What Is The Shape Of The Cross Section Formed When A Cone Intersects A Plane As Shown In The Drawing

Explain xkcd: It'southward 'cause you're dumb.

Explanation [edit]

This explanation may be incomplete or incorrect: Created past a STRIPED AND DOTTED TEXTBOOK ILLUSTRATOR. Explain the formulas for each of the areas, and likewise the correct formula for the 3D object they seem to represent. Consider whether to add a tabular array with the formula given and the right formula for the 3D shape. Do NOT delete this tag likewise presently. If you lot can accost this outcome, please edit the page! Thanks.

This comic showcases area formulas for the areas of four ii-dimensional geometric shapes which each have extra dotted and/or solid lines making them look like illustrations for 3-dimensional objects. The first, a simple equation for the expanse of a circle, the second an equation for the surface area of a triangle with a semi-elliptic base, the third an equation for the area of a rectangle with an elliptical base and tiptop, and the 4th an equation for the area of a hexagon consisting of two opposing correct-angled corners and two parallel diagonal lines connecting their sides. In each case, only the area formed by the outline of each shape is calculated.

Similar illustrations are commonly plant in geometry textbooks, which are used to depict three-dimensional figures on a two-dimensional page. They commonly brand utilise of slanted lines to indicate edges receding into the altitude and dashed lines to indicate an edge occluded by nearer parts of the solid. The joke is that the formulae given here are for the surface area of each 2-dimensional shape inside its outer solid lines, not for the surface surface area or volume of the illustrated 3D object (every bit would exist shown in the geometry textbook). The title text continues the joke by challenge that the dotted lines are only decorative.

The illustrations describe the following plane or solid figures, depending on the interpretation.

Pinnacle Left - Circle with an inscribed ellipse, or Sphere

This illustration is commonly used to depict a 3-dimensional sphere, with the ellipse representing a "horizontal" or axial cross-section through the center; the solid lower half of the ellipse represents the "front" of the circumference of this cantankerous-section, while the dotted upper half represents the "back" of the same section, which would be occluded from view if this were a solid shape.

The radius of the circle, from the center to the right edge where information technology meets the ellipse, is labeled 'r'. In a textbook diagram of a sphere, the radius might be instead labeled with a diagonal line from the centre to a different betoken on the ellipse, implying the generality that all points on that cross-section, and indeed on the whole spherical surface, are at the same radius from the center. Even so, this line would be shorter on the page than the actual radius, making it useless for the formula of the area of the 2D outer shape.

The area of the 2D shape on the page is the surface area of the circle, which is A = πrⁱⁱ. This is captioned below the figure.

Coincidentally the area of the horizontal cross-section of the 3D sphere, every bit depicted by the ellipse, is also πr^two, and a reader familiar with such diagrams might initially assume that this is what was meant. However, this does non extend to the other figures.

The 3D sphere commonly depicted by this drawing would accept a volume of ⁴/_three πr³ and a surface area of 4πr².

Top Right - Ellipse with symmetrical diagonal lines, or Cone

This illustration is commonly used to depict a three-dimensional right circular cone, with the lower half of the ellipse representing the "front border" of the bottom surface, and the upper half representing the occluded "back edge". Still such drawings would commonly not employ both 'a' and 'b' to describe the radius of the base of the cone, which is drawn as an ellipse due to foreshortening. Alternatively, the cartoon could depict a right elliptical cone.

Randall approximates the surface area of the 2nd shape on the page as the sum of the surface area of the triangle formed by the major axis of the ellipse and the two lines, and half of the area of the ellipse (^π/₂ ab) since most of the upper half of the ellipse overlaps the triangle. The equation for this area is A = 1/2 πab + bh. This is captioned below the effigy.

The actual area of a picture of a cone is not Randall's approximation, because the sides connect at the points on the ellipse where they can spread widest and form tangents to the ellipse, and such points are a footling higher than those which define the major axis. This is most obvious in cases when h is simply a piffling larger than a. The surface area can be computed to be exactly A = b (a arccos(-a/h)) + √(h²-aⁱⁱ)).

The 3D right circular cone usually depicted by this drawing would take a book of πr^twoh/3 where r=a=b. The area of the "lower" surface would be πr², while the surface area of the upper conical surface would be πr√(h² + r²). Neither of these areas can represent with the explanation in the comic, nor does the total surface surface area (the sum of these two).

If we do not assume that a = b, this drawing could also draw a right elliptic cone. The volume of the elliptic cone would be ^π/₃ abh. The area of the lower surface would exist πab and the surface area of the curved upper surface would be
2a√(b² + h^two)₀∫¹ √(^{a²h²(t²-1) - b²(a²+h²t²)}/_{a²(t²-one)(b²+h²)}) dt.

Bottom Left - Two ellipses joined vertically, or Cylinder

This illustration is unremarkably used to depict a 3D cylinder or right round prism. In this case, the upper ellipse represents the "visible" office of the top circular surface, with its "depth" shorter than its "width" due to foreshortening, and the lower part of the lower ellipse represents the "front" edge of the lower surface; the dotted half of the lower ellipse represents the occluded "back" edge of the lower surface.

To add to the confusion, the upper ellipse has its major axis labeled 'd' which unremarkably denotes the diameter of a round surface, while the lower ellipse has its semimajor axis labeled 'r' which similarly denotes a radius, fifty-fifty though the ellipses drawn have neither diameter nor radius. The 'h' denoting elevation is likewise used for both rectangles and solid objects. While 'd' in this case is required for the area adding of the second shape, in textbooks just 'r' may be marked and the arrow may be offset at a diagonal rather than in line with whatever figurative axis, to imply its applicability to whatsoever angle of radius.

The non-overlapping parts of the 2nd shape are composed of the rectangle formed by the major axes of the 2 ellipses and the vertical lines, plus one-half of the summit ellipse and half of the lesser ellipse. The surface area of the rectangle is dh, and the area of an ellipse with semimajor centrality d/2 and semiminor axis r is πrd/2. The total area is A = d(πr/two + h), which is captioned below the figure.

A 3D correct circular prism (cylinder) would have a book of πr²h and a surface expanse of 2πr² + πdh, or 2πr(r + h) since in this case d = 2r. The area of each flat surface would be πr². If we do not presume d = 2r, so the lateral surface area of the right elliptic cylinder is 4h₀∫¹ √(^{1 - t²(1-4r²/d²)}/_1 - t²) dt. The book is ^π/₂ rdh.

Bottom Right - Parallel Hexagon, or Prism

This illustration is usually used to depict a rectangular prism, with 'b' cogent the 'breadth', 'd' the 'depth' and 'h' the 'height'. However, the labeled angle θ, which is necessary for the surface area adding of the 2nd shape, would not normally be used in a diagram of a rectangular prism, as all angles are assumed to be right angles. A rhomboidal prism could exist accurately described past this diagram with the supposition that the 'base of operations' parallelogram is perpendicular to the 'front' and that the only non-right bending is θ. In that example 'd' would not accurately describe the depth of the solid, which would exist d sin θ.

The area of the 2D shape is comprised of the rectangle at the lower left, the parallelogram above information technology, and the parallelogram on the right. The expanse of the rectangle representing the front face of the prism is bh. The area of the upper parallelogram is db sin θ. The area of the right parallelogram is dh cos θ. The equation for this area is A = bh + d(b sinθ + h cosθ) as is given below the effigy.

The surface surface area of the prism would exist 2bh + 2db sin θ + 2dh. The volume is bdh sin θ. Assuming a 3D shape, θ tin can be artificially contradistinct by the projection; the assumption could be made that θ is 90 degrees, and sin θ is one (and therefore can be eliminated from the formulas), but since θ is marked, such an assumption might not be valid.

In the history of the development of estimator-generated 3D graphics, calculations of the apparent visual area taken up by the projection of a volume may take been useful in occlusion-like optimizations, where each fatigued pixel may be passed through many fragment shaders.

Transcript [edit]

This transcript is incomplete. Delight assist editing information technology! Cheers.

[Iv figures in two rows of two, each beingness a mutual ii-dimensional representation of a three-dimensional object, with solid lines in front and dotted lines behind. Each effigy has some labeled dimensions represented with arrows and a formula underneath indicating its surface area. Above the iv figures is a header:]

Useful geometry formulas

[Peak left; A circle with an inscribed concentric ellipse sharing its horizontal diameter. The border of the ellipse above the major axis is drawn with a dotted line, while the lower edge is fatigued with a solid line, similar to textbook depictions of a 3D sphere. The shared radius/semi-major axis to the right of the centre is drawn as an arrow and labeled 'r'. ]

A = πr²

[Top right; An ellipse with horizontal major axis, plus two straight lines: ane from each cease of the major centrality, up to a point vertical to the eye of the ellipse, so that the major axis of the ellipse (not drawn) and the two lines would course an isosceles triangle with a vertical axis of symmetry. The upper border of the ellipse above the major axis is drawn with a dotted line, while the lower border is drawn with a solid line, similar to textbook depictions of a right elliptical cone, or more ordinarily a right circular cone. The semi-small centrality of the ellipse is fatigued with an pointer down from the heart and labeled 'a' and the semi-major axis is similarly drawn to the correct of the center and labeled 'b'. To the right of the shape, the peak of the isosceles triangle is drawn using arrows, and labeled 'h'.]

A = 1/2 πab + bh

[Bottom left; Two ellipses of the aforementioned dimensions, with major axes horizontal, drawn vertically one above the other, with vertical lines connecting each cease of the major centrality of the top ellipse to the respective points on the bottom ellipse. The upper edge of the bottom ellipse above the major axis is drawn with a dotted line, while the lower edge is drawn with a solid line, similar to textbook depictions of a right elliptical prism or, more than commonly, a right cylinder (circular prism). Inside the shape, the major axis of the upper ellipse is drawn as a double-concluded arrow and labeled 'd'. The semi-minor centrality of the lower ellipse is fatigued as an pointer down from the center and labeled 'r'. To the right of the shape, the length of the vertical lines is replicated using arrows and labeled 'h'. ]

A = d(πr/2 + h)

[Bottom right; Two rectangles of the aforementioned vertical and horizontal dimensions, drawn with one offset diagonally to the upper right of the other, with diagonal lines connecting the corresponding vertices, forming a hexagon with opposite sides parallel. The upper right rectangle has its left and bottom sides fatigued with dotted lines, and a like dotted line is used connecting the bottom left corner of the ii rectangles, like to textbook depictions of rhomboid-based right prisms, or more than commonly rectangular prisms. Exterior the shape, the bottom border of the lower rectangle is redrawn below the shape with arrows and labeled 'b'. The length of the left edge is similarly redrawn to the left and labeled 'h'. The length of the diagonal line connecting the upper left corners of the ii rectangles is similarly redrawn on the top left using arrows and labeled 'd'. The acute angle between the bottom border of the lower rectangle, and the dotted diagonal connecting the two lower left corners, is labeled 'θ']

A = bh + d(b sinθ + h cosθ)

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Discussion

Area formulas are for 2D object equally seen instead of surface of a projected 3D object. Sebastian --162.158.89.200 02:36, 31 August 2021 (UTC)

The "decorative stripes and dotted lines" are the parts of the diagrams that are intended to indicate the 3rd dimension. The conceit of the comic is that these are superfluous. Barmar (talk) 02:56, 31 August 2021 (UTC)

Can you "see" this as second?

One aspect of this comic that has non been mentioned is how potent the dashed line convention is if you are very familiar with these sorts of representations. I constitute it nearly impossible to force my brain to see this every bit a second diagram with solid and (superflous) dashed lines. That 3rd dimension merely keeps popping out, although I could occasionally reverse the convention and see the dashed lines as in front rather than in dorsum. Arl guy (talk) xiv:27, 2 September 2021 (UTC)

Tin can someone explicate how the terminal one works? GcGYSF(asterisk)P(vertical line)eastward (talk) 04:28, 31 August 2021 (UTC)

bh is the area of the front face. The top face is a parallelogram with sides d and b, with an bending of θ between them, and so its expanse is d b sin(θ). The right face is a parallelogram with sides d and h, with an angle of 90º - θ between them, so its area is h d sin(90º - θ) = h d cos(θ). And then the area of the whole picture is bh + d b sin(θ) + d h cos(θ).

--172.68.24.165 04:46, 31 Baronial 2021 (UTC)

In case you lot don't know the area of a parallelogram by middle, you tin read d b sin(θ) every bit b * d sin(θ), where d sin(θ) is the superlative of the parallelogram; if yous cut the right corner of the parallelogram off and add it on the left, you get a rectangle where the bottom side is b and the height is that d sin(θ), so it works out. The other parallelogram'south area is h * d cos(θ), with the aforementioned reasoning. 162.158.ninety.241 05:00, 31 August 2021 (UTC)

Funnily enough, both this comic and 2506 are about projection. CRLF (talk) 05:11, 31 August 2021 (UTC)

I had considered working that into the explanation, just that needs to account for the fact that the indicated measurements (e.1000. the angle θ) accept to be read in second, not in 3D and projected. Merely it would exist correct to say that the second shapes are projections of simple 3D objects. 162.158.90.149 05:23, 31 August 2021 (UTC)

Betwixt this, 2506, and all the ones about Mercator and other map projections ... "project" is a very large discussion in Randall'south brain's word deject. 172.69.63.viii xv:29, 31 August 2021 (UTC)

Feels to me similar every comic since 2500 could be tagged "projection" in one sense of the word or another. --172.69.69.225 21:55, 31 August 2021 (UTC)

Does the bottom-left formula have a error?</s>

It seems like the bottom-left formula should exist A=d(πr+h) rather than A=d( ^πr /₂+h), considering in that location are two half-ellipses that add together up to a consummate ellipse. Am I missing something? (This doesn't seem like an extra joke, does it?) 162.158.106.179 05:28, 31 Baronial 2021 (UTC)

No, information technology'due south correct. d is all of the major centrality, non but one-half, so we accept to dissever that by two. 162.158.92.83 05:51, 31 August 2021 (UTC)

Oh, right; good call! 162.158.106.179 06:49, 31 August 2021 (UTC)

Does the top-correct formula have a mistake?

I recollect it should be in brackets, the top triangle surface area needs the ^one/₂ also, so information technology should be: A= ¹/₂(πab + bh)

No, it's correct. The bottom is a half ellipse, with area 1/2 π a b, and the peak is a triangle with base of operations 2 b and height h, and then its area is 1/2 2b h = bh. The total surface area is 1/2 π a b + b h.

--172.68.25.144 06:49, 31 August 2021 (UTC)

3D formulae for reference

4πr^two

πb(a+√(b^ii+h^two)) if a=b

πr(2r+h)

two(bd+bh+dh)

162.158.107.80 09:54, 31 August 2021 (UTC)

It would exist clarifying to add these to the comic, merely of course they are flagrantly wrong. Baffo32 (talk) 09:57, 31 August 2021 (UTC)

Surely ripe for a tabular array, in place of much of the longhand paragraph spiel (which could exist kept, just simpler for just the narrative but otherwise non-technical details)... "Shape (2nd)", "Area", "Pretended Shape (3D)", "Surface Surface area", "Volume", ¿"Notes"? (Not certain about specific Notes, some things could/should be said beneath the formulae/descriptions in the relevent prison cell to which that matters, in special cases where necessary, which might exist better than a Notes either empty or jammed up with all the combined row-specific corollaries, etc, that I can imagine.) Anyhow, an idea. 141.101.76.11 xi:56, 31 Baronial 2021 (UTC)

I think the formulas are correct. Those given should be from the text book, not for those with ellipse bases. Someone has put a lot of work into giving these complicated formulas for the cone and cylinder. Just I think that is overkill. I accept added to the caption the simple versions before, and would advise deleting the complicated, which was never the intention of either text book or Randall! ;-)--Kynde (talk) 12:36, 31 August 2021 (UTC)

What complicate formulae?? --GcGYSF(asterisk)P(vertical line)east (talk) twenty:28, 4 September 2021 (UTC)

Surface area. Not volume. My bad. I ordinarily consider volume associated with pics like like that. Don't apply surface area much. Baffo32 (talk) 22:22, one September 2021 (UTC)

add an extra edited image that is the comic without dotted lines to make information technology easier to see the 2d shapes? 172.69.71.177 12:46, 31 August 2021 (UTC)Bampf

And an animated GIF of the 3D solid objects rotating to show their real shapes. At different speeds. If you have the time.  :-) Robert Carnegie [email protected] 141.101.76.11 16:31, 31 August 2021 (UTC)

Image hither: https://i.imgur.com/dq7VmnK.png Editing done myself, experience costless to upload it to this wiki if you have an account on this wiki. :) --162.158.88.29 17:22, ane September 2021 (UTC)

Delight do bank check my (additional) changes to the lesser-right detail (hexagon-cum-prism) in both master and transcript texts. As hinted in my edit notes, cos-theta is of import considering the skewed tetrahedron (rhomboid, whether in program or the true area of the 'fake' perspective) is not d*b in area. The fact that without the theta information technology would look like a standard oblique orthographic projection with entirely right-angled corners is peradventure part of the (intended?) confusion, although we tin can probably assume that all unmarked (and, of course, uncongruent/uncomplimentary) angles are 90° so that it isn't a full on parallelepiped with an additional phi-angle on an side by side confront and a complicated tertiary dependent-angle somewhere upon the remaining face-airplane. Equally such, I put in the cosine element to both the 3d surface formula (information technology simply affects the bd-shape, the both of them) and the 3d volume (from this shape, extrudes without further adjustment straight up the h-centrality), only I e'er accept to second guess if I've done this uncomplicated bit of trig correct, it seems, even though I should know better and only trust to SOHCAHTOA... ;) 162.158.158.146 13:24, 31 August 2021 (UTC)

(Case in indicate: I idea I'd added cosines, and I'd put sines anyway, when fussing about copying the clipboarded theta-grapheme into the right place! Re-read, seen, corrected(?) this myself. Unless I thought I was was wrong; only I was incorrect, I was right!) 162.158.155.145 13:33, 31 Baronial 2021 (UTC)

I believe both of those prism formulas should use sine theta. If theta is 90 degrees, and so sine theta will be 1 (thus reducing to the rectangular instance), whereas cosine of 90 degrees is cypher.Tovodeverett (talk) xv:xix, 31 August 2021 (UTC)

Y'all're correct (me over again, from but higher up), I was rushed and had been correct showtime fourth dimension, I realised while I was off-filigree and it was nagging away at the back of my caput. I'm ameliorate on paper (or when I can sanity-test real code, just for some reason tapping it in like this just screws my listen up, taking away/inverting my technical ability and reason. (I arraign the microwaves emitting from my tablet... pass the tinfoil hat!) 162.158.158.178 16:29, 31 August 2021 (UTC)

Unconvinced by the cone! The equation shown, is correct for an isosceles triangle with a half-ellipse on its base. But that shape has 'corners' where the sides meet that half-ellipse. In a 3D projected view of an actual cone, the sides volition meet the base ellipse at a tangent, pregnant that information technology is more than than a half-ellipse. But I suppose it's close enough every bit an approximation...172.69.55.131 15:57, i September 2021 (UTC)

I verified your claim by imagining the surface of the cone as formed by a set of lines extending from the dissimilar points on the ellipse to a single fixed indicate at the tip. No matter where you put that tip betoken, the outermost lines seem tangent to the ellipse. Seems information technology works for both perspective and orthographic projections. Updated the caption. Randall's formula is wrong, especially for very short cone projections. Baffo32 (talk) 22:46, 1 September 2021 (UTC)

I expanded the text slightly, and worked out the correct formula: it should exist (2π - 2arctan(h/a))ab + b sqrt(h^two - a^2). Can someone verify that, format it properly for the wiki, and add it? The easy way to bank check it is to shrink the diagram horizontally then the ellipse is a circle of radius a. (Does the wiki not take MathJax or similar installed? Seems odd, given Randall Munroe's interests.) Information technology'southward perhaps also worth mentioning that looking closely at the picture at a pixel level shows that he did draw the tangents, rather than do the half-ellipse + triangle that the formula suggests. --172.69.90.75 xvi:48, 2 September 2021 (UTC)

It'due south 3am (okay 5am) and I fabricated it really long!

I just followed the directions in the "incomplete" which said to add in explanations of the formulae ... Delight experience free to edit to have out redundancy. However I did add together in the following explanations: - the fact that the formula in the 3rd effigy is actually the aforementioned as the cross-section represented by the ellipse, which is why you lot may not become the joke afterwards reading the first picture; - the utilise of 'd', 'r' and 'h' in the third effigy, which adds to the confusion as they imply "diameter", "radius" and "height" - the fact that the area calculations must have into account the overlapping shapes (there were previously references to "semi-ellipses" which are extrapolations, not what'due south fatigued in that location) Haven't still done the last effigy - pretty sure 'b' 'd' and 'h' are for 'breadth', 'depth' and 'height' and while 'height' is as well used for second rectangles, 'latitude' less then in maths textbooks (usually 'width') - whoever pointed out that in that location is a theta too, pretty certain it's only at that place because it'south necessary for the surface area adding, as 'depth' only really applies as labelled to rectangular prisms - if the base were not rectangular, 'd' would not be equal to the 'depth' Volition try to come dorsum later and shorten... 162.158.166.40 18:56, 1 September 2021 (UTC)

Someone thought that "formulae" was a typo for "formulas" (which it might hands be, on a QWERTY or similar layout). Non going to revert, but note that (for a mathematical formula, if maybe not a chemical one/etc, but there's enough of mixed employ) this is actually quite correct. If information technology were up to me solitary (I didn't write that 1, orother mentions like in the above Talk contribution), for the tape, I'd probably accept used "formulæ" myself. ;) 162.158.155.145 20:28, 1 September 2021 (UTC)

If you don't presume that the lesser right effigy is 3D, what'southward the justification for projecting upwards and assuming that the angle theta is also the angle of the top parallelogram? Arl guy (talk) 02:25, 2 September 2021 (UTC)

If yous assume that the two rectangles take equal width and height, then information technology can exist mathematically proven that the angles must be equal (probably using congruent triangles). However this assumption is not stated on the figure. That said, you lot would make the same supposition for the 3D figure, along with a whole bunch of other assumptions of course.172.lxx.147.23 06:31, two September 2021 (UTC) Edit: The two rectangles must have equal width and height to brand the rest of the shapes parallelograms in the get-go place. If they aren't identical yous get trapezoids all round and possibly a bunch of different angles. 172.70.143.22 06:43, ii September 2021 (UTC)

Aye, I think you are mostly correct, but you don't necessarily get trapezoids all effectually. Yous tin can move the top solid line down and to the correct and rotate the acme left and elevation right slanted lines to keep the top shape a parallelogram, only the right shape will now be a trapezoid. I call up the key assumption is that all the solid and dashed lines that loop parallel are, in fact, parallel. Arl guy (talk) fourteen:xx, 2 September 2021 (UTC)

There's e'er many possible unindicated perspective tricks (or axonometric/othographic ones, as might be more than appropriate) in which the components of lines and relationships towards/abroad from the viewer are not-zippo (or non non-zero in the style expected). Diverse of the shapes involved could be infinitely warped with leans, curves or even highly inflected wiggles to allow some 'as expected' profile even as others that should be continued are not. (Though it would make the surface 'planes' warped and distorted, even more against the conventions of wireframe diagrams.)

Because of dependencies, as the theta moves the other like angles might exist thetas too, or they could be kept every bit RAs and (?)three other chosen angles could exist distorted (in or out of the page, by a calculable and entirely derivative amount) to compensate. Or, like the Triangle illusion, there'due south overlaid duplicate edges with non all/any vertices really being the same, just in the same illustrated place. (Again, making a mockery of the 'uncomplicated' diagram, but we're already way past that. ;) ) 162.158.155.247 08:56, two September 2021 (UTC)

Sure, just we were talking most the 2D figure and taking it at "face" value - the internal bending is divers while the ones you'd really use for the parallelogram area calculations aren't, and neither are the dimensions of the upper rectangle. In a proper geometric figure y'all'd have markings showing which lengths are equal and which angles are correct angles.172.70.143.22 05:38, three September 2021 (UTC)

Subtext??: I wonder if this comic implies that we are being transformed from 3d creatures into 2d creatures... hadaso --162.158.91.184 07:04, five September 2021 (UTC)

Cross-department - reword please I made a wording error in describing all points on the "cross-section" of the sphere as being equidistant from the centre, and I'm stuck on how to gear up information technology concisely. Problem is that "cantankerous-department" is ambiguous - I think mathematically it could mean the circumvolve at the intersection of the hollow sphere and the aeroplane, merely to me it usually ways the part of the aeroplane where information technology intersects the solid sphere, which makes the statement false. I want to supplant it with "circumvolve", with the quote marks, but we're really talking about the ellipse-cum-circle so I'thousand not certain if that's adept enough. 172.lxx.142.239 xi:00, seven September 2021 (UTC)

Source: https://www.explainxkcd.com/wiki/index.php/2509:_Useful_Geometry_Formulas

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