Mensuration (3D)

medium importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Cone

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A cone of radius r and height h has slant height root of r squared plus h squared. Its volume is one third of a cylinder on the same base; its curved surface is pi r l. Radius-height-slant triples are Pythagorean triplets, most often 3-4-5 and 7-24-25.

The three lengths first

\ell=\sqrt{r^2+h^2}

Radius $7$ and height $24$ give slant $25$ : the $7$ - $24$ - $25$ triplet. Check for a triplet before squaring anything. Radius $21$ with height $28$ gives slant $35$ , and $\dfrac{22}{7}\times21\times35=2310$ sq cm of curved surface.

The three lengths make a right triangle inside the cone: radius and height at right angles, slant along the surface. Families to know: $3$ - $4$ - $5$ and its multiples, $7$ - $24$ - $25$ , $8$ - $15$ - $17$ , $12$ - $35$ - $37$ .

A classic ratio: cylinder, hemisphere and cone on the same base with cylinder height $=$ radius hold volumes $3:2:1$ .

Rule: Find $\ell$ first. Both surface formulas need it, and triplet spotting avoids the root.

Volume and the one-third

V=\frac{1}{3}\pi r^2h

Exactly one-third of the cylinder with the same base and height. A cylinder of volume $120$ pairs with a cone of $40$ . Radius $7$ with height $9$ gives $V=\dfrac13\times154\times9=462$ cu cm.

The factor $\dfrac13$ also runs in reverse: cone volume $132$ with radius $3$ gives height $\dfrac{132\times3}{\dfrac{22}{7}\times9}=14$ cm.

Tip: Multiply by $3$ before dividing by $\pi r^2$ . Keeping the fraction until the end invites slips.

Surfaces

Curved surface: $\pi r\ell$ . Radius $7$ , slant $25$ : $\dfrac{22}{7}\times7\times25=550$ sq cm.
Total surface: $\pi r\ell+\pi r^2$ , the skirt plus the circular base. Same cone: $550+154=704$ sq cm.

A conical tent uses the curved surface only: no floor. Canvas for radius $7$ m and slant $25$ m is $550$ sq m. A heap whose base circumference is $44$ m has radius $\dfrac{44\times7}{44}=7$ m, so slant $25$ m again means $550$ sq m of sheet.

Watch: Tents and caps are curved surface; ice-cream cones and solid cones are total surface. Read what is covered.

Melting a cone

Volume survives. A cone of radius $6$ , height $24$ has volume $288\pi$ ; as a sphere:

\frac{4}{3}\pi R^3=288\pi \Rightarrow R^3=216 \Rightarrow R=6

The sphere comes out with the same $6$ cm radius, a clean coincidence worth noticing. And $216=6^3$ came straight from the cubes table, not a calculator.

The reverse melt works the same way: a sphere of radius $6$ recast as a cone of radius $6$ needs $\dfrac13\times36\times h=288$ , so $h=24$ cm.

Heaps and tents

A heap of rice is a cone: base circumference gives the radius, slant gives the canvas or sheet. Cloth width questions divide the curved area by the width.

Remember: Cone problems are two-part: the inner right triangle for lengths, then one formula for the asked quantity.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Slant height and CSA/TSA with a triplet

How to spot it:

Radius and height given, slant or a surface asked.

Method

Look for a triplet among $r$ , $h$ , $\ell$ .
Compute $\ell$ if missing.
CSA $=\pi r\ell$ ; TSA adds $\pi r^2$ .

Why it works:

The inner right triangle settles the lengths, then one multiplication finishes.

Try this

The radius of a cone is 7 cm and its height is 24 cm. Its curved surface area is:

Show solution

$\ell=\sqrt{49+576}=25$ .
CSA $=\dfrac{22}{7}\times7\times25=550$ sq cm.

Answer

550 sq cm

Type 2very common3 practice Q

Volume of a cone (and reverse)

How to spot it:

Volume asked from dimensions, or a dimension from the volume.

Method

$V=\dfrac{1}{3}\pi r^2h$ .
Reverse: multiply the volume by $3$ .
Divide by the known factors to isolate the unknown.

Why it works:

The one-third factor runs both directions with one multiplication.

Try this

The volume of a cone is 132 cu cm and its base radius is 3 cm. Its height is:

Show solution

$3V=396$ .
$\pi r^2=\dfrac{198}{7}$ .
$h=396\times\dfrac{7}{198}=14$ cm.

Answer

14 cm

Type 3common2 practice Q

Recasting: cone into sphere or spheres

How to spot it:

A cone melted and recast; a radius of the new solid asked.

Method

Write the cone volume.
Set it equal to the target solid's volume.
Cancel $\pi$ , solve the remaining equation.

Why it works:

Melting conserves volume, and the pi cancels, leaving integer arithmetic.

Try this

A solid metallic cone of radius 6 cm and height 24 cm is melted into a sphere. The radius of the sphere is:

Show solution

$V=\dfrac13\pi\times36\times24=288\pi$ .
$\dfrac43\pi R^3=288\pi\Rightarrow R^3=216$ .
$R=6$ cm.

Answer

6 cm

Type 4common2 practice Q

Cone vs cylinder and hemisphere ratios

How to spot it:

Two solids on the same base; a volume ratio asked.

Method

Same base and height: cone $=$ one-third of cylinder.
Hemisphere on the same radius: volume $\dfrac23\pi r^3$ .
Compare by dividing the formulas.

Why it works:

With equal bases the pi and radius cancel, so ratios come from the constants alone.

Try this

A cylinder and a cone have equal bases and equal heights. If the cylinder's volume is 120 cu cm, the cone's volume is:

Show solution

Cone $=\dfrac13$ of the cylinder.
$\dfrac{120}{3}=40$ cu cm.

Answer

40 cu cm

Type 5common2 practice Q

Conical heap or tent word problems

How to spot it:

A tent, heap or cap; canvas, cloth or sheet area asked.

Method

Extract $r$ (from circumference if needed) and $\ell$ .
Tent: CSA $=\pi r\ell$ .
Cloth width: divide the area by the width.

Why it works:

Real-world wording hides the same two formulas; the curved surface is almost always the target.

Try this

A conical tent has base radius 7 m and slant height 25 m. The canvas required to make it (sq m) is:

Show solution

CSA $=\dfrac{22}{7}\times7\times25$ .
$=550$ sq m.

Answer

550 sq m

Formula sheet

Slant height

\ell=\sqrt{r^2+h^2}

Radius, height, slant: a right triangle.

Cone volume

V=\frac{1}{3}\pi r^2h

One-third of the same-base cylinder.

Cone surfaces

\text{CSA}=\pi r\ell,\quad \text{TSA}=\pi r(\ell+r)

Skirt alone, or skirt plus base.

Same base ratios

V_{\text{cone}}=\frac{V_{\text{cyl}}}{3}

Equal base and height.

Shortcuts that save time

⚡ Hunt the triplet

7-24-25, 3-4-5 and their multiples cover nearly every cone. Two lengths known, read the third.

Example

The radius of a cone is 7 cm and its height is 24 cm. Its curved surface area is:

Show solution

$\ell=\sqrt{49+576}=25$ cm (triplet).
CSA $=\dfrac{22}{7}\times7\times25$ .
$=550$ sq cm.

Answer

550 sq cm

⚡ One-third both ways

Cone to cylinder: divide by 3. Cone volume given: multiply by 3 before dividing by the base area.

Example

The volume of a cone is 132 cu cm and its base radius is 3 cm. Its height is:

Show solution

$132\times3=396$ .
$\pi r^2=\dfrac{22}{7}\times9=\dfrac{198}{7}$ .
$h=396\div\dfrac{198}{7}=14$ cm.

Answer

14 cm

⚡ Tent = curved surface

Canvas touches only the slanted side, so use pi r l. Add the base circle only for a solid cone.

Example

A conical tent has base radius 7 m and slant height 25 m. The canvas required to make it (sq m) is:

Show solution

CSA $=\dfrac{22}{7}\times7\times25$ .
$=550$ sq m.

Answer

550 sq m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $\pi rh$ for the curved surface.

It is $\pi r\ell$ with the slant. For 7 and 24 that means 550, not $\pi\times7\times24$ .

Mistake 02

Treating the slant as the height.

Height is the drop down the middle; slant runs along the surface. $\ell=\sqrt{r^2+h^2}$ .

Mistake 03

Forgetting the $\dfrac13$ in the cone volume.

A cone is one-third of its cylinder: $120$ cylinder means $40$ cone.

Mistake 04

Adding the base to a tent's canvas.

Tents need the curved surface $\pi r\ell$ only.

Mistake 05

Cancelling $\pi$ but not the $\dfrac13$ when melting.

Carry the one-third: $288\pi$ as a sphere gives $R^3=216$ , $R=6$ .

Quick revision

Read this the night before the exam.

Slant $\ell=\sqrt{r^2+h^2}$ ; hunt for triplets first.
$V=\dfrac13\pi r^2h$ : one-third of the same-base cylinder.
CSA $=\pi r\ell$ ; TSA adds the base circle.
Tents and caps: curved surface only.
Melting keeps volume: equate $\dfrac13\pi r^2h$ with the new solid's formula.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.

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