubled at fixed base radius.

However, its curved surface area changes from15π to 3π√73cm², not necessarily doubles. This is not a similar enlargement because radius remained fixed.

Find an unknown height from a cone’s volume

Suppose a separate cone has radius3cm and volume 18πcm³. Use V=(1/3)πr²h.

Then 18π=(1/3)π×9×h=3πh. Dividing by3π gives h=6cm.

This is formula rearrangement using correct dimensions. The slant height would be √(3²+6²)=√45=3√5cm if subsequently requested, but it is not part of the volume equation.

Find an unknown cylinder height from a total volume

Imagine another composite with radius3cm, a cone of height4cm and a cylinder of unknown perpendicular height H. Suppose the total solid volume is 75πcm³.

The cone contributes12π, so the cylinder contributes75π−12π=63π. Since its base area is9π, its height is H=63π/(9π)=7cm.

The overall height then becomes7+4=11cm. This reversed problem tests whether students can isolate one component before rearranging a formula.

An exterior-area formula for variable cylinder height

Keep radius3cm and the cone height4cm, so slant height remains5cm. Let the cylinder’s height be Hcm rather than ten.

Exposed area is curved cylinder2π×3×H, plus curved cone15π, plus bottom9π, giving A(H)=6πH+24π cm².

Substitute H=10 to recover A=60π+24π=84πcm². A simple change of H leads to a linear increase in this particular surface-area model.

A target exterior area: reverse the formula

Suppose the variable-height composite must have exposed area 72πcm², with the cone and radius unchanged.

Set 6πH+24π=72π. Subtract24π and divide by6π to obtain H=8cm.

Check directly: curved cylinder48π, cone15π and bottom9π sum to72π. The full height would be8+4=12cm.

An area bound and a whole-centre interpretation

Suppose the fictional design condition is external area at most 90πcm². With radius and cone fixed, solve6πH+24π≤90π.

Then6πH≤66π, so H≤11cm for a nonnegative cylinder height. The original H=10cm satisfies this mathematical limit.

This is a length inequality, not a claim about actual product safety. If a separate minimum height were required, both constraints would need to be combined.

Compare a cone cap with a flat cap

Consider two ideal composite variants sharing the same cylinder r=3cm, h=10cm. The first has the cone cap already studied; the second has a simple flat circular upper face instead.

The flat-capped cylinder has total outside area 2πrh+2πr²=60π+18π=78πcm². The cone-capped composite has84πcm².

The cone cap increases exterior area by 6πcm² in this comparison, because its curved 15π replaces the flat upper9π. But it also adds12πcm³ of solid volume and four centimetres of height.

The exact difference between the two caps

The cone’s curved cap area is15π, while the flat circular upper face would be9π. Subtract to find 15π−9π=6πcm² additional exterior area.

The percentage increase relative to the flat-capped cylinder’s78π exterior area is 6π/78π×100%=100/13≈7.69%.

If the question instead asks the extra area relative to the cone-capped result84π, the percentage base changes. An answer must identify which design represents the original whole.

A composite solid need not be filled uniformly

Our primary volume calculation treats both components as ideal solid shapes. If the cone and cylinder had different material densities, their masses should be computed separately.

For example, if the cylinder density is0.8g/cm³ and the cone density is0.5g/cm³, mass becomes 0.8×90π+0.5×12π=72π+6π=78πg, about245g.

Multiplying the total102π volume by a single density0.8 would now be wrong because the cone was assigned a different density. The stated material property must accompany its component.

A composite-solid three-line final answer

Write the volume as 90π+12π=102πcm³. State the exposed-area inventory explicitly: cylinder curve60π, cone curve15π and bottom9π.

Then report 84πcm² for the exterior area. Include a sentence explaining that the circular joining faces are internal and therefore excluded.

This compact mathematical chain makes the most important reasoning visible: component choice, formula substitution, exposed-face counting and final units.

A seven-day composite-solids revision sequence

Day one: distinguish radius, diameter, vertical height and slant height. Day two: calculate cylinder and cone volumes. Day three: find slant height using a 3–4–5 triangle.

Day four: list exposed and hidden faces before computing external area. Day five: practise whole-unit coating coverage and allowances. Day six: use similarity ratios for scaled solids. Day seven: solve a new composite with changed radii and heights independently.

Keep an error log labelled radius not squared, cone one-third missing, slant versus perpendicular height, hidden interface counted, area versus volume unit, whole-pack threshold or invalid physical assumption.

Frequently asked questions about composite solids

Why is the composite volume 102πcm³? The cylinder contributes90π and the cone12π, with no interior overlap. Why is outside area 84πcm²? Add the two curved areas60π and15π plus the bottom9π.

Should the internal joining discs be included? No, not in exterior area. Why is slant height five? Pythagoras gives √(3²+4²)=5. Does a 1.5 length scale multiply volume by1.5? No, for similar solids it multiplies volume by1.5³.

Are the filters and coating quantities real? No. This is a fictional geometry model. Is composite-solid mensuration covered by K310? Yes, the official 2027 G3 Mathematics syllabus explicitly includes it.

Three-pupil Mathematics tutorials near Sixth Avenue MRT

The unchanged eduKateSG Mathematics small-group reference describes premium three-pupil tutorials lasting 1.5 hours weekly, with deliberate practice, close corrections and individual feedback, at 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. This article does not announce a separate centre or guarantee a seat.

A learner who understands volume may still count two hidden circles in exposed area; another may use the slant height in the cone-volume formula, while a third may mishandle a similar-solid volume ratio. A tutor can identify their distinct first mistakes and check them through changed, independent problems.

Enquire about an eduKate Singapore Mathematics parent–student consultation with the actual examination year and an independently completed school question.

Which Mathematics syllabus applies?

The first Singapore–Cambridge SEC sitting takes place in 2027, with G3 Mathematics code K310 on SEAB’s official subject list. The 2027 K310 syllabus includes volumes and surface areas of composite solids. A 2026 GCE O-Level Mathematics candidate must use the applicable 4052 syllabus and exam instructions instead.

Official sources and connected guides

All cartridge dimensions, density, coating and pricing are fictional Mathematics teaching data, not actual engineering or product safety specifications. The article does not predict an exam question or guarantee a grade.