Alliance Form 3 P1 Q22 — Cone with a Conical Hole

KCSE (Form 3) Form 3 Measurement

Published

The Question

“A cone of height 12 cm and base radius 9 cm has a smaller similar cone removed from its base, leaving a conical hole 4 cm deep. (a)(i) Find the radius of the hole. (a)(ii) Find the volume of the small cone that was removed, in terms of pi. (b) Find the slant height of the big cone. (c) Find the total surface area of the solid that remains, in terms of pi.”

1

Find the radius of the hole

The small cone that was scooped out is similar to the big cone, so all their matching lengths are in the same ratio. The hole is 4 cm deep against the big cone's height of 12 cm, so the radius of the hole compares to the base radius of 9 cm in that same ratio. Setting the two ratios equal and solving gives the hole's radius.

r9=412\frac{r}{9} = \frac{4}{12}
r=9×412=3 cmr = 9 \times \frac{4}{12} = 3 \text{ cm}
2

Find the volume of the small cone removed

Use the standard cone volume formula with the hole's radius of 3 cm and depth of 4 cm. Keeping the answer in terms of pi avoids rounding and matches how the question is asked.

V=13πr2h=13π×32×4V = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi \times 3^{2} \times 4
V=12π cm3V = 12\pi \text{ cm}^{3}
3

Find the slant height of the big cone

The radius, height and slant height of a cone form a right-angled triangle, with the slant height as the hypotenuse. Apply Pythagoras to the base radius of 9 cm and the height of 12 cm.

l=92+122=81+144l = \sqrt{9^{2} + 12^{2}} = \sqrt{81 + 144}
l=225=15 cml = \sqrt{225} = 15 \text{ cm}
4

Add up the three surfaces that remain

The remaining solid has three exposed surfaces. First is the curved outer surface of the big cone. Second is the base, which is now a flat ring because the hole opens through the middle of it, so subtract the hole's circle from the full base circle. Third is the inner cone surface of the hole itself, which needs its own slant height from Pythagoras on the hole's radius of 3 cm and depth of 4 cm.

Big curved surface=πrl=π×9×15=135π\text{Big curved surface} = \pi r l = \pi \times 9 \times 15 = 135\pi
Flat ring base=π(92−32)=72π\text{Flat ring base} = \pi(9^{2} - 3^{2}) = 72\pi
Hole slant=32+42=5 cm\text{Hole slant} = \sqrt{3^{2} + 4^{2}} = 5 \text{ cm}
Inner cone surface=π×3×5=15π\text{Inner cone surface} = \pi \times 3 \times 5 = 15\pi
Total=135π+72π+15π=222π cm2\text{Total} = 135\pi + 72\pi + 15\pi = 222\pi \text{ cm}^{2}

The pointed tip of the big cone is still solid, so there is no extra top circle to add.

Final Result

The hole has radius 3 cm, the small cone removed has volume 12 pi cubic centimetres, the big cone's slant height is 15 cm, and the total surface area of the remaining solid is 222 pi square centimetres.

Why this method works

The whole question turns on the two cones being similar. Because the small cone is a scaled copy of the big one, every length shrinks by the same factor of 4 over 12, which is one third, so its radius must be one third of 9, giving 3 cm. That single ratio unlocks both the volume of the removed piece and the slant height of the hole. For the surface area you must picture what the eye would actually see on the finished solid: the outer slope stays, the flat base loses a disc and becomes a ring, and the scooped-out space adds a new inner cone surface. Slant heights, not vertical heights, are used for curved surfaces because a cone's skin runs along the slope, which is why Pythagoras appears twice.

✓

Check the ring: pi(81 - 9) = 72 pi, and 135 pi + 72 pi + 15 pi = 222 pi, confirming the total surface area.