Frustum of a Cone

Worked example

1. 1

   $\frac{1}{3}\pi \times 9^2\times 15$ — Big cone: put r = 9 and h = 15 into the cone formula $V=\frac{1}{3}\pi r^{2}h$.

2. 2

   $=\frac{1}{3}\pi \times 81\times 15=405\pi$ — 9² = 81, then 81 × 15 ÷ 3 = 405. Leave π as it is, so nothing is rounded yet.

3. 3

   $\frac{1}{3}\pi \times 3^2\times 5=15\pi$ — Small cone: the same formula with r = 3 and h = 5. 9 × 5 ÷ 3 = 15.

4. 4

   $405\pi -15\pi =390\pi$ — Take the small cone away. What is left is the frustum.

5. 5

   $390\pi \approx 1225.2{\text{ cm}}^3$ — Only now use the π button: 390 × π = 1225.22…, so 1225.2 to 1 d.p.

Worked example

1. 1

   $\frac{3}{9}=\frac{1}{3}$ — Small radius ÷ big radius. Every length of the small cone is a third of the big one: this is the scale factor.

2. 2

   $h=\frac{1}{3}H$ — Call the big cone's height H. The small cone's height h is a third of it.

3. 3

   $H-\frac{1}{3}H=10$ — The frustum's height is the gap between the two cone heights.

4. 4

   $\frac{2}{3}H=10$ — A whole H take away a third of H leaves two thirds of H.

5. 5

   $H=10÷\frac{2}{3}=15$ — Divide both sides by ⅔: 10 × 3 ÷ 2 = 15. The big cone is 15 cm tall.

6. 6

   $h=15-10=5$ — The small cone is the big cone minus the frustum: 5 cm. Check: 5 is a third of 15.

Worked example

1. 1

   $12÷2=6$ — The question gives the diameter. The formula needs the radius, so halve it: R = 6 cm.

2. 2

   $16-12=4$ — The small cone's height: the big cone's height minus the frustum.

3. 3

   $\frac{4}{16}=\frac{1}{4}$ — Small height ÷ big height gives the scale factor, a quarter.

4. 4

   $r=6\times \frac{1}{4}=1.5$ — The small radius is the big radius times the same scale factor: 1.5 cm.

5. 5

   $\frac{1}{3}\pi \times 6^2\times 16=192\pi$ — Big cone: 36 × 16 ÷ 3 = 192.

6. 6

   $\frac{1}{3}\pi \times 1.5^2\times 4=3\pi$ — Small cone: 2.25 × 4 ÷ 3 = 3.

7. 7

   $192\pi -3\pi =189\pi$ — Take the small cone away to leave the frustum.

8. 8

   $189\pi \approx 593.8{\text{ cm}}^3$ — Round once, at the end: 189 × π = 593.76…

Worked example

1. 1

   $\frac{12}{36}=\frac{1}{3}$ — Small height ÷ big height. The small pyramid is a copy a third the size: the scale factor.

2. 2

   $30\times \frac{1}{3}=10\text{ cm}$ — The small base side. Every length shrinks by the same scale factor.

3. 3

   $30^2=900{\text{ cm}}^2$ — The big base area. The base is a square, so side × side.

4. 4

   $\frac{1}{3}\times 900\times 36=10800$ — Big pyramid: $V=\frac{1}{3}\times \text{base area}\times \text{height}$.

5. 5

   $10^2=100{\text{ cm}}^2$ — The small base area, side × side again.

6. 6

   $\frac{1}{3}\times 100\times 12=400$ — Small pyramid: the same formula with its own base and height.

7. 7

   $10800-400=10400{\text{ cm}}^3$ — Take the small pyramid away. What is left is the frustum.

Worked example

1. 1

   $k=\frac{h}{3h}=\frac{1}{3}$ — Small height ÷ big height. The h's cancel, leaving a third.

2. 2

   $k^3=\frac{1}{27}$ — Cube it, because volume grows in three directions: ⅓ × ⅓ × ⅓.

3. 3

   $V_{\text{big}}=\frac{1}{3}\pi r^2\times 3h=\pi r^{2}h$ — The big cone. The ⅓ and the 3 cancel.

4. 4

   $V=\pi r^{2}h\times \left(1-\frac{1}{27}\right)$ — The small cone is 1/27 of the big one, so the frustum keeps the rest.

5. 5

   $V=\frac{26}{27}\pi r^{2}h$ — 1 − 1/27 = 26/27. If the question says this equals another volume, write the two equal and simplify.

Worked example

1. 1

   $R^2+Rr+r^2=81+27+9=117$ — The frustum from before: R = 9, r = 3, h = 10. Work out the bracket: 9², 9 × 3 and 3².

2. 2

   $\frac{1}{3}\pi \times 10\times 117=390\pi$ — The same 390π as big minus small, without finding the 15 cm or the 5 cm.

Worked example

1. 1

   $L=\sqrt{6^2+8^2}=\sqrt{100}=10$ — Radius, height and slant make a right-angled triangle, so Pythagoras gives the big slant.

2. 2

   $l=\sqrt{3^2+4^2}=\sqrt{25}=5$ — The small slant, the same way.

3. 3

   $\pi \times 6\times 10=60\pi$ — Curved area of the big cone: π × radius × slant.

4. 4

   $\pi \times 3\times 5=15\pi$ — Curved area of the small cone, the same way.

5. 5

   $60\pi -15\pi =45\pi \approx 141.4{\text{ cm}}^2$ — Take away. This is the frustum's curved surface.

6. 6

   $\pi \times 6^2=36\pi$ — The flat circle at the bottom: $\pi r^2$ with r = 6.

7. 7

   $\pi \times 3^2=9\pi$ — The flat circle at the top, with r = 3.

8. 8

   $45\pi +36\pi +9\pi =90\pi \approx 282.7{\text{ cm}}^2$ — Total surface: the curved part plus both circles.

Worked example

1. 1

   $s=\sqrt{4^2+3^2}=5$ — The frustum is 8 − 4 = 4 cm tall and its radii differ by 6 − 3 = 3 cm. These make a right-angled triangle with s.

2. 2

   $\pi \times (6+3)\times 5=45\pi$ — The same 45π as before, in one step.

Worked example

1. 1

   $\frac{4}{16}=\frac{1}{4}$ — Small height ÷ big height gives the scale factor, a quarter.

2. 2

   $24\times \frac{1}{4}=6\text{ cm}$ — The top square's side.

3. 3

   $\sqrt{16^2+12^2}=\sqrt{400}=20$ — The slope of a big face, from the tip to the middle of a base edge. The pyramid's height and half the base (12) make a right angle with it.

4. 4

   $20\times \frac{1}{4}=5$ — The slope of a small face, by the same scale factor.

5. 5

   $\frac{1}{2}\times 24\times 20=240$ — One big triangular face: ½ × base × slope.

6. 6

   $\frac{1}{2}\times 6\times 5=15$ — The small triangle cut off that face.

7. 7

   $240-15=225$ — One sloping face of the frustum: big triangle minus small triangle.

8. 8

   $4\times 225=900$ — There are 4 identical sloping faces.

9. 9

   $24^2+6^2=576+36=612$ — The bottom and top squares.

10. 10

    $900+612=1512{\text{ cm}}^2$ — Total surface: sloping faces plus both squares.