Exercises — 3D shapes — cube, cuboid, cylinder, cone, sphere, prism, pyramid

$\pi$" and "cubic vs square units" mean $\pi \approx 3.14159$ is the fixed ratio (circumference $÷$ diameter) of any circle — see Circle Geometry. A cubic unit (cm³) measures how much space fits inside — three lengths multiplied. A square unit (cm²) measures skin — two lengths multiplied. See Units and Measurement. If your answer for volume comes out in cm², you multiplied the wrong number of lengths.

WHAT we do: substitute $a=7$ straight into the two cube formulas — nothing to build. Volume: $V=a^3=7^3=343$ cm³. Surface area: $SA=6a^2=6\times 7^2=6\times 49=294$ cm². Why cube for volume, square for area: volume stacks length three times ($7\times 7\times 7$); each of the 6 faces is a flat $7\times 7$ square.

$SA=4\pi r^2=4\pi (3^2)=4\pi (9)=36\pi \approx 113.10$ cm². Why: surface area of a sphere is exactly four "great circles" of area $\pi r^2$ each.

Volume: $V=lwh=25\times 15\times 10=3750$ cm³. Surface area: three pairs of faces. $SA=2(lw+lh+wh)=2(25\cdot 15+25\cdot 10+15\cdot 10)$ $=2(375+250+150)=2(775)=1550{\text{ cm}}^2.$ Why the three products: top/bottom are $lw$, front/back are $lh$, sides are $wh$; each appears twice.

Volume: $V=\pi r^{2}h=\pi (5^2)(12)=\pi (25)(12)=300\pi \approx 942.48$ cm³. Surface area: two circular lids + one unrolled rectangle. $SA=2\pi r(r+h)=2\pi (5)(5+12)=2\pi (5)(17)=170\pi \approx 534.07{\text{ cm}}^2.$ Why "unroll": the curved wall, cut and flattened, is a rectangle of width $2\pi r$ (the base circumference) and height $h$.

Step 1 — slant height (WHY Pythagoras here): the radius, the height, and the slant form a right triangle (look at the amber triangle in the figure). The right angle sits where the height meets the base. The slant $l$ is the hypotenuse, so Pythagorean Theorem gives $l=\sqrt{r^2+h^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ cm}.$ Step 2 — surface area: $SA=\pi r(r+l)=\pi (6)(6+10)=\pi (6)(16)=96\pi \approx 301.59{\text{ cm}}^2.$

WHY a cube root: $V=a^3$, so to undo "cube it" we take the cube root — the inverse operation. $a=\sqrt[3]{512}=8\text{ cm}(\text{since }{8}^3=512).$ $SA=6a^2=6(8^2)=6(64)=384{\text{ cm}}^2.$

KEY IDEA: melting changes shape but not volume — the same metal, rearranged. So set sphere volume equal to cylinder volume. $\underset{\text{sphere}}{\underbrace{\frac{4}{3}\pi r^3}}=\underset{\text{cylinder}}{\underbrace{\pi R^{2}h}}.$ Cancel $\pi$ from both sides (it appears on both), then substitute $r=R=6$: $\frac{4}{3}(6^3)=(6^2)h\Rightarrow \frac{4}{3}(216)=36h\Rightarrow 288=36h.$ $h=\frac{288}{36}=8\text{ cm}.$

New edge $=3a$. Surface area: $6(3a{)}^2=6\cdot 9a^2=9\times$ the old $SA$ → factor 9. Volume: $(3a{)}^3=27a^3=27\times$ the old $V$ → factor 27. Why 9 and 27: area is two lengths so it scales by $3^2=9$; volume is three lengths so it scales by $3^3=27$. This is the general square–cube scaling rule (see Real-World Applications).

WHAT: split the object into two known pieces and add their volumes (a hemisphere is half a sphere). Cone: $V_{\text{cone}}=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (9)(4)=12\pi$ cm³. Hemisphere: $V_{\text{hemi}}=\frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi (27)=18\pi$ cm³. Total: $V=12\pi +18\pi =30\pi \approx 94.25$ cm³. Why add and not subtract: the two solids sit side-by-side (no overlap), so their spaces simply combine.

KEY: metal fills the outer cylinder minus the hollow inner cylinder. $V=\pi R^{2}h-\pi r^{2}h=\pi h(R^2-r^2)=\pi (20)(25-9)=\pi (20)(16)=320\pi \approx 1005.31{\text{ cm}}^3.$ Why $R^2-r^2$ and not $(R-r{)}^2$: we subtract the two areas, and $(R-r{)}^2$ is not $R^2-r^2$. Compute each area, then subtract.

Find heights from $V=\pi r^{2}h$: Tall: $1000\pi =\pi (25)h\Rightarrow h=40$ cm. Wide: $1000\pi =\pi (100)h\Rightarrow h=10$ cm. Surface areas $SA=2\pi r(r+h)$: Tall: $2\pi (5)(5+40)=2\pi (5)(45)=450\pi \approx 1413.72$ cm². Wide: $2\pi (10)(10+10)=2\pi (10)(20)=400\pi \approx 1256.64$ cm². Conclusion: the wide can uses less metal ($400\pi <450\pi$). The very tall can wastes metal on a long thin wall. (The true minimum sits between these, where height equals diameter — a calculus result, see Integration.)

Set volumes equal (metal conserved): $\frac{4}{3}\pi r^3=\frac{1}{3}\pi R^{2}H.$ Use $R=H$, so $R^{2}H=H^3$: $\frac{4}{3}\pi r^3=\frac{1}{3}\pi H^3.$ Multiply both sides by $3$ and cancel $\pi$: $4r^3=H^3\Rightarrow H=\sqrt[3]{4r^3}=r\sqrt[3]{4}.$ Evaluate $r=3$: $H=3\sqrt[3]{4}\approx 3(1.5874)\approx 4.762$ cm. Why cube root at the end: $H$ appears cubed, so we undo it with the cube root — the inverse of "raise to the 3rd."

Volume of cone is what fraction of the cylinder with same base/height? ::: One third, $\frac{1}{3}\pi r^{2}h$. To go from a sphere's volume to metal reused as a cylinder, what stays equal? ::: The volume — melting conserves it. Tripling every edge multiplies volume by what? ::: $27$ (that is $3^3$). Slant height of a cone in terms of $r,h$? ::: $l=\sqrt{r^2+h^2}$ (Pythagoras on the right triangle). A hemisphere scoop has what fraction of a full sphere's volume? ::: One half, $\frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$.