Exercises — Nets of 3D shapes

WHAT the pieces tell us. Two circles means two flat circular faces. A cube has only squares — out. A cone has only one circle plus a pie-slice sector — out. A square pyramid has a square plus triangles — out. A cylinder has a top circle, a bottom circle, and a curved wall that unrolls into a rectangle. In figure s01 (described above) the tall blue rectangle is that unrolled wall, capped by two yellow circles. Answer: (b) cylinder.

Six equal squares that fold up with no gaps → a cube. As the parent note states, there are exactly 11 distinct nets for a cube. Answer: cube; 11 distinct nets.

False. When you fold a straight strip, the two end squares would need to meet the same edge but they are too far apart — the fold overlaps rather than closing (see the parent note's "Non-Valid Net" example). A valid cube net never has more than 4 squares in one straight line. Answer: False.

WHAT. The net is 6 squares, each of area $s^2$. WHY. A cube has 6 faces; laying them flat does not change any area, so the net area equals the surface area. $\text{Area}=6s^2=6\times 4^2=6\times 16=96{\text{ cm}}^2$ Answer: $96{\text{ cm}}^2$.

WHAT. Three pairs of rectangles: $l\times w$, $l\times h$, $w\times h$. WHY. Opposite faces of a cuboid are identical, so each rectangle appears twice. $\text{Area}=2lw+2lh+2wh$ $=2(5)(3)+2(5)(2)+2(3)(2)=30+20+12=62{\text{ cm}}^2$ Answer: $62{\text{ cm}}^2$.

WHAT. Two circles $2\pi r^2$, plus a rectangle (the unrolled wall) of width $2\pi r$ and height $h$, giving $2\pi rh$. In figure s01 (described at the top) the rectangle's width equals the circle's circumference — that is the whole trick. $\text{Circles: }2\pi r^2=2\pi (49)=98\pi$ $\text{Wall: }2\pi rh=2\pi (7)(10)=140\pi$ $\text{Total}=238\pi \approx 747.70{\text{ cm}}^2$ Answer: $238\pi \approx 747.70{\text{ cm}}^2$.

WHAT the slant height is. In figure s02 (the pyramid/slant idea described at the top) each triangle stands on a base of length $a$ and its height is the slant height $s$ — the distance up the sloping face, not the pyramid's vertical height. (a) One triangle: $\frac{1}{2}as=\frac{1}{2}(6)(5)=15{\text{ cm}}^2$. (b) Base square $a^2=36$; four triangles $4\times 15=60$. $\text{Total}=36+60=96{\text{ cm}}^2$ Answer: (a) $15{\text{ cm}}^2$; (b) $96{\text{ cm}}^2$.

WHY a sector. Unrolling the cone flattens it into a pie-slice (the blue sector in figure s02). The arc of that slice used to wrap around the base circle, so $\text{arc length}=\text{base circumference}=2\pi r.$ Arc length also equals $l\theta$ (radius × angle, in radians). Setting them equal: $l\theta =2\pi r\Rightarrow \theta =\frac{2\pi r}{l}=\frac{2\pi (4)}{12}=\frac{2\pi }{3}\text{ rad}.$ Convert: $\frac{2\pi }{3}\text{ rad}\times \frac{180}{\pi }=120^{\circ }$. Answer: $\theta =\frac{2\pi }{3}\text{ rad}=120^{\circ }$.

WHAT. Base circle $\pi r^2$ plus sector (curved surface) $\pi rl$. WHY the sector area is $\pi rl$. A full circle of radius $l$ has area $\pi l^2$ and angle $2\pi$. Our sector uses fraction $\frac{\theta }{2\pi }=\frac{2\pi r/l}{2\pi }=\frac{r}{l}$ of it, so its area is $\frac{r}{l}\cdot \pi l^2=\pi rl$. $\text{Total}=\pi r^2+\pi rl=\pi (16)+\pi (4)(12)=16\pi +48\pi =64\pi {\text{ cm}}^2$ Answer: $64\pi {\text{ cm}}^2\approx 201.06{\text{ cm}}^2$.

WHY Pythagoras enters. The slant height, the vertical height, and the base radius form a right triangle (the green triangle in figure s02): $r$ along the ground, $H$ straight up, $l$ as the hypotenuse. See Pythagoras Theorem. $l=\sqrt{r^2+H^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ cm}.$ Now the surface area: $\pi r^2+\pi rl=\pi (36)+\pi (6)(10)=36\pi +60\pi =96\pi {\text{ cm}}^2.$ Answer: $l=10\text{ cm}$; area $=96\pi \approx 301.59{\text{ cm}}^2$.

WHAT we do. Write the net-area formula and solve backwards for $h$. $2\pi r^2+2\pi rh=150\pi$ Divide every term by $\pi$: $2r^2+2rh=150\Rightarrow 2(25)+2(5)h=150\Rightarrow 50+10h=150.$ $10h=100\Rightarrow h=10\text{ cm}.$ Answer: $h=10\text{ cm}$.

WHY Pythagoras. The slant height runs from the apex down the middle of a triangular face to the midpoint of a base edge. That midpoint sits a distance $\frac{a}{2}$ from the pyramid's centre. So $s$ is the hypotenuse of a right triangle with legs $H$ and $\frac{a}{2}$: $s=\sqrt{H^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ cm}.$ Net area: $a^2+2as=10^2+2(10)(13)=100+260=360{\text{ cm}}^2.$ Answer: $s=13\text{ cm}$; net area $=360{\text{ cm}}^2$.

WHAT is missing. A closed cuboid has 6 faces; "open top" removes one $l\times w$ face. So we take the full formula and subtract one top: $\text{closed}=2lw+2lh+2wh.$ Remove one $lw$ face: $\text{open}=lw+2lh+2wh.$ With $l=8$, $w=6$, $h=5$: $=(8)(6)+2(8)(5)+2(6)(5)=48+80+60=188{\text{ cm}}^2.$ Answer: $188{\text{ cm}}^2$.

WHY some faces vanish. Where the cone's flat base meets the cylinder's top, both of those circular faces are sealed inside — neither is part of the outer surface. So we count:

• cylinder bottom circle: $\pi r^2=9\pi$
• cylinder curved wall: $2\pi rh=2\pi (3)(4)=24\pi$
• cone curved surface: $\pi rl=\pi (3)(5)=15\pi$
• cone base: excluded (hidden), cylinder top: excluded (hidden) $\text{Total}=9\pi +24\pi +15\pi =48\pi {\text{ cm}}^2.$ Answer: $48\pi \approx 150.80{\text{ cm}}^2$.

Set up both areas. Cube net: $6s^2$. Pyramid net: base $s^2$ plus 4 triangles each $\frac{1}{2}s\cdot s=\frac{1}{2}{s}^2$, so $s^2+4\cdot \frac{1}{2}{s}^2=s^2+2s^2=3s^2$. Compare. We need $6s^2=3s^2$, i.e. $6s^2-3s^2=0\Rightarrow 3s^2=0\Rightarrow s=0$. WHAT this means. The only solution is the degenerate $s=0$ (no shape at all). For every positive edge the cube's net is exactly twice the pyramid's ($6s^2=2\cdot 3s^2$), so they can never match. Answer: impossible for any real solid ($s>0$); they are equal only in the degenerate case $s=0$.