can an equilateral triangle have the same area and perimeter (ignoring the unit)?

[@joshuahillerup4290]

I mean, a triangle with a perimeter of 0 would also indeed have an area of 0

[@Tletna]

True, except, would one still consider that a triangle?

[@joshuahillerup4290]

@@Tletna yes, except it's also a sphere and a square

[@vladislavanikin3398]

@@Tletna depends on your definition, actually. In general by triangle one means some figure (clarification needed) with exactly three vertices, which don't lie on one line. A triangle with perimeter zero surely doesn't meet the requirements. However, you can define a triangle slightly broader, so that even a single point can be considered a triangle. In such case you will call this nonstandard triangles degenerate. And sometimes they behave really weirdly.

[@kayoucat6352]

To have a perimeter of zero, x must be equal to 0, but I think a segment can't be egal to 0cm or 0m

[@forcelifeforce]

@ Joshua Hillerup -- Thumbs-down. It is not a triangle. It does not exist. There is nothing there. It would be better for you to go back and delete this thread and help clear some of the clutter among the threads.

[@jamiewalker329]

This literally works for any shape. Imagine the shape having negligible size, and then gradually scaling the shape (to create enlargements). The perimeter increases proportionally to the scale factor, and the area proportionally to the square of the scale factor. A line y =ax intersects y = bx^2 for exactly one positive value of x.

[@markstahl1464]

Exactly. The graphs must cross at the optimized side length.

[@TonyHammitt]

I wonder what shape has that come out to a whole number?

[@youtubersingingmoments4402]

@@TonyHammitt A square, for one. 4x4=4+4+4+4.

[@jamiewalker329]

@@TonyHammitt Or a 6, 8,10 right angled triangle.

[@alexhenson]

Nice

[@quantumelle]

One interesting corollary of this is that you can make any physical equilateral triangle have this property by simply choosing units such that the side length is 4 root 3 of those units.

[@johncrwarner]

I used Heron's formula to calculate the area of the triangle A = sqrt ( s . (s - a) . (s - b) . (s - c)) so I didn't have to think about the "height" and use Pythagoras' Theorem.

[@mashrafimorshed18]

That's a great way to solve.

[@johncrwarner]

Forgot to say that s = the semiperimeter = 0.5 (a + b + c)

[@yusufat1]

I used ½ a b sin c for triangle area

[@johncrwarner]

@@yusufat1 That works but I was trying to avoid the use of angles and trigonometric functions by using Heron's formula.

[@ommakadia3815]

thats actually a longer way.. still its cool... i just love it when people have their own way of solving .

[@statsy150]

I also got to the same solution but I switched some things around. I started from the fact that P=A so 3x=xh/2: divide both sides by x and then multiply by two and you get that h=6 from that you can use the special 30-60 right triangle that creates with sides lengths of x/2, x and h. From that you get that if you divide 6 by the square root of 3 you get half of the side and then you just multiply by two, so you get 12/square Root of 3 a that you can simplify in the result you got

[@tavasp]

Tbh that’s actually a lot fast way to do it

[@EggZu_]

you can say sqrt(3) instead of "square root of 3"

[@masonseminario7435]

I did the same thing but using sin and got the same number in the form of 6/sin60

[@WaluigiisthekingASmith]

Theres an interesting way to think about this. The area will always look like ax^2 and the perimeter will always look like bx so setting ax^2=bx, ax =b, x=b/a or trivially 0. So you can find the perimeter and area of one instance and scale by b/a

[@jabarzua]

if the question is presented as in the title, you can just notice that the area is a function of the side squared (a positive parabola), and the perimeter is 3x. So the two must intersect in x=0 and some other positive value, and the answer is yes.

[@sethb3090]

That was my thought. I honestly didn't even watch the video, just solved it and then came to look at the comments. What I came out to is you split this in half, you get a 30/60/90 triangle which means the base is x/2 and height is x/2 root 3, and you have two of those so the area is (x^2 root 3)/2. The perimeter is 3x, so you solve and get (x^2 root 3)/2 = 3x x^2 root 3 = 6x x root 3 = 6 x = 6/root 3

[@ijarbis187]

@@sethb3090 you made a small mistake when finding the area, since A=1/2*bh and you found the height to be x/2*sqrt(3) and we know the base is just x then A=1/2*x/2*sqrt(3)*x which simplifies to A=x^2/4*sqrt(3). And since P=3x and A=P then x^2/4*sqrt(3)=3x which equates to x=12/sqrt(3). I think you just didn’t add the 1/2 in the 1/2bh equation

[@JashXD]

i knew that the area of regular triangle is s^2sqrt(3)/4 where s is the side length and the perimeter is 3s so you can use quadratic formula to solve for s when the quadratic is s^2-4sqrt(3)s=0 and solutions are s=0 and s=4sqrt(3) and take the positive result and you get s=4sqrt(3)

[@mrosskne]

Length and area are incomparable.

[@trueriver1950]

@@mrosskne which is exactly why the q includes the important words "excluding the units"

[@harmannatsingh3933]

I just found the perimeter as 3x, and then used the sine rule for area of a triangle and got 1/2(x² sin60). When you set then equal to each other you get 4√3 and it's much simpler.

[@matthewkeenan3003]

That’s what I did :)

[@ldc0322]

Me too!

[@peterpan408]

..because sin60 is just a number. This is practical math vs clever math.

[@technoeevee6969]

As a bonus, x=0 does still have a geometric reasoning for its existence. That solution is simply telling us that a triangle with length side 0, in other words basically a single point or an infinately small triangle, will have its area and sides equal, because they are both 0.

[@lool8421]

you can as well go for 3x=x^2*sqrt(3)/4 immediately because it's one of those basic formulas, then use some basic operations 1 = x * sqrt(3) / 12 1/x = sqrt(3)/12 x = 12/sqrt(3) x = 4 * sqrt(3) but whatever, the results is what matters, it's always good to explain why some formulas actually exist

[@nilsalmgren4492]

The reason it works is that the height of an equilateral triangle cuts the triangle into two 30-60-90 triangles, both with hypotenuse x and height side opposite 60 degrees, so root 3 over 2 times x. Now use your basic triangle area formula with base being x. 1/2 x times root 3/2 times x . That results in root 3/4 times x squared.

[@philipyoung3028]

It can be done much more easily in trigonometry. The height H of the triangle is Xsin60, and thus H=6. The answer to the problem is a triangle with height of 6.

[@finntrautermann3451]

Still requires you to calculate X to get H.

[@jmjm9294]

Solving for x though, not only h

[@philipyoung3028]

No need to solve for X. The height of the triangle that complies with the terms of the problem must be 6.

[@Misteribel]

I must be missing something. Yes, the height is x∙sin(60º), which equals x∙√3/2, as in the vid. But we need to find a height where area (product) and circumference (sum) are equal. I don’t see where your ‘6’ here comes from. I don’t follow the part ‘and thus H=6’. My trig is rusty at best, could you elaborate how the sin of the angle relates to the area and gets you ‘6’?

[@philipyoung3028]

@@MisteribelThe area of either of the inner triangles formed by a height line is (X/2 times H)/2. The area of the big triangle is twice this, or (XH)/2. This must equal 3x, so H=3X/(X/2), which is 6. I just like the surprisingly neat value of H.

[@alonamaloh]

If the question is simply if this is possible, as in the title of the video, the answer is trivially yes, because of calculus. The area is proportional to x^2 and the perimeter is proportional to x. So for tiny triangles, the perimeter is much larger, and for large triangles, the area is larger. Now because of the mean value theorem, there will be a value of x where they are equal.

[@Yougottacryforthis]

Bringing the heavy canons for a simple question

[@andrewkarsten5268]

Technically the mean value theorem is analysis, not calculus. Many learn the theorem in calculus, but it’s a well known real analysis result, and typically discussed/proved in a real analysis course. But I get your point. Good observation

[@alonamaloh]

@@andrewkarsten5268 Ah, you are probably right. I took classes in college which had "analysis" in the name, but they also covered calculus, so the distinction between analysis and calculus is not very clear to me.

[@andrewkarsten5268]

@@alonamaloh well, in a sense, the “point” of an analysis course is to prove and give the fundamentals, or groundwork, for calculus. It gives the rigor needed for the foundation of calculus, which the some of the rigor in a calculus class is usually skipped in favor of doing actual calculus

[@markstahl1464]

I was initially going to say no, but I did the Algebra; and of course, yes!

[@dannypipewrench533]

1:08 I tried it as soon as I saw the title of the video. I checked my math with some triangle calculator I found online, and I did get the correct answer.

[@sanchaymuralikrishnan283]

For an Equilateral Triangle, Let: Area = Perimeter ½.b.h = 3b ½.h = 3 h = 6 (In terms of height)

[@bobh6728]

You still need the Pythagorean theorem to find b.

[@RanEncounter]

@@bobh6728 Or use the memory triangle.

[@ingenuity23]

@@bobh6728 nah, it simplifies a lot by trigonometry, as h=b sin60

[@Penrose707]

The solution for the same class of problem generalized to any number sided polygon is readily given by considering the area formula for a regular polygon when given a side length. With side length = x, and number of sides of regular polygon given by n then A = (x^2)*n/(4*tan(pi/n)). Let 4*tan(pi/n) = D then: A = (x^2)*n/D P = nx 0 = A - P 0 = ((x^2)*n/D) - nx 0 = x(xn - nD) Therefore x = 0 and x = D, but D = 4*tan(pi/n) and so x = 4*tan(pi/n). Which means that any regular polygon with n sides and side length 4*tan(pi/n) will exhibit the same area as it's perimeter length, negating units.

[@charlesmaillho367]

Knowing how to solve problems are great and you are a master of it. Being able to convey that method which includes the whys and the how's is something that I can see you still can improve in. Finding H first works for you but can easily confuse others because it has no bearing on the initial question without proper context. Start with your A = P, substitute those formulas in and then explain why we should solve for H right there. We have 2 unknowns in 1 equations, math says we need 1 more equation. Putting H in terms of X allows us to turn this problem into a system of two equations.

[@curtiswfranks]

This was a fun problem. If I ever teach algebra again, I will try to remember to give this as an exercise. I like the mention of dividing by x being a bad habit.

[@ReetaSingh29323]

We know that area of equilateral triangle is also √3a²/4 and perimeter is same i.e. 3a Now, If we have to find value of 'x' so that Area = Perimeter (of triangle) Then, Substitute 'a' by 'x' Then we get, 3x = √3x²/4 Dividing 'x' both sides 3 = √3x/4 Multiplying 4/√3 both sides 4√3 = x Side of Equilateral triangle = x = 4√3

[@cparks1000000]

You can prove the same result for a triangle with side lengths ax, bx, and cx. Hint: examine the function A(x)-P(x). You can easily prove a solution exists without finding it.

[@LeeCanPotato]

I choose the unit such that when ignored the perimeter equals the area. Works for every triangle 😅

[@diggoran]

The answer is exactly one hudfisjerml perimeter and one square hudfisjerml area of course!

[@dw8931]

With Heron's formula for a triangle's area it would even be easier: Let s be half the perimeter, so in our case s = 3x/2 A = sqrt( s*(s-a)*(s-b)*(s-c) ) A(equil. triangle) = sqrt( 3x/2 * x/2 * x/2 * x/2 ) = sqrt ( 3*(x^4)/16 ) = sqrt(3)*(x^2)/4

[@elanosinho3936]

The triangle is equilateral so you can match the formula of its perimeter [3x] to the formula of its area based on its own side [(x².√3)/4] getting [3x = (x².√3)/4] being "x" its side lenght, then you solve it. Multiply each side by [4] -> x².√3 = 12x Divide each side by [√3] -> x² = (12x)/√3 Correct [(12x)/√3] -> x² = (12x.√3)/3 Simplify [(12x.√3)/3] -> x² = 4x.√3 Divide each side by [x] -> x = 4.√3 That way of doing it is really easy because it only uses 1 icognito (the side of the triangle), so you just need to rearrange it.

[@ephraimb8671]

If P=3x=1/2bh=A, and we know b=x, we can divide both sides by 1/2x and we get h=6. To solve for x, use the Pythagorean Theorem: 6^2+(1/2x)^2=x^2. 36+x^2/4=x^2. Multiplying both sides by 4, 144+x^2=4x^2, or 144=3x^2. Dividing both sides by 3, 48=x^2, and x=sqrt(48). sqrt(48) can be written as 4sqrt(3).

[@SmokeyBear04]

I think x is actually 8 root(3) because you forgot that the base of the triangle is (1/2)x so Area = (1/2)[(1/2)x][(root(3)/4) x]

[@taranmellacheruvu2504]

You can generalize this to all regular polygons using the generalized area formula: A = .25ncot(pi/n)s^2 n is the number of sides, s is the side length. The perimeter would be n*s. .25ncot(pi/n)s^2 = ns .25cot(pi/n)s^2 -s = 0 s(.25cot(pi/n)s -1) = 0 s = 0, 4tan(pi/n). From this you get that the side length of any regular polygon whose area is equal to its perimeter is four times the tangent of the quantity pi divided by the number of sides.

[@klepikovmd]

And what is cot?

[@taranmellacheruvu2504]

@@klepikovmd The cotangent of an angle. 1 divided by the tangent of an angle. I used radians here.

[@klepikovmd]

@@taranmellacheruvu2504 ok 🤣 I've been more attuned to "ctg".

[@taranmellacheruvu2504]

@@klepikovmd Ah, I see.

[@jaydenyer6470]

I used the area formula, 1/2absinC, it gets the same answer as the angle has to be 60° in an equilateral triangle and sin(60°) is sqrt(3)/2 multiplied by 1/2 and x^2 (as for an equilateral triangle a=b which is x in the example) you get (sqrt(3)/4)x^2 and then it is solved the same way by setting against 3x for the perimeter :) Overall, great video and a fun problem to try, good job

[@0VexRoblox]

We are taught area of an equilateral triangle is [sqrt(3)/4]a^2, so I equated [sqrt(3)/4]a^2 = 3a [sqrt(3)/4]a = 3 a/4 = sqrt(3) a = 4 sqrt(3)

[@nobodynowhere8061]

I didn’t do any algebra but my intuitive solution was to compute the area and perimeter for 3 equilateral triangles. One with a side length of 1, one with a side length of 100, and the third was 10000, the math isn’t too hard to do in your head and you see that at 1 the area is lowers than the perimeter, and at 10000, the perimeter is lower, thus, at some size in between there’s a value where the two are equal

[@RufusLibra]

There's an area formula specially for equilateral triangle: sqrt(3) * a^2 / 4. If we equal this to 3a, then get two solutions: a = 0 and a = sqrt(48).

[@robertlunderwood]

The only thing I did different was finding the height of the triangle. Since we have an equilateral triangle, all the angles are the same. By evenly dividing the angle to get the height, a 30-60-90 triangle is formed with a base of x/2 and a hypotenuse of x. This the height must be x*sqrt(3)/2.

[@snowjix]

i suck at math, but i managed to squeeze out about 7 when i gave it a go. And you know what? Im happy with that.

[@Mathematical-Mind]

We know the formula for a equilateral triangle is (sqrt3)(x^2)/4. Therefore (sqrt3)(x^2)/4=3x. (sqrt3)(x^2)/4-3x=0. Factoring out x, we get x=0 as an answer. But a triangle’s side can’t be 0. So our only other option would be (sqrt3)(x)/4=3. Working out the algebra, we can get x=4sqrt3. I think this way is also very efficient.

[@arcadeleon6077]

“Please Pause the video and try this first…………… Done? Aight cool!” 1:09 had me laughing for no reason

[@radhairadhai1907]

As area of a triangle is (b*h)/2 and perimeter of an equilateral triangle is x+x+x =3x so our equation is Area=Perimeter 3x=(b*h)/2 Notice that base of the equilateral triangle is always x. So substituting we get 3x=(x*h)/2 6x=x*h h=6 So this shall be true in an equilateral triangle with height=6

[@flleaf]

i loved the ending.. because i hate when people have like a minute long outro and im so annoyed

[@ThePalsanich]

In a spherical geometry it is possible. For example on a sphere with r=sqrt(3), triangle with a=b=c=1 will have S=3.

[@andy02q]

I want it to have the same side length too, but that's kinds ridiculous. :( An equilateral Tetrahedron could have the same side length, area and volume though.

[@trueriver1950]

Not sure about that... What side length do you get?

[@zachansen8293]

Obviously, if the unit of area has the right scaling factor to the unit of length. In fact, any two positive numbers can be the correct answer if we don't care about which units are involved.

[@eyalharel810]

I just did (x^2*sqrt3)/4=3x. Worked fine!

[@ProfessorHoax]

Valid only for equilateral triangle, with the semiperimeter area formula you can do it valid for all triangles; if a, b, c are the sides of the triangle then the semiperimeter is s = (1 + b + c) /2 and the area formula is A = √s(s - a)(s -b)(s-c) then √s(s - a)(s -b)(s-c) = a + b + c, solve it and you will have a general solution.

[@soumyaranjansahoo4231]

My school math teacher gave our class the same exact question. Have to find sides. Just take formula for area of EQ triangle= perimeter of EQ triangle. Then solve it.

[@louisvictor3473]

Went a bit different route. Since I know the angles and sides: x.3 = x.(sin(tau/6) * x/2) [so both sides are x.someJunk, I just equated the junks] 3 = [ (√3)/2 ] * (x/2) | and the rest is basically samey

[@geppettocollodi8945]

The question is "Can" is not necessary to calculate the solution. Obviously zero is a rogue solution we are not accepting. But if we grow the side we notice that the perimeter grows linearly but the area grows as a quadratic (a parabola) The parabola has zero slope at the origin so initially the area is smaller than the perimeter. Eventually the quadratic area grows faster and is guaranteed to cross the linear perimeter. So the answer is YES, no algebra, no Pitágoras.

[@AManOfMusic]

The same answer can be found with Heron's formula. 3x = √((3x/2)((3x/2)-x)^3) x = 0 or 4√3

[@BatchDrake]

Any shape with area A and perimeter P has a unit u such that A/u² = P/u. If you choose u = A/P, ta-da! The area of your shape _equals_ to its perimeter.

[@sagov9]

in other words, you can always choose the units such that the perimeter and area have the same numerical value.

[@Erlewyn]

Ignoring the units means that for the purpose of the exercise, you don't care that, for example, the perimeter is in cm while the area is in cm². Otherwise they're obviously different values.

[@dannypipewrench533]

@@Erlewyn Imagine trying to measure length with units of area. Scrooge McDuck has 3 cubic acres of money.

[@christopherramsey7027]

Yes by the intermediate value theorem. An incredibly small triangle would have a much larger perimeter than area. An incredibly large triangle would have a much larger area. Since both of these values increase continuously, there must a triangle where the perimeter and area are equal.

[@du42bz]

A square whose perimeter is the same as its area has an x equal to 4

[@reTache4955]

And a circle needs a radis of 2 or a diameter of 4

[@_-___________]

You've convinced me to go back and take some more math courses.

[@marcushendriksen8415]

Without actually working through it, yes there is such an x. That's because the area scales in proportion to x^2, while the perimeter scales in proportion to just x. Set these equal to each other and all you have is a quadratic in x

[@robertarmagost9908]

It might have been easier taking a side as 2x. Then the height is √3 x and we get perimeter = area as 6x = √3 x^2 which quickly reduces to x = 0 and x = 2√3. Since a side is 2x we get the two triangles, one with sides 0 and the other with sides 4√3

[@informatimago]

Yes, if p = a , with p = 3c and a = ½c²sin(π/3). (c = side of equilateral triangle). This solves to c = 12/√3 ≅ 6.93 ■

[@gumball6804]

I used the formula ((sin(x)ab))/2 as to not worry about the height so that ((sin(60)x^2))/2=3x (sin(60)x^2)=6x sin(60)x^2-6x=0 *use the quadratic formula and ignoring the negative value since a side cannot be negative* a=sin 60 b=6 so the final answer would be 6/sin 60 after simplification which equals to 4 sqrt(3)

[@alansmithee419]

To answer the title you can observe that since perimeter varies with x and area varies with x^2, they must at some point cross (since one has constant gradient and the other has 0 gradient at x=0 and has linearly increasing gradient - so the latter must initially fall behind the former since it has lower gradient before eventually catching up as its gradient grows without bound). This doesn't tell you where they cross but it does answer the question.

[@geppettocollodi8945]

I did not read your post before posting my identical observation.

[@anonymouscheesepie3768]

Prove that an equilateral triangle with 3 side lengths n and an area 1/2 * h * n can exist, where h is the height of the triangle. 1. P = 3n, A = .5hn 2. P = A 3. In this case, h as the height of the triangle is the straight line between the top point of the triangle and the bottom base, perpendicular to the bottom base. 4. This means that h forms a perpendicular bisector and thus a two right triangles out of the original triangle with one leg .5n and another leg h. The hypotenuse of the triangle is n. 5. Then, to find h, we can use pythagorean theorem: n^2 = h^2 + (.5n)^2 = h^2 + .25n^2 Thus, h^2 = n^2 - .25n^2 = .75n^2, and h = sqrt(.75n^2) = sqrt(3n^2)/2. 6. Plugging back into our equation P = A, we find 3n = .5n * (sqrt(3n^2)/2). 7. Now, solve for n. 3n = .5n(sqrt(3n^2)/2) 3 = .5sqrt(3n^2)/2 6 = sqrt(3n^2)/2 36 = (3n^2)/4 144 = 3n^2 48 = n^2 n = +-sqrt(48) = sqrt(48) Thusly, one equilateral triangle exists such that the magnitude of its area is equal to the magnitude of its perimeter, such that it has a side length n = sqrt(48).

[@robertpearce8394]

Does this make sense? Surely all equilateral triangles are similar.

[@Damjes]

ez. just draw whatever (even pentagram bruh), divide perimeter p over area a to get x (so p = ax). Scale original figure by scale of x. Perimeter will be px = ax * x = ax2. Area will be ax2, because it scales by square factor. QED.

[@alansmithee419]

[insert some pedantry about just using different units for the area and perimeter such that any size triangle can have equal area and perimeter here]

[@mathgeeksrus9399]

Depending on how you interpret (ignoring the units), one could also state that an equilateral triangle with perimeter of1728 sqrt(3) inches, has an area of 1728 sqrt(3) square feet.

[@protocol6]

Just a right triangle with equal base and height is kind of interesting, too. x=4+√8

[@BeaDSM]

If you're ignoring the units then it's trivially true by just defining your units such that the area of the triangle is equal to thrice the side length.

[@seroujghazarian6343]

3x=(x^2)sqrt(3)/4 Since x is a measure, we know x≠0 xsqrt(3)/4=3 x=4sqrt(3)

[@vincentweatherly9991]

Rather than using pythag, I knew that an equilateral triangle has angles of 60° so therefore I used sin(60°) or sin(π/2) which equals √3/2 rather than bringing in powers. A bit easier to do in my head

[@baumeisterjack9281]

Just make up a new unit of like 1m^2 equals 4 smallsquarefeet… Boom square with side legth 1 works: 4 meter perimeter and an area of 4 smallsquarefeet

[@GirishManjunathMusic]

Find the side length, x, of an equilateral triangle such that the perimeter is equal in value to the area: Perimeter of equilateral triangle: 3x Area of equilateral triangle: (√3/4)·x² For equivalence: 3x = (√3/4)·x² √3x·((¼x - √3) = 0 As x has to be positive, x = 4√3 units.

[@gdtargetvn2418]

it's 3x, not x

[@GirishManjunathMusic]

@@gdtargetvn2418 oh right

[@superiontheknight963]

Before watching the video: Let x = side length of equilateral triangle. Assume area of that same equilateral triangle is equal to its perimeter. 3x = (1/2) ⋅ sin(60°) ⋅ x² 3x = √(3)/4 ⋅ x² 4√(3)x = x² x² - 4√(3)x = 0 (x - 4√(3))x = 0 x₀ = 0 && x₁ = 4√(3) As it doesn't make too much sense to have a triangle with side lengths of 0(although I suppose you could do that), I would say the answer to the videos title is yes if the side lengths of the equilateral triangle are equal to 4√(3).

[@superiontheknight963]

After watching the video: Cool to see this can be done rather easily algebraically as well! :)

[@manganesegoblin981]

Surely 1/2 abSinC is easier here 3x = 1/2 x^2 sin 60 3 = 1/2 x * sqrt(3)/2 12/sqrt(3) = x x = 4sqrt(3)

[@pixl_xip]

i love how i tried, got the same result, but when watching the video, i did it totally differently. math :)

[@kfftfuftur]

yes, proof: P = 3x, easy. A = kx^2, no need to work out what k is. P = A 3x = kx^2 x = 3 / k. (or x = 0)

[@МішаПідлісний]

it aslo appoximates to 6.9 which I think is pretty nice

[@lisamariefan]

I believe that the answer after doing the math but before watching the video is 4√3. It actually seems up check out too assuming no mistakes anywhere.

[@user-fv6jp7xo4x]

the easiest solution is: 3x=(hx)/2 6x=hx h=6 6^2+(1/2x)^2=x^2 36=3/4x^2 6=(√3/2)x 12=√3x 144=3x^2 x^2=48 x=√48 and just to prove that its the same solution as his you can do: x=4√3 x=√16*√3 x=√48

[@ethandavis7310]

All shapes have area equal to their perimeter as long as you define the length of a unit properly. Strange to think about. A small circle of radius 2cm has area=perimeter (ignoring unit disparity), and a big circle of radius 2 miles also has area equal to perimeter.

[@thehungarywaffleinc.7775]

(1/2)bh = 3x (1/2)x^2=3x x^2=6x x^2-6x=0 x(x-6)=0 x=0,6 When x=0, area is (1/2)0^2=0 And perimeter is 3*0=0 When x=6, area is (1/2)6^2=18 And perimeter is 3*6=18 Does this make sense?

[@PubicGore]

We can make use of Heron's formula here. The semiperimeter of this triangle is 3x/2 So its area will be, after simplification x^2sqrt(3)/4. So, solving 3x=x^2sqrt(3)/4, we get x = 12 / sqrt3 = 4sqrt3, or x=0.

[@slug4057]

This made me think of the same problem but with isosceles - and I found an interesting fact. There seem to be no isosceles triangles with areas the same as their perimeters until the leg size reaches about 6.6605 units (not sure of the exact form), and then there are two triangles (2 angles) that work until the leg size reaches ~6.83 and then there is only 1 triangle (angle) that works past then as the leg gets bigger to infinity. Very curious. I'd be interested if someone could verify this and find the exact form of those numbers.

[@ZombieKillerThe]

sin60 = (3^(1/2))/2 area of triangle = 1/2 ab sinC (a and b are side length and C is the angle where side a and b meet) a=b=x and C =60 1/2 (x^2) (3^(1/2))/2 = (x^2)(3^(1/2))/4 = 3x (x^2)(3^(1/2)) =12x x(3^(1/2)) =12 x =12/(3^(1/2)) = 4(3^(1/2)) sinX = opp/hyp hyp sinX = opp Here are all 6 trig functions (opp = opposite, adj = adjacent and hyp = hypotenuse of a right angle triangle) sinX = opp/hyp cosX = adj/hyp tanX = sinX/cosX = opp/adj cscX =1/sinX = hyp/opp secX = 1/cosX = hyp/adj cotX = 1/tanX = cosX/sinX = adj/opp

[@DASPRiD]

Well, if you ignore the unit, you can make up any unit for the area and it will match ;)

[@materiasacra]

The ratio of the area and the perimeter depends on the length unit used. If it happens to be 1 when using meters, it will become 100 when switching to cm. Therefore the ratio is not a property of the triangle by itself. The devious purpose outlined near the end of the video can only be accomplished by adopting, and specifying, a length unit to be used. If I'm the intended victim, and the specification is indeed incomplete, I will protest by pointing out that the question is meaningless and instruct my lawyers accordingly :+)

[@jpolowin0]

I would argue that an area and a length are fundamentally different properties, and cannot be equal or "the same". The question would need to be restated to be meaningful. EDIT: "An equilateral triangle has a perimeter of X units and an area of X square units. What is X?" would be a meaningful question. Obviously, the answer differs from that of the original problem by a factor of 3.

[@Luca-sl9ot]

…you aren’t right. Using any measurement will get the same ratio, as long as you use it the whole time. If you switch it up halfway through the equation then you’ll get some funny answers, but he clearly didn’t.

[@materiasacra]

​@@Luca-sl9ot If you repeat the whole exercise with different units, you get a different ratio.

[@vivekgaur8915]

Bro you calculated the formula for area of equilateral triangle but in India in our schools it is taught to kids as direct formula and derivation is also taught in higher class So in every mathematical problem we do not have to find formula The formula is √3/4*a^2 Here a is side of triangle BTW Your method is also appreciable ☺️☺️

[@mekkler]

Area and perimeter app ≈ 20.78. Because Sketchup. ≈ Alt+247

[@eddyhedy5173]

pretty cool keep making vids!!!

[@nielle1963]

The actual triangle does depend on the unit. If the unit was 1m, then the triangle world have sides 6.928… m And if the unit was 1 foot, then the triangle world have sides 6.928… feet The result would have been so much nicer if the actual result was independent of units. As it is now, every equilateral triangle is a candidate, and you just have to decide what your unit should be.

[@MuffinsAPlenty]

Everything involving physical measurement works this way.

[@pikkoval5122]

Instead of using Pythagoras to find h couldn't you use Opposite = Sin(angle) x Hypotenuse to get h = x(sin60) since you know that all the angles are 60. I get that both work I just think its easier this way.

[@damiennortier8942]

H = sqrt(x^2 - x^2/4) = sqrt 3 x/2 And we want h = 3x or h = 6. So x = 12/ sqrt 3

[@pranavrs184]

I solved it using Area of Equilateral Triangle formula , root(3)x^2/4 = 3x, and got x

[@xanderlastname3281]

I keep forgetting you can delete fractions by multiplying them by their reciprocal (i.e dividing them by themself... The same way you eliminate any number. Seriously bro how did I forget that)

[@ZaeraeWasTaken]

There is a simpler way that I have done to solve this. A = P (sqrt3/4)x^2 = 3x 3x is the sum of the 3 sides of the triangle (sqrt3/4)x^2 is the specific formula for an equilateral triangle Divide x by both sides (sqrt3/4)(x^2/x) = 3x/x (sqrt3/4)(x) = 3 Multiply 4 by both sides (sqrt3/4)(4)(x) = 3(4) (sqrt3)(x) = 12 Divide the Square root of 3 by both sides (sqrt3/Sqrt3)x = 12/Sqrt3 x = 12/Sqrt3 Rationalize the fraction x = (12/Sqrt3)(sqrt3/sqrt3) x = 12(sqrt3)/3 Simplify the fraction x = 4(sqrt3)

[@derblaue]

This video also shows that sin(60°)=sqrt(3)/2 and cos(60°)=1/2

[@j.s.ospina9861]

If there's a triangle with an area bigger than its perimeter, and there's a triangle with a perimeter bigger than its area, then there's something in between.

[@felzm]

It is a less precise answer, but this fact follows imediately from the mean value theorem

[@cas3724]

Oh damn this is the guy with the whiteboard pens

[@MisterIncog]

Did you just said “let’s move it to the other side”? Wow. I’ve never heard that way of thinking in American education, it’s always subtract from both sides kinda thing. I’m just surprised because this “moving” with changing the sign is how we taught where I live

[@MuffinsAPlenty]

When I was in the education system in the US, I had plenty of teachers say "move it to the other side", and I used to say it that way too. And while that is a common way of thinking, it could potentially be confusing to people who don't grasp the basics of algebra, which is why I think it is often frowned upon.

[@d0pomein]

But what if it wasn't equilateral? Could some other triangle have the same area and perimeter ignoring units? Area increases exponentially but perimeter increases linearly. 3 4 5 right triangle has a of 6 and p of 12. 6 8 10 has a and p of 24 I would guess that any triangle with angles x y z could have equal a and p if you adjust the lengths of the sides correctly.

[@JonHaa87]

If we can ignore the unit, we can take any equilateral triangle and define it's parameter as 1 length unit and it's area as 1 area unit. Done.

[@sodiboo]

well no, because it's implied that the area unit and length unit are from the same, *coherent* system of measurements (which should've been stated in the problem tbh), but other than the units having a different exponent, they should be equal

[@MrBeiragua]

All triangles have area=perimeter. All you need to do is creat a unit system for that triangle in which this coincidence occurs.

[@Golololololo]

Wow this is really awesome👍🏼🤔 Thanks💖

[@meenatisahoo947]

I think its side should be 4√3 units Yess! To be fair it was a very easy question

[@johneonas6628]

Thank you.