What is the total area of ABC? You should be able to solve this!

[@mike1024.]

Wow Presh, that 2.5 minute tangent on the area of a triangle formula was worthy of a college professor. They love to tell stories in the middle of a lesson!

[@harrysakata3082]

An alternative solution with a bit less calculation: Given ABG = 70 and DBG = 35, we know that AG = 2GD. So if CGD = y, then CGA = 2y. Since CGA/CGB = 4/3, we have 2y/(y+35) = 4/3, which yields y = 70. You can now add everything up or note that CD = 2BD (and thus ABC = 3ABD) and arrive at 315 for the entire area.

[@Tiqerboy]

Nice. That's a better solution.

[@mahmoudalaaeldin3318]

Nice, quicker solution but same idea which is the ratio of areas is equal to the ratio of bases in case of common height, or the ratio of heights in case of common bases

[@soundsoflife9549]

I thought that G spot needed going over some more!

[@jeremiahlyleseditor437]

Great Video. This I followed completely and thank you for pointing out the history of the area of the triangle.

[@TotensBurntCorpse]

read CEVA's theorem and apply to this problem.. its doesnt need to make a cheat assumption to work.

[@NonFatMead]

Diagramming the triangle with right angles at D, E and F can lead to some confusion.

[@BigDoodles]

To use the area formula in the way you did the angles need to be 90 degrees. For example CF needs to be perpendicular to AB. However this information is not stated in the problem outline.

[@kilroy987]

Is that really true? His formulas make no mention of the length of any of the internal segments, only that there is some base and some height and some total area involved.

[@mglenadel]

@@kilroy987 Yes. This solution only works if the internal segments are perpendicular to the outer sides (so the segments are the height).

[@mglenadel]

Yeah, that caught my eye as well. And you definitely need that info to solve the problem.

[@vcvartak7111]

Saw your problem after so many days .nice to see you again

[@henk-ottolimburg7947]

The hight of ABG does not necessarily go through point F, this is only the case when GF is perpendicular, which is not stated anywhere in the video. Let the hight h intersect with AB on point P. That is not (necessarily) point F The beginning of the video is still true, dividing AB up in two 4:3 ratio pieces . (But BFG=30, not BPG) But dividing the big triangle up into two pieces with the same 4:3 ratio does not hold, for we could be talking about a hight H not intersecting at the same point P

[@rolandfisher]

It doesn't have to be perpendicular and he covers it, albeit indirectly.

[@ekatvakushvaha1814]

Welcome back!!

[@vcvartak7111]

Proof of area of triangle formula is simple. Any triangle can be divided into two right angle triangle by dropping a perpendicular from vertex. Any right angle triangle can be extended to rectangle by it's base and height. Now area of rectangle is l×b (length and breadth) since l×b unit squares form in rectangle. Hence right angle triangle area is l×b/2 .Then you can get area of any triangle by dividing in two right angle triangles

[@tomdekler9280]

"any triangle can be divided into two right angle triangle by dropping a perpendicular from vertex" is true, but only if the base is not adjacent to an obtuse angle. Otherwise the perpendicular doesn't intersect with the base and doesn't divide the triangle. Of course, this still works if you rotate the triangle so that the obtuse angle is the one opposing the base.

[@vcvartak7111]

@@tomdekler9280 Actually no need to rotate the triangle. You can draw perpendicular from obtuse vertex to slanted (largest) side. Base of triangle is not necessarily a horizontal line.

[@tomdekler9280]

@@vcvartak7111 Naturally, but I feel a proof is easier to visualize by adjusting the orientation to match the natural associations of the terms "base" and "height".

[@SimsimaFo]

We missed you

[@jeffhall7262]

The problem can be solved by using a comparison of ratios. The total left and right sides have a 4:3 ratio, as determined by the base with the given areas of 40 and 30. There is also a ratio of 1:2 across AD with 35:70 below the line and y:x+84 above the line. So 2Y =X+ 84. So, by creating the ratios 3/4 = 35+Y/X+84 and since we know 2Y =X+ 84, we replace X + 84 to get the equation 3/4 = 35+Y/2Y. Y=70. Using this Y value on the equation. 3/4 = 35 +70/ X + 84, we get that X=56.

[@ArabianShark]

2:30 I was explained in excruciating detail with visual and tactile aides why the formula of the area of a triangle holds true early on in the 7th grade. Score one for 1997 Portuguese school system!

[@hasibulislamshanto1190]

He made another video with similar problem like this one which involved the ratio of area. I couldn't find this trick then. So I solved that problem with coordinate geometry. But this time I used that trick.

[@peterlittlewood9810]

You did not make clear that the lines AD, BE and CF are perpendicular to the sides of the triangle but you assumed this in your proof.

[@braydentaylor4639]

Beautiful problem

[@Prisauria]

YOU'RE BACKKKKK!!!!!!!!!!

[@PugganBacklund]

Missed the reason why FG was the height, AFG looks like 90°, but did we confirm it was 90°? Or is the answer the same even if h & H isn't at the point F?

[@MindYourDecisions]

The length h is the length of the altitude from point G to the base AF. The altitude may be FG or not, so yes the answer is the same whether FG is the height or not. It does look like FG is the height in the diagram but that does not have to be the case for the answer or method to work.

[@smylesg]

I had the same question. Although the altitudes start from points G and C, it is not necessary or specified for them to intersect the base at point F.

[@deerh2o]

FG is not necessarily an altitude. Pedagogically, it takes something away with CF so close to being perpendicular to AB. The solution is so pretty on its own; it would have been nicer if CF and AB were clearly not perpendicular.

[@odourpreventer]

Edit: FG is not the height. The height is unknown, but it doesn't matter since it can be removed from the equations.

[@mike1024.]

​@@deerh2oAgreed, the math still works, but I would have liked to have seen the problem drawn where h and H are clearly not necessarily the lengths of FG and FB, respectively. But I don't think that was explained well when the area formulas were mentioned either.

[@vxllabh09]

i did it by establishing that since they all were cevians to the same triangle passing through the same point, and applied menealus theorem to prove that they were indeed perpendicular bisectors, i applied law of sin and cosines

[@Darisiabgal7573]

Cannot be true. CF is not 90° to AB, but instead has an Angle about 85°. The rest of what you said was true. If you use the law of congruent angles on 30 and 84 assuming GF and GE are perpendicular to their respective sides then we get EG is 1.6732 * GF BG = 5/4 GE and CG = 7/2 GF CG^2 = 84/30 BG^2 GE^2 = 84/30 GF^2 7/2)^2 GF^2 = 84/30 5/4)^2 GE^2 49/4 GF^2 = 70/20 GE^2 7/10 * 5 = 3.5 GF^2 = GE^2 84/30 = 2.8 Therefore EGC and FGB are not similar triangles therefore either FC is not perpendicular, EB is not perpendicular to AC or neither is perpendicular.

[@vxllabh09]

@@Darisiabgal7573 Yeah my professor explained the exact same thing to me after I showed him the video. Thank you so much for your help though!!

[@WRSomsky]

Interestingly enough, you can omit giving the area of CEG and it is still determinable. The areas of AFG, BFG, and BDG are sufficient to determine the results. (And CEG is indeed 84. This can exist.) I'm doing this algebraically in Mathematica, so not sure right off if it is solvable by simple geometric relations.

[@WRSomsky]

Mathematica 13.2.1 Kernel for Linux x86 (64-bit) In[1]:= PA:={Ax,Ay}; In[2]:= PB:={Bx,By}; In[3]:= PC:={Cx,Cy}; In[4]:= PG:={Gx,Gy}; In[5]:= PD:=ResourceFunction["LineIntersection"][{PA,PG},{PB,PC}]; In[6]:= PE:=ResourceFunction["LineIntersection"][{PB,PG},{PC,PA}]; In[7]:= PF:=ResourceFunction["LineIntersection"][{PC,PG},{PA,PB}]; In[8]:= AFG:=ResourceFunction["SignedArea"][{PA,PF,PG}]; In[9]:= FBG:=ResourceFunction["SignedArea"][{PF,PB,PG}]; In[10]:= BDG:=ResourceFunction["SignedArea"][{PB,PD,PG}]; In[11]:= DCG:=ResourceFunction["SignedArea"][{PD,PC,PG}]; In[12]:= CEG:=ResourceFunction["SignedArea"][{PC,PE,PG}]; In[13]:= EAG:=ResourceFunction["SignedArea"][{PE,PA,PG}]; In[14]:= ABC:=AFG+FBG+BDG+DCG+CEG+EAG; In[15]:= Sol=Solve[{ PA=={0,0}, (* A at origin *) PB=={Bx,0}, (* B on x axis *) AFG==40, (* area of AFG *) FBG==30, (* area of FBG *) BDG==35 (* area of BDG *) }]; In[16]:= Sol//InputForm (* Triangle Solution *) Out[16]//InputForm= {{Ax -> 0, Ay -> 0, Bx -> 140/Gy, By -> 0, Cx -> (-560 + 9*Gx*Gy)/(2*Gy), Cy -> (9*Gy)/2}} In[17]:= Simplify[CEG/.Sol[[1]]] (* Area of CEG *) Out[17]= 84 In[18]:= Simplify[ABC/.Sol[[1]]] (* Area of ABC *) Out[18]= 315 Note that if we put point A on the origin and point B on the x-axis, we have complete freedom as to where we put point G (as long as it is off the x-axis) with the x-coordinate of B and the location of C hence being determined from there.

[@byczakrew1093]

true. CEG is irrelevent. even easier to solve geometrically with thales theorem

[@Sphinxinator]

does the perpendicularity (or how perpendicular) of the middle segments that intersect point G affect how the triangle is solved? Because FG may not be perpendicular to AB; and same for other sides.

[@EdNinja-o6h]

The graph of the triangle given in the problem is not right: it was drawn that AD, BE and CF as if they are the heights of ABC triangle. I know that it was never mentioned that they are the heights, but the initial graph is misleading. Other than that, great solution!

[@christopherwellman2364]

If they were the heights, wouldn't AFG + BDG + CEG = AEG + BDG + CDG?

[@nightatgrambys2340]

The sum of the angles at point g equal 360*. there are 6 angles total, so 1 of the angles must equal 60*. you can be sure of this by drawing a bisector at the G point of the triangle AGE and reasoning that only the half and whole pairing of 60 and 30 can be summed with the 90* to be a total of 180* on the triangle AGF thanks to the z rule. this confirms that the triangle ABC is an equilateral one, with all the baby triangles inside of it having 30*,60* and 90* respectively. this means the area and side lengths of all the triangles are proportional. Now take triangles AEB and CEB. you can see that they both posses an equilateral side length y and the height EB therefore making their EA and CE side lengths must be equals, also making their areas as such. after this revelation we can deduce that AEG and CEG are equal for the same reason which all boils down to the equation 84+40+30 = 84+35+CGD which makes the area of triangle CGD in all realm of possibility be 35. 84+40+30+35+35+84 = 308. Your solution was right until I proved it wrong and mine is right until I am proved wrong.

[@mrfork1176]

It is not possible to know the area of triangle ABC because there are 4 available configurations for the triangle! 1) G inside ABC (no angle greater than 90)=> area = 315 2) G outside ABC, (one angle greater than 90 ) 3 cases: - angle A > 90 : invert G and A, => area = 105, - angle B > 90: invert G and B => area = 140, - angle C > 90: invert G and C => area = 70

[@hankmax8899]

Hey, I just have one question. Why can you take the ratio of the two triangles? Are they similar? and if so, how do we know?

[@msnirajagrawal]

NCERT publication (most common maths textbook in Indian school) has this proof in the school maths textbook. I don't exactly remember the class but I have read that proof of area is half base times height

[@rennleitung_7]

As far as I remember, but that's some 50 years ago, the proof of the area of a triangle was easy as soon as you learned point reflection. You started with an arbitrary triangle, created the parallelogram and finished with a rectangle. You didn't find the proof in the book, because it was an exercise for the students.

[@luizrobertocamargo5864]

I need help with a question. Please. Knowing that lim(x-->-2)f(x)/(x^2)=1, which is the value of lim(x-->1)f(x^2-3x)/[(x+8)^(1/2)]? Answer bellow. Answer: 5/3

[@samkarn]

I guess answer is 4/3, f(-2) = 4 from first limit and second limit reduces to f(-2)/3 = 4/3

[@afzal1626]

correct me if I'm wrong, but in 8:00, can't we just solve the equation by expressing y in terms of x (I learned it as simultaneous equations in school)

[@samkarn]

The way of your solving Linear equation in two variable is substitution methord.

[@1a1u0g9t4s2u]

Is it safe to conclude that problems which include diagrams is to never assume it is to scale? I have notice many comments where the viewer appears to be depending of the diagram being to scale, i.e an angle is at 90 degrees. The math works not knowing any dimensions, thus a scale diagram is not required, in this case the angles. I too enjoy learning the history of math. I need to refresh my memory on why the norm is to divide a circle into 360 degrees versus another number. Thanks for sharing.

[@SRAVANAM_KEERTHANAM_SMARANAM]

Your Channel is watched by Many People. Kindly Create Video on 1) Without using Scale and Compass, Can you Create a Equliateral Triangle. 2) Using only Traingles, Can you show the difference between Squares, Rectangles, Parellograms, Rhombus andTrapezium.

[@darshanlal263]

Or probably email if you are a very old viewer you may have his email id

[@robertveith6383]

* triangles * parallelograms

[@CatMC1030]

In 9:07 should 35/y = BD^2/CD^2 correct me if I am wrong

[@مرادمحمدصبري]

Thanks a lot

[@eliseuportes1260]

I really like your videos! Even when I can't solve.

[@epovo]

Beautiful

[@WhiteGandalfs]

You could have simplified the approach because the equivalence in area-split for two triangles with same baseline means as well the same aera-split for the remaining triangles (f.e. ACG / BCG). Thus, your equations could have been made a little more simple. Or how i like to express it: Another example of "many ways to Rome" :D

[@jvarunkumar]

Is this information enough to find the length of sides of the triangle?

[@sundareshvenugopal6575]

84 + 30 = 40 + Y = 35 + X, where Y = area of CGD, X = area of AGE.

[@Tiqerboy]

Your diagram is not to scale. I have a *much* harder time solving problems when they are presented as such. Solving it one way, I got an impossible result, one of the triangles had to have a negative value. I solved it another way and then found an answer that makes sense. **EDIT** I solved it exactly how Presh did it at first, but I made a mistake in my arthmetic and found one of the triangles have negative area. Comparing the ratio of the areas of ABE and BCE I ended up a quadratic formula which gave one real positive root which I was happy about that and that gave me the same answer as Presh.

[@langasdeloslangas2238]

I usually love this channel's problems, but this one had a HUGE flaw: at no point it is said that these are the heights! No mention of perpendicularity is made (I know they "seem" perpendicular, but any Math student knows they shouldn't trust appearances). Therefore, the whole explanation is flawed. Don't get me wrong, the final solution is correct, and I figured it out, without loss of generality, because this is not relevant, but Presh makes a big (and WRONG) assumption while solving this problem.

[@thorntontarr2894]

I totally agree but as I just posted the same conclusion and now I have read your comment, I am pleased that finally someone else saw this absolutely HUGE flaw in MindYourDecisions work.

[@rickychoy1030]

@@thorntontarr2894 For the same height triangle, ratio of area = ratio of base length. it is not required CF is perpendicular to AB. The proof is 100% correct.

[@racpa5]

How did you like living in Palo Alto? My great aunt was a professor there way back in the day.

[@madaunn42221]

DEAR PARESH HOW DIDYOU PRESUME THE LINE CF IS A PERPENDICULAR ON AB? Dr. MADAN ARORA

[@WhiteGandalfs]

He didn't, but the depiction could have been a little bit better if the triangles would have been a little more sheared (or the Point G put a little more excentric). So that the arbitrary lines would not coincidentally look like being perpendiculars.

[@robertveith6383]

Stop yelling your post in all caps. It is rude.

[@DominicLeung87]

Wasn't explicit in the video but from the areas given (and the assumption that they're all constructed in the same way) it can be deduced that the line segments must be perpendicular bisectors as opposed to angle bisectors or median bisectors. Proving why they cannot be median bisectors is trivial, proving why they cannot be angle bisectors is slightly more (though not at all) difficult.

[@realnazarene5379]

You're assuming that AD, BE, and CF are perpendicular to the sides they intersect. You must justify their use as heights in the formula.

[@JonathanKahan-q2g]

You actually can solve the given problem without knowing that one of the triangles area is 84. I solved it slightly differently than presh, but my method did not identify the areas x and y he references in the video

[@PetelProduction]

How do you know GF is the height of GFA? It wasn't stated anywhere that angle GFA is 90 degrees.

[@mathiaskjeldgaardpetersen5926]

You don't know. It´s just assumed. Because all the lines meet at G, and all the triangles are inside the internal system of triangle ABC, any ´´inaccuracy´´ of height GF would be cancelled by the ´´inaccuracies´´ of GE and GD. But all that doesn´t matter, only the ratios matter here. -"If two triangles have the same height, then the ratio of their areas equal to the ratio of their bases"

[@libranden1333]

Dovetail theorem makes this so much easier, btw, when I was in China, I remember this type of question being given to year 5's.

[@TotensBurntCorpse]

Its a cheat -- dove tail theorem makes the same assumption that the author did.. the G to flat lines are NOT necessarily perpendicular.

[@yanggang4352]

@@TotensBurntCorpse They are. The height is the same as shared by the pair of triangles, and by definition they are perpendicular. It doesn't mean they are necessarily GE or GD, etc. but they are perpendicular.

[@libranden1333]

@@TotensBurntCorpse There's a concept in Chinese geometry called 一半模型 which states that on parallel lines, if a triangle with the base and tip always on the parallel lines that always have the same base and height will always have the same area. The triangles in this question and in the typical dovetail theorem can have parallel lines drawn to them. I recommend looking at this Chinese book:高思学校数学竞赛本. Unfortunately, this book is only available in Chinese but it covers a lot of useful content including the dovetail theorem and butterfly model.

[@nitelikli]

birde bunu dene istersen IAGI / IGDI = 2 A(AGC) = 2.A(CGD) and A(AGC) = 2y 2y / 40 = (y+35)/30 y = 70 A(ABC)= 315

[@willem3990]

That is only true if G is the center of mass or equivalently if each side of the triange is cut in two equal parts.

[@nitelikli]

lAGl / lGDl = 2 You perceived it as the center of gravity because 70 / 35 = 2 due to the given areas. If the ratio was another number instead of 2, the solution would not change.

[@Happy_Abe]

I thought this was going to go into the proof for A=1/2bh Maybe in a future video?

[@JonathanKahan-q2g]

I could be wrong, but i do not believe such a triangle can actually be constructed. Taken by themselves to produce a triangle of 40 next to a triangle of 30 next to a triangle of 35, i believe angle ABD would have to be greater than 90 degrees, and it would be impossible for a point C to exist

[@delenk]

Helo, i read many years befor Wikpedia exist that the Sumerians knew the area of triangles, Pitagoras formel etc....

[@kilroy987]

It looks like the segments have a right angle to the sides they intersect, but that doesn't have to be true here.

[@christopherwellman2364]

I figured AFG + BDG + CEG would be equal to AEG + BFG + CDG, so I quickly concluded that the answer would be 318. I guess I was wrong.

[@banibratamanna5446]

this trick is used to prove Ceva's Theorem

[@flickingbollocks5542]

Please provide a link to interesting...

[@mohitrawat5225]

Alternate title - Can you find the areas in this picture?😂😂😂😂

[@Vecto314]

I want to also make vdo like this in my channel how to ?

[@davidkahnt2632]

Do we have enough information to get the base and height of the triangle?

[@OnwardToMail]

A square is a parallelogram, so it would be the formula of finding the area of a square.

[@Ivan-fc9tp4fh4d]

Ok. But, at the first glance, one could assume, that x must be greater than y.

[@thorntontarr2894]

You have proceeded using the assumption that each line from a vertex is an altitude, i.e. a perpendicular. Yet, you NEVER stated that and the drawing does not state although the drawing appears to be that way. Without that "given", show me the solution please.

[@RiggedDummy]

But its G, and thats for altitudes

[@Utkarsh.22013]

Answer 315 correct 🎉🎉😂😂❤❤

[@gkotsetube]

Was it necessary to do a search to find that the father of geometry gave the area of a triangle? He just said it with words, as he classically did, without using formulas.

[@climbJackbean]

Sorry, I think the method shown in the video is too slow. Look at this : AG:GD=2:1. Let a be the area of triangle CGD. Then the area of triangle AEG is 2a-84 which tells EG:GB=(2a-84):70, that is, (a-42):35. At the same time, it equals to 84:(a+35). So (a-42)(a+35)=(35)(84) which gives you a = 70 or a = -63(rejected). Therefore, the area of triangle ABC is 70+140+105=315. I only need to introduce one variable and a simple quadratic equation to solve it.

[@willem3990]

AG:GD is not necesarily 2:1.

[@climbJackbean]

​Hi. AG:GD must be 2:1. The reason of it is because triangle BDG and triangle BGA have the same height if the height is drawn from B. Therefore, the ratio of area of triangle BAG to the area of triangle BDG equals to the ratio of AG to GD. Since the ratio of their areas is 70:35 which is 2:1, so as AG:GD. I hope the explanation is clear. Thanks

[@chewy232]

My brain hurts now. But in a good way.

[@serhanyegen]

It's a very long solution, there is a short solution

[@Ark.AI.]

Hello sir. Would you be merciful on me to elaborate the short answer of yours

[@wesss9353]

Just for fun : Fast inverse square root function of quake 3 arena

[@manwaiho4821]

this one is a cinch

[@gaspardcaux5294]

100*pi

[@eldoctoroso]

BFG is always equal to 9000

[@zdrastvutye]

it is remarkable that in my solution the result won't depend on xg=... in line 30: 10 print "mind your decisions-what is the total area of abc?" 20 dim x(5,2),y(5,2):lab=20:a1=40:a2=30:a3=35:a5=84:sua=a1+a2+a3+a5:sw=.1 30 yg=2*(a1+a2)/lab:xf=2*a1/yg:yf=0:xg=.55*lab:yg=2*(a1+a2)/lab 40 yd=2*(a1+a2+a3)/lab:xd=yd*xg/yg:yc=yg+sw:goto 90 50 dxc=(xg-xf)*(yc-yf)/(yg-yf):xc=xf+dxc:ny=yg*xc/yc:ny=ny+lab-xg:ye=yg*lab/ny:xe=ye*xc/yc 60 a4=yc*(lab-xf)/2:a4=a4-a3-a2:a6=yc*xf/2:a6=a6-a5-a1 70 dgu1=ye*lab/2:dgu1=(dgu1-a1-a2)/sua:dgu2=yc*xf/2:dgu2=(dgu2-a5-a1)/sua 80 dg=dgu1-dgu2:return 90 gosub 50 100 dg1=dg:yc1=yc:yc=yc+sw:if yc=100*lab then stop 110 yc2=yc:gosub 50:if dg1*dg>0 then 100 120 yc=(yc1+yc2)/2:gosub 50:if dg1*dg>0 then yc1=yc else yc2=yc 130 if abs(dg)>1E-10 then 120 140 print yc,"%",a4,"%",a6:ages=yc*lab/2:print "agesamt=";ages 150 x(0,0)=0:y(0,0)=0:x(0,1)=xf:y(0,1)=0:x(0,2)=xg:y(0,2)=yg 160 x(1,0)=xf:y(1,0)=yf:x(1,1)=lab:y(1,1)=0:x(1,2)=xg:y(1,2)=yg 170 x(2,0)=lab:y(2,0)=0:x(2,1)=xd:y(2,1)=yd:x(2,2)=xg:y(2,2)=yg 180 x(3,0)=xd:y(3,0)=yd:x(3,1)=xc:y(3,1)=yc:x(3,2)=xg:y(3,2)=yg 190 x(4,0)=xg:y(4,0)=yg:x(4,1)=xc:y(4,1)=yc:x(4,2)=xe:y(4,2)=ye 200 x(5,0)=0:y(5,0)=0:x(5,1)=xg:y(5,1)=yg:x(5,2)=xe:y(5,2)=ye 210 masx=1200/lab:masy=900/yc:if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

[@iamarpit.]

3rd

[@rudrapratapsingh983]

There is another formula tha triangles with same base and which lie between same parallel line have equal are we learned this in 9th standard in india

[@rudrapratapsingh983]

Area*

[@hyperboloidofonesheet1036]

Driving me nuts. Area of ABC is 3 times area of ABD; is it just coincidence?

[@prithwirajroy3915]

Hello

[@hvnterblack]

cool :)

[@wesss9353]

9th grade math was 2 decades ago...

[@funwithiti_2012]

I was thinking that we should remove the area of any one triangle. Then add the remaining after that multiply it by 2 What can a 7th grader think 💀

[@gm_matthew]

“You should be able to solve this”? I’m pretty good at mathematics but I don’t have a clue how to solve this thing

[@T0Mvdt]

1000’s like person

[@kam7056]

❤

[@rauls4972]

Anyone can write anything they want to in Wikipedia.

[@baron_navapol7835]

I think it is not that hard I learn that in grade 6 and i am asain

[@JonathanKahan-q2g]

ignore my last comment

[@huzefa6421]

Early

[@samtigernotiger3886]

This solution is incorrect because the question was not asked correctly. Quote "construct 'D' were 'D' is in the side 'BC'". The question never mentions a right angle. Therefore later 'h' is not necessarily the height and the shape of the area does not apply either. I am very disappointed to see such a fundamental mistake on this channel!

[@EricOneOneNine]

The angles don't need to be right angles for the argument to work.

[@samtigernotiger3886]

Please inform yourself. In 'F = 1/2 (AB)h', 'h' is the height and by definition is perpendicular to '(AB)'. This is also shown in detail in the video, but the right angle is never drawn. And that is exactly the criticism.

[@EricOneOneNine]

@@samtigernotiger3886 'h' is a numerical quantity and hence cannot be said to be perpendicular to anything. 'h' is the length of the altitude which is indeed perpendicular to AB. But the argument does not require the altitude to coincide with CF. There is no missing hypothesis in the question.

[@flickingbollocks5542]

Connect F-U then C-K Pause this comment, only then ask for the solution.

[@climbJackbean]

OMG, I can figure out another way.... Maybe too many ways to solve it by using one variable only