Great, now engineers will start using the area of that triangle for the gravitational constant. Edit: guys, I get it I mixed up the gravitational constant and average gravitational acceleration on earths surface. I'm sorry, I was very tired when I posted this
:))yep i was surprised that i got 9.8 as a result even tho i went with the classic x,l-x aproach and then equations to calculate l But nevertheless since i study engeenring,i therefore am allowed to use the triangle as g:))) Ps:pi squared=g
Uhm, the gravitational constant is about 6.674 * 10^(-11) m³/(kg*s²) en.wikipedia.org/wiki/Gravitational_constant Too me that looks like you confused "Big G" (the gravitational contstant) with "little g" (the gravitational acceleration at the surface of earth). The 9.81 m/s² is not really a constant. It differs quite a bit even at the surface of earth. Further more it of course decreases the further away from earth you get. The gravitational acceleration can be calculated by g = G * M / r² (where "r" is the distance from the earth center and M is the mass of earth)
There's a simpler and more elegant solution. Note that the rectangle's area (A), is comprised of 2 pieces each of the smaller triangles MINUS the upper-left smaller rectangle (B) formed from the basis of the 3 and 4- triangle. More formally, using Presh's notation: A = 2*(3+4+5)-B. ac/2*bd/2=3*4 -> B:=cd = 48/(a*b) = 48/A So A = 24 - 48/A, with solution A=4(3+sqrt(6)). Subtracting the smaller triangles yields the answer *4*sqrt(6)*
@Everybody Nobody solving a second-degree equation is straight forward and not covered here. My solution is simpler because it introduces fewer unknowns.
my method for intermediate learners: let: A= area of rectangle x= width of rectangle y= height of rectangle a= height of triangle with area=3 b= base of triangle with area=4 ... eq1. ax/2 = 3 hence; ax =6 eq2. by/2 = 4 hence; by =8 eq3. (y-a)(x-b)/2 = 5 hence, xy-by-ax+ab = 10 . take eq.1 and eq.2, multiply to get: abxy = 48 hence ab=48/xy which is now eq.4 . take eq 3. and substitute ax,by, and ab. xy - 8 - 6 + 48/xy = 10. since A = xy , substitute A^2 - 24A + 48 = 0 you get, A= 2.2 and 21.8 we use 21.8 as the logical answer for the area of rectangle . area of middle triangle = A - 3 - 4 -5 which is 9.8 answer is 9.8. .
This is one of my favorite videos you made. You took a nice problem and solved it in a simple yet satisfying way, showing the general case than applying the numbers. Cheers!
Just have to be careful what areas you label as "x" and "y" since u can rotate the rectangle, esp if its a square. You need x and y to be the areas of the 2 triangles with a side of the rectangle as a leg
Thank you. The triangles are not all drawn the same. x and y have a whole side length, but z does not. I was wondering what would keep the formula from working with any order of the triangles, and that's why. Thank you. Z is the unique one here.
for a less complicated quadratic equation : Notice that ab is the area of the rectangle. Therefore ab=x+y+z+w. I also called "t" the quantity x+y+z. If you do this changes in the quadratic equation at 2:58, it becomes less complicated : (t+w)² - 2t(t+w) + 4xy = 0 and then, by using distributivity w² - t² + 4xy = 0 that leads directly to w = sqrt(t²-4xy) = sqrt((x+y+z)²-4x)
At time 0:19, Let AF=d and CE= 3d/4 since the areas of triangle BAF=4 and triangle BCE =3. Let side of square ABCD =x and area of square is therefore x^2 Consider triangle BAF, 4= xd/2 d=8/x...............(1) Consider triangle FDE, 5=(x-d)(x-3d/4)/2............(2) Substitute d=8/x into (2), We end up with 10=x^2-14+48/x^2 Let y=x^2, 0=y-24+48/y Multiply through by y, y^2-24y+48=0 y= 12+4sqrt(6) (-ve sqrt rejected in quadratic formula) Area of square ABCD =x^2 = y =12+4sqrt(6), Area of triangle BEF= area of ABCD-area of ABF-area of BCE-area of FED =(12+4sqrt(6)}4-3-5 =12+4sqrt(6)-12 = 4sqrt(6) square units as required. Note the -ve value of the sqrt in the quadratic formula was rejected because the area of BEF would have been -ve.
I had a little different approach. I called AB and BC as x. Then using a system of equations, I expressed FD and ED in terms of x and plugged those into the equation for the area of FDE. This ended up giving me x^4-24x^2+48=0. Solve for x then just calculate the area of the square and subtract the three triangles. Same answer in the end.
This is what I was trying to do, but went wrong somewhere. Was a bit annoyed that the video went straight to demonstrating the general case instead of the specific one so I wasn't able to figure out my mistake! Thanks for explaining how you did it, I managed to get there in the end.
At time about 1:53, one can view the area (ab) of the big rectangle as the sum of three rectangles' areas (2x, 2y, 2z) minus the overlapped rectangle's area [ (2x/b)(2y/a)]. Therefore, ab = (2x+2y+2z)-(2x/b)(2y/a).
Hmm, I kinda brute-forced this one. I wrote 5 equations to calculate the length of he side first then subtracted the areas of the triangles. But your approach is more elegant.
The 5 equations model gives something quite amusing because at the end, you should have something like (l²-8)(l²-6)=10l² (l => length of the square) But, l² = Area of the square ! So (A(square)-8)(A(square)-6)=10A(square) A(square)=12+4sqroot(6).
Looking at the thumbnail, I thought you could use Heron's Formula to find the area: s = (3+4+5)/2 = 6 s-a = 6-5 = 1 s-b = 6-4 = 2 s-c = 6-3 = 3 Area = √s(s-a)(s-b)(s-c) = √6*1*2*3 = √36 = 6 But this doesn't work because the AREA's of the outer triangles are 3, 4, 5. They're not the sides of the outer triangles.
My solution also started with denoting the sides and coming up with systems that will end up being quadratic in form. I’m glad I got another ome of these with almost the same approach you did.
Problem: Resistors 1 and 2 in parallel yields total current of 2A Resistors 1 and 3 in parallel yields total current of 3A Resistors 2 and 3 in parallel yields total current of 4A The power supply is 12V Calculate: a. Total current when all 3 resistors are connected in parallel b. Total current when all 3 resistors are connected in series
Same problem: a and b can do a work in 2 hours a and c can do the same work in 3 hours b and c can do that work 4 hours then how long it would take if a, b and c all work together. Ans: in 2 hours a,b completes 1 in 1 hour a,b complete 1/2 of the work Same way in 1 hour a,c completes 1/3 b,c complete 1/4 of the work in 1 hour a+b+a+c+b+c can do 1/2+1/3+1/4 of the work = 13/12 in 1 hour 2a+2b+2c can do 13/12 in 1 hour a+b+c can do 13/24 =>the a+b+c completes whole work In 24/13 hours.
Good generalization. For specific case, I used the constraints and set one of short bases as a fraction of length of square. That simplified to a quadratic homogeneous equation of the ratio. Rest of it is algebra, getting the expected result.
here is my solution to the problem with the square. Just a bunch of equations that rely on "area of a triangle" and "area of the square"-function a = Distance from A to F b = Distance from C to E c = length of each edge of the square x = area of triangle 3 = 1/2 * c * b; 4 = 1/2 * c * a; 5 = 1/2 *(c-b) * (c-a); x = c² - 3 - 4 - 5 This easy approach does only work with the square and not with the more general rectangle
I'm happy that I managed to solve this one orally without much effort, although took 15 odd minutes to correct my earlier equation. Honestly it's a very easy problem as other lengths can be easily obtained after assuming square side length as say 'x'. Solving the equation x^4 - 24 (x^2) + 48 = 0, x^2 = 12 + or - 4 √6 Obviously area sum of 3 triangles is 12. So x^2 will be greater than 12. Hence, Required area = 4 √6 square units
Elegant solution. 4:57 I did it differently and got the same answer, 9.798. I ended up with A^4 -24A^2 +48=0 (A raised to the power of 4 minus 24A raised to the power of 2 plus 48=0) and factoring for A get the length of the square as 4.668, and so the AREA square is then 21.798 and thus by subtracting the TOTAL length of the three triangles 12 the result is therefore 21.798-12=9.798. To show how I arrived at A^4 -24A^2 +48=0 Let A = the length of the square, let B= the base of the triangle with area 3, and let C the height of a triangle with area 5. Then (1) A x B=6 or B= A/6 since the area of triangle = 1/2 base x height (2) (A-B)x C=10 C=10/A-B or C= 10/A-6/A or AC-AB =10 (3) (A-C) x A=8 or A^2 - AC=8 (4) substituting equation (1) into (2)= (A-6/A)x C=10 or AC-6/AC=10 Adding equations (3) and (4) = A^2-6/AC=18 (5) substituting equation (1) into equation (3) = A^2-A(A/6)= 18 ........ .... I ended up ( as said above) with A^4 -24A^2 +48=0 and solving for A resulted in 4.668 What is the length of the square? Therefore, the area is 4.668 x 4.668 or 21.798 and 12 (the sum of the three triangles) from this yields 9.798. Answer: let's plugin 4.668 in the above equation to check. 4.668^4- 24 x 21.798 +48=0 475.152-523.152+48=0 or 523.152-523.152=0 So, yes, 4.666 satisfies this equation. 1
I found it out similarly but It's much easier if you apply a trick: when you get to the equation A^4 -24A^2 +48=0 just simply introduce a new "'x" which is a^2. This way you get that x is 12+4sqrt6 but the trick is that this way x is the area of the entire square so you just deduct 3+4+5 from it and you get 4sqrt6.
Simpler solution: Define base b from left to right and height h from top to bottom b = b1 + b2 h = h1 + h2 Rectangle areas are twice the given triangle areas b.h1 = 2.3 = 6 b1.h = 2.4 = 8 b2.h2 = 2.5 = 10 Define areas A = b.h Q = b1.h1 S = (b.h1 +b1.h + b2.h2) / 2 = 12 W = A - S Product (b.h1)(b1.h) = 6.8 = 48 = AQ Solve for W in terms of known S and AQ. Add the three rectangles (2x, 2y and 2z in your scheme) and notice a double counting of Q = b1.h1. Thus the key insight by inspection... 2S = A + Q Subtract S from both sides S = W + Q (or W = S - Q) Add the definition of W 2W = A - Q Subtract the squares of the 2S and 2W equations, and only the cross terms remain, 4S^2 - 4W^2 = 4AQ Divide by 4 and rearrange W^2 = S^2 - AQ = 12^2 - 48 = 96 W = 4 Sqrt(6) For completeness A = S + W = 12 + 4 Sqrt(6) Q = S - W = 12 - 4 Sqrt(6) FYI, these solutions also result from the quadratic equation obtained by multiplying the 2S equation above by A, and rearranging... 0 = A^2 - 2SA + AQ.
First case: The area we're looking for = A Square sides: x: one formed from x and x-a The other from x and x-b ax=2×3=6 bx=8 (x-a)(x-b)=10 ==> x²+ab=24 means x²
At 0:20 use Heron's Formula : Which states Area of a triangle = √{S(S-A)(S-B)(S-C)}, where S is the semi - perimeter of the riangle which is the sum of all all side divided by 2. So, here S will be (3+4+5)/2=6 So, therefore area will be, √{6*(6-3)*(6-4)*(6*5)} = √6*3*2*1 Which can be written as √3*2*3*2 = (*Taking off the square root*) 3*2 = 6unit^2
Let x be the side of the sqr be x So the other side of triangle ( that area is 4 ) be 8/x The other side of triangle (that area is 3) be 6/x So the area of third triangle 1/2(x-8/x)(x-6/x)=5 =>X^2=12+-4✓6 Hence W=x^2-(3+4+5) =12+4✓6-12=4✓6 (ngtv vlu not acceptable)
Thanks for pointing out! There was another comment talking about using Heron's formula which i kind agree but the final answers are different and now i know why, due to the error you have mentioned
Let s = side length of the outer square = a + b, the side lengths = c + d, the bottom lengths Then rectangles sa = 2*3 = 6 sc = 2*4 = 8 bd = 2*5 = 10 Sum and product 2S = sa + sc + bd = s^2 + ac (S = 3+4+5 = 12) P^2 = sa*sc = s^2 * ac (= 6*8 = 48) Desired area W = s^2 - S 2W = s^2 - ac W^2 = S^2 - P^2 = 12^2 - 48 = 96 W = 4 Rt(6) For completeness s^2 = S+W = 12 + 4 Rt(6) ac = S-W = 12 - 4 Rt(6) ad = sa-ac = -6 + 4 Rt(6) bc = sc-ac = -4 + 4 Rt(6) sb = bc+bd = 6 + 4 Rt(6) sd = ad+bd = 4 + 4 Rt(6) Get a, b, c, d by dividing sa, sb, sc, and sd by s.
2:08 , if you still in the square situation (a^2 and not ab) without formula, you can say that a^2 = w + x + y + z = w + 12, so you will get after the distribution and simplication : w - 12 + 48/w + 12 = 0 (w - 12)(w + 12) + 48 = 0 w^2 = 96 w = sqrt(96) = 4sqrt(6)
Elegant solution for the general case of a rectangle. There is an easier approach for this example of a square, however. Setting X as the length of one side of the square, using the triangle formula 1/2(b)(h) and the given areas one finds X^4 - 24X^2 = -48 after some basic algebra. Complete the square and find X^2 (the total area of the square) is 12 + 4(6^1/2). Subtract the given areas 3 + 4 + 5 and this also solves w as 4(6^1/2). Thank you for the problem!
Should not assume it is a square. For versatility, set X for one side of the rectangle and Y for the perpendicular side. If X =Y then it is a square otherwise it is a rectangle. One will realise it does not matter if it is a square or rectangle, the key thing is X.Y = overall area of the square or rectangle and > (3+4+5 = 12).
There is a much simpler way without having to get to quadratic equation. Complete parameterization: a x b is the area of rect. ab = 12 + z (assume area of inner triangle/inscribed wrong word as no circle) Area of ∆ = 1/2 X base X height ∆1 = a x X = 8 ∆2 = b x Y = 6 ∆3 = (a-Y)(b-x)=10 Now expand ∆3 and solve collect all variables in one side, We have: XY + z = 10+8+6-12 »12 z = 12-XY----------------(main equation) Now find XY X = 8/a Y=6/b Therefore XY = 48/ab But we know ab = 12+z So XY = 48/(12+z) Now sub this value in main equation z = 12 - (48/(12+z)) Upon rearranging we have z = 4√6 Area of inner triangle is 4√6 sq units
Let x, a and b be side of the square, bases of right-angled triangle-(4) and right-angled triangle-(3) respectively. My solution with simplified steps: xa = 8, xb = 6 leading to ab = 48/x^2, (x - a)(x - b) = 10; combining the latter two gives x^4 - 24x^2 + 144 = 96 or x^2 - 12 = 4√6, i.e. area of triangle BEF = 4√6.
Hay I first found the side length of square x(a)= 4.67 by solving polynomial quadratic equation x^4-24x^2+48=0 (actually simple but also got √2.28 as an extra answer)
Именно так сторону квадрата за х, основание треугольника 4 за у, основание треугольника 3 за z, выписать для всех трёх треугольников площадь вот и система, выразить все через х и вот вам уравнение см выше....оно решается и ответ....
a=4.668828437 units,AF=c=1.713491962 units,CE=d=1.285118972 units,FD=2.955336475 units and ED=3.383709465 units if ABCD was a square.All the other numbers w,x,y and z would all be the same.Note that d/c=0.75 because if they are 3/4 in area the bases would have the same ratio since they have the same height,the side of the square ABCD.The triangles simply change their shape to suit the dimensions of the rectangle.ab always has the same value and in the case of a square a^2=ab with ab having the same value given the area of the triangles. Check this out; CE=6/a=1.285118972 AF=4/3*CE=1.713491962 FD=a-AF=2.955336475 ED=a-CE=3.383709465. Area triangle FED=FD*ED/2=5.000000001 w=a^2-12.000000001=9.797958973=4*sqrt6=9.797958971 error 0.000000002 The error is caused by calculator rounding and rounding surds.
wow thanks presh. Very nice. I solved it the hard way (ugly). Reduced the problem to 3 *(x^4) - 96*(x^2) +256 =0. where x = base of triangle with A=4. With other information given , The side of the square = 8/x. Setting u = x^2. we can solve for x = sqrrt ( (48 +/- 16*sqrrt(6)/3). Side of the Square = 8 * sqrrt (3 / (48+/- 16 * sqrrt(6) ) ) . Area of the square = 12 /(3 - sqrrt(6)) . We can ignore the solution for 48 + 16* sqrrt (6) as it will result in a negative area for the 'W' in your example. W = 12 * (sqrrt(6) - 2) / (3 - sqrrt(6)) ~= 9.8
Using the notation of Presh, x=blue=3,y=green=4 and z=yellow=5 c=2y/a and d=2x/b. a=width,b= length. Area ABCD=ab= 2x+2y+2z-cd=6+8+10-4xy/ab=24-48/ab (the sum of the blue and green areas overlap by the area of rectangle cd hence the -cd) Therefore, ab=24-48/ab...………...……………….(1) Area ABCD=ab=x+y+z+w=3+4+5+w=12+w Therefore, ab=12+w …………………………………..(2) Substitute (2) into (1) 12+w=24- 48/(12+w) (12+w)^2-24(12+w)+48=0 144+w^2+24w-288-24w+48=0 -144+w^2+48=0 -96+w^2=0 w^2=96=16*6 w=sqrt(16*6)=4*sqrt(6) square units as required. To find ab, ab=12+w from (2) ab=24-48/ab from (1) =24-48/(12+w) from (2) =24-48(12-w)/{(12+w)(12-w)}=24-48(12-w)/(12-w^2)=24-48(12-w)/(144-96) =24-(12-w)=12+w,therefore the answer checks out. ab=12+w from (2) =12+4sqrt(6) square units. If you work out ab from the quadratic in (1) you get ab=12+4sqrt(6) square units. In the case of a square, a=b and therefore, a^2=12+4sqrt(6) Giving a=sqrt{12+4sqrt(6)} >>>>>> 4.668828437
We can look for integer triplets x,y,z giving integer w. First solutions by increasing x+y are 2,2,1 ; 2,3,2 ; 2,4,3 ; 2,5,4 ; 3,4,1 ; 3,4,6 ; 2,6,5 ...
People that allow themselves to be ruled by a communist dictatorship that harvests organs from their political opposition and uses terrorism against their own population can't be all that clever......
@@ansalan7734 China is certainly into manipulating the figures. Your list is a country's average over the population as a whole, China's 1.4 odd billion population wasn't tested overall, they would have restricted areas of the country to the tests and only allowed the tests to be done to a preselected portion of the country only in order to make it seem China's scores are higher on average than they in fact are in order to 'save face' before the world and to deceive the world into believing they have the best system. A computer glitch here, a few thousand lost test results there. How do I know this? It's what they do with everything.........lie.
Let the side of square to be y and the short side of triangle with area of 4 to be 4x with xy=2. The area to be calculated is y^2-12: We have (y-4x)*(y-3x)=10 => y^2 - 7xy +12x^2=10 => y^4 -24y^2 + 48 =0 => y^4 -24y^2 +144 = 96 Area = y^2-12 = 4 sqrt(6)
There is a much simpler way. Let the side of the square be a. Then the square area A = a^2. Let the short side of the 3 triangle be b Let the short side of the 4 triangle be c Then ac= 8 (1) and ab = 6 (2). Finally (a-b)(a-c) = 10. (3) Multiply out (3) and substitute (1) and (2) so we get a^2 +bc = 24 (4) MULTIPLYING (1) and (2) gives a^2bc = 12 or bc = 12/a^2 (5) Substituting (5) in (4) gives a^2 + 48/a^2 = 24. Or A + 48/A = 24. Multiply this by A to get the quadratic for the square's area: A^2 - 24A + 48 = 0. Soving the quadratic gives A = 12 + 4√6. Subtracting the known areas (3+4+5) gives your answer.
I had the same answer, but derived it differently. I solved these 4 equations: (1) a(c+d)=8 --> a = 8/(c+d) (2) b*c=10 --> b = 10/c (3) d(a+b)=6 --> 6 + 5*d/c = 3*c/d (substituting a and b) and then solving for d/c using the quadratic equation. (4) b(c+d) -4 = w --> 6 + 10*d/c = w solving (3) gives d/c = (-6 + sqrt(96)) / 10 Substituting d/c in (4) results in w = sqrt(96)
2:26 The area of the rectangle is larger than the area of the triangle, so we keep the positive and discard the negative. Understood, but the negative also solves the equation. What would the triangle look like if we went with the minus sign? Could this even be drawn?
Let CE=a, AF=b, square side length = L. Then aL=3/2, bL=2, (L-a)(L-b)=10. Substitute for a, b , and letting L^2 = u produces: u^2 -24u +48 = 0. Solving the quadratic: u= 12 + or - 4sqrt(6). The required area is A = L^2-12 = u-12 = 4sqrt(6), dropping the -ve root. The more general rectangle case with sides: L, L' follows the same kind of way, only there L*L' = u in the same quadratic.
Also; 5 = (1/2) (a - 8/b )(b - 6/a ) ; simplifying, (ab)^2 - 24(ab) + 48 = 0 , But, (ab) is the area of rectangle, (A). Then substitute, A^2 - 24A + 48 = 0, by Quadratic formula, we get A = 21.80 Thus, the remaining area of triangle (w) is, A = 3 + 4 + 5 + w = 21.80 w = 21.80 - 12 = 9.80 sq. unit. Ans.
I want to ask a geometric problem, which has made me confused for a long time. In a triangle ABC, AB=AC, point D is at AB, then angle ACD=angle BCD, and AD+DC=BC.Find angle ABC My English is not very well, so I don't know if you can understand what I mean. I have solve the problem with sin,cos but I want to know a better solution. I can tell you angle ABC equals 40°. I hope you can see this comment.
There's actually another solution, but since there is a formula for this kind of questions, this way will actually be quite inefficient. Assump the area of the rectangle as x. First add all the triangle areas, and then ×2, and we have 24. There will be an extra area A which is consisted of the two short sides of the two triangles (3 and4) that had the full length of the the rectangle's sides as one of the sides. Then multiple the two triangles mentioned previously, which is 3×4=12. 12 actually contains x and A Now we can write a function which is 48÷A=24-A -> A²-24A+48=0 (24±√24²-4×1×48)2×1 (24±8√6)/2=12±4√6 Since the area of the rectangle must be bigger than 12, 12+4√6 will not be a considerable answer for A Therefore A=12-4√6 The area of the rectangle will be 24-(12-4√4)=12+4√6 The area of triangle in the middle will be (12+4√6)-3-4-5=4√6
We can skip the quadratic formula by rearranging the process. Let me name s=x+y+z, a=AD, a′=AF, b=CD, b′=CE. We have (1) ab=s+w (2) ab'=2x, a'b=2y (3) 2z=(a-a')(b-b')=ab-ab'-a'b+a'b'=s+w-2x-2y+a'b' , resulting in (4) a'b'=s-w Now we have 2 expressions for aa'bb' : from (2) : aa'bb'=4xy and from (1) and (4) : aa'bb'=s²-w² Hence w²=s²-4xy, and the conclusion since w>=0.
Indeed he does. This kind of mistake is often what prompts him to upload corrected videos. It's minor but he strives to be correct when it comes to the math.
Array of Squer is x^2 (x-8/x)(x-6/x)=10 . After substitution a=x^2 we have x^2=12± 4 sqrt6, there Is ONLY solution x^2= 12+4 sqrt6, we have the same result 4 sqrt6 for unknow array of triangle.
Lol I solved a lot of these in a 5xxx level university geometry course. We also did Inscribed Circles in triangles. Later I realized I didn’t need to take that class since it was mostly for people who were going to teach math, not applied mathematics. I enjoyed differential geometry much more than geometry.
Very nice problem! Please note that from an algebraic point of view, one should point out that \sqrt{(x+y+z)^2-4xy}>=0. Namely, by inserting 0=-(x+y)^2+(x+y)^2 under the root, one has \sqrt{(x+y+z)^2-4xy}=\sqrt{(x+y+z)^2-(x+y)^2+(x-y)^2} (1) since (x-y)^2=(x+y)^2-4xy holds. Now the right hand side of (1) is clearly nonnegative, hence we're all safe ;-). Of course, from a geometric point of view, this problem does not occur since, as Talwalkar shows, w=\sqrt{(x+y+z)^2-4xy}, so the left hand side is nonnegative by definition.
The area of the triangle that is to be determined equals sqrt((A+B+C)^2-4AB) with A, B and C the areas of the three given triangles and C the triangle in the bottom-right corner of the rectangle.
Here's the complimentary solution using side lengths that exposes the underlying geometry. The key is to factor (ab/2) out of everything. From the problem: ab=x+y+z+w, a=a_x+a_z, b=b_y+b_z x=(1/2)a_x*b, y=(1/2)b_y*a, z=(1/2)a_z*b_z =(1/2)(a-a_x)(b-b_y) =ab/2+2xy/ab-x-y Let's factor the general area of a triangle and solve for w. x=(ab/2)(a_x/a) y=(ab/2)(b_y/b) z=(ab/2)(1+(a_x/a)(b_y/b)-(a_x/a)-(b_y/b)) Notice how z cancels the area contribution from x and y. Now we can solve for w: ab=x+y+z+w=(ab/2)(1+(a_x/a)(b_y/b)+w(2/ab)) (ab/2)(1-(a_x/a)(b_y/b))=w I find the underlying ratios and difference between product and sums of these ratios fascinating. Thanks Presh for the great video 😁🙏🙋♂️
Easier solution without (complete) quadratic equation: a*b=x+y+z+w b*d=2*x (I) a*c=2*y (II) (a-d)*(b-c)=2*z (III) from the last equation one gets a*b-a*c-b*d+c*d=2*z where the product c*d is unknown. From (I)*(II) one gets a*b*c*d=4*x*y or c*d=4*x*y/(a*b). Substituting back in (III) this and the other products: x+y+z+w-2*y-2*x+4*x*y/(x+y+z+w)=2*z or w-x-y-z+4*x*y/(x+y+z+w)=0 then (w-x-y-z)*(x+y+z+w)+4*x*y and finally w^2-(x+y+z)^2+4*x*y=0 since the first two terms in parents simplify to a notable product. From this last equation we get the final answer without baskara...
Got this one! I defined the left side as Y and the top side as X, this gave me 3 formulas using triangles (8 = ay, 6 = bx, 10 = (x-a)*(y-b)). Plugging into the last eqn, I got: 0 = (xy)^2 - 24(xy) + 48. Since xy is the area of the square, I don't need to independently solve X and Y, I instead just solved for the total area which is 12 + 4sqrt(6) and then subtracted the three triangles to get the answer.
@@naturelover4148 May be you solved it really, but I have noticed that in other videos where Presh has given very complicated problem, so many Indians say that they have solved the problem within a minute or simply say we have done those problems when we were studying in 4th or 5th grades.
@@sabyasachirimpa Calm down man....you are showing your ego by posting such comments....Its not about Indians, it's about individuals....I mean ego is not associated with a nation but individuals....
Almost the same solution, but a bit more visual: Flip the three triangles on the long sides. Then the complete rectangle a*b is filled one time - except the upper left rectangle c*d, which is filled two times. So ab+cd=2*(3+4+5)=24 Because of (ac)/2=4 and (bd)/2=3 we have ac*cd=ab*cd=48. Calling p=ab and q=cd you have to solve the eqations p+q=24, p*q=48 which leads to your result.
@@leif1075 Draw two lines into the given figure: a vertical line through the touching point of the green an yellow triangles and a horizontal line through the touching point of the yellow and blue triangle. This way you split the given rectangle into four smaller rectangles. For example the lower right one is just the yellow triangle doubled. If you double the green and blue triangles in the same way, the whole figure is covered, but the upper left rectangle is covered two times. Using the notation of the video the whole figure a*b plus the upper left rectangle c*d must be twice the area of the area of the tree triangles. So ab+cd=24. To compute c*d multiply the areas of the green triangle ac/2=4 and the blue one bd/2=3 to get ac*bd=48 or ab*cd=48. Rename p:=ab, q:=cd we get p+q=24 and p*q=48.
Staying in a square, x(x-y)=8 [Eq1]; yz=10 [Eq2]; x(x-z)=6 [Eq3]; x^2-12=A [Eq4]; Putting Eq1, Eq3 and Eq4 in Eq2 we get (A+4)*(A+6)=10A+120 leading to A^2=96=16*6 so A=4*Sqrt(6).
By the given formula the answers may vary depending upon which triangle we take as z, because (x+y+z)is getting squared but -4xy is subtracted there is no variable z. As a result answers vary accordingly
@Adam Romanov okay so while computing the values to the equation we have to keep in mind that triangle x and y should b the triangle sharing the common vertex. 🙂
If we draw a horizontal line through "E" and another vertical through "F", the original rectangle is divided into four rectangular cells that, ordered from left to right and from top to bottom, have areas of value (a), (6- a), (8-a) and (10). Since the relationship between cells in the same column or row is constant, the following equality is verified: a/(8-a)=(6-a)/10 ⇒ a²-24a+48=0 ⇒ a=2.2 ⇒ (6 -a)=3.8 and (8-a)=5.8 ⇒ The total area of rectangle ABCD is: 2.2+3.8+5.8+10=21.8 ⇒ The area of triangle FBE = 21.8-3-4-5=9.8
Great, now engineers will start using the area of that triangle for the gravitational constant.
Edit: guys, I get it I mixed up the gravitational constant and average gravitational acceleration on earths surface. I'm sorry, I was very tired when I posted this
:))yep i was surprised that i got 9.8 as a result even tho i went with the classic x,l-x aproach and then equations to calculate l
But nevertheless since i study engeenring,i therefore am allowed to use the triangle as g:)))
Ps:pi squared=g
Uhm, the gravitational constant is about 6.674 * 10^(-11) m³/(kg*s²)
en.wikipedia.org/wiki/Gravitational_constant
Too me that looks like you confused "Big G" (the gravitational contstant) with "little g" (the gravitational acceleration at the surface of earth). The 9.81 m/s² is not really a constant. It differs quite a bit even at the surface of earth. Further more it of course decreases the further away from earth you get. The gravitational acceleration can be calculated by g = G * M / r² (where "r" is the distance from the earth center and M is the mass of earth)
Acceleration due to gravity, and gravitational constant are two different things.
@@Bunny99s trueee. Gravitational constant is different from acceleration due to gravity
@@therandomnomad435 truth
Tip: It is just about seeing the sides, and naming them
Exactly
that's it,
i love it when geometry combines with algebra
Thanos Becomes Darkseid name one time that geometry didn’t get involved with algebra
@@inakibolivar664 limits
Omer Keskin how are limits not related to algebra
There's a simpler and more elegant solution. Note that the rectangle's area (A), is comprised of 2 pieces each of the smaller triangles MINUS the upper-left smaller rectangle (B) formed from the basis of the 3 and 4- triangle. More formally, using Presh's notation:
A = 2*(3+4+5)-B.
ac/2*bd/2=3*4 -> B:=cd = 48/(a*b) = 48/A
So A = 24 - 48/A, with solution A=4(3+sqrt(6)). Subtracting the smaller triangles yields the answer *4*sqrt(6)*
Presh was just showing us the more general one, he wasn't only trying to show us the square solution but also the rectangle one
@@kylesheng2365 My solution is general. Just replace 24 with 2(x+y+z) and 48 with 4xy.
Where did 48 come from
@Everybody Nobody Its actually simpler and more intuitive imo. I found the same solution
@Everybody Nobody solving a second-degree equation is straight forward and not covered here. My solution is simpler because it introduces fewer unknowns.
my method for intermediate learners:
let:
A= area of rectangle
x= width of rectangle
y= height of rectangle
a= height of triangle with area=3
b= base of triangle with area=4
...
eq1. ax/2 = 3 hence; ax =6
eq2. by/2 = 4 hence; by =8
eq3. (y-a)(x-b)/2 = 5 hence, xy-by-ax+ab = 10
.
take eq.1 and eq.2, multiply to get:
abxy = 48 hence ab=48/xy which is now eq.4
.
take eq 3. and substitute ax,by, and ab.
xy - 8 - 6 + 48/xy = 10.
since A = xy , substitute
A^2 - 24A + 48 = 0
you get, A= 2.2 and 21.8
we use 21.8 as the logical answer for the area of rectangle
.
area of middle triangle = A - 3 - 4 -5
which is 9.8
answer is 9.8.
.
This is one of my favorite videos you made. You took a nice problem and solved it in a simple yet satisfying way, showing the general case than applying the numbers. Cheers!
I want to know how tall is presh TALLwalkar
At least as tall as AT Walkers from Star Wars xD
Tall
I did not laugh 😏
For me, Presh Tallwalker is nearly as tall as Luke Skywalker.
Kyon.. Shadi ke liye rishta bhejna hai kya?
Just have to be careful what areas you label as "x" and "y" since u can rotate the rectangle, esp if its a square. You need x and y to be the areas of the 2 triangles with a side of the rectangle as a leg
Good point, you need a way to distinctively identify the quantity 4xy, otherwise the result would change.
I noticed the same thing, and I was about to write a comment to alert the author of this video. Thumbs up!
he labeled it in the diagram. the reader should be aware of the cyclic asymmetry
Thank you. The triangles are not all drawn the same. x and y have a whole side length, but z does not. I was wondering what would keep the formula from working with any order of the triangles, and that's why. Thank you. Z is the unique one here.
Thank you for this comment this had helped me...😊
for a less complicated quadratic equation :
Notice that ab is the area of the rectangle. Therefore ab=x+y+z+w. I also called "t" the quantity x+y+z. If you do this changes in the quadratic equation at 2:58, it becomes less complicated :
(t+w)² - 2t(t+w) + 4xy = 0
and then, by using distributivity
w² - t² + 4xy = 0
that leads directly to w = sqrt(t²-4xy) = sqrt((x+y+z)²-4x)
I can't solve any of the questions, but these videos are enjoyable to watch
Same lol
I could solve only one
*Mertcan Bal* Why are you unable to solve any questions?
At time 0:19,
Let AF=d and CE= 3d/4 since the areas of triangle BAF=4 and triangle BCE =3.
Let side of square ABCD =x and area of square is therefore x^2
Consider triangle BAF,
4= xd/2
d=8/x...............(1)
Consider triangle FDE,
5=(x-d)(x-3d/4)/2............(2)
Substitute d=8/x into (2),
We end up with 10=x^2-14+48/x^2
Let y=x^2,
0=y-24+48/y
Multiply through by y,
y^2-24y+48=0
y= 12+4sqrt(6) (-ve sqrt rejected in quadratic formula)
Area of square ABCD =x^2 = y =12+4sqrt(6),
Area of triangle BEF= area of ABCD-area of ABF-area of BCE-area of FED
=(12+4sqrt(6)}4-3-5
=12+4sqrt(6)-12
= 4sqrt(6) square units as required.
Note the -ve value of the sqrt in the quadratic formula was rejected because the area of BEF would have been -ve.
I had a little different approach. I called AB and BC as x. Then using a system of equations, I expressed FD and ED in terms of x and plugged those into the equation for the area of FDE. This ended up giving me x^4-24x^2+48=0. Solve for x then just calculate the area of the square and subtract the three triangles. Same answer in the end.
Got it by the same way .
Even easier to solve for x2 instead of x, because that is the area of the entire square. You just don't need x at all.
This is what I was trying to do, but went wrong somewhere. Was a bit annoyed that the video went straight to demonstrating the general case instead of the specific one so I wasn't able to figure out my mistake! Thanks for explaining how you did it, I managed to get there in the end.
At time about 1:53, one can view the area (ab) of the big rectangle as the sum of three rectangles' areas (2x, 2y, 2z) minus the overlapped rectangle's area [ (2x/b)(2y/a)]. Therefore, ab = (2x+2y+2z)-(2x/b)(2y/a).
Hmm, I kinda brute-forced this one. I wrote 5 equations to calculate the length of he side first then subtracted the areas of the triangles. But your approach is more elegant.
Post it
The 5 equations model gives something quite amusing because at the end, you should have something like (l²-8)(l²-6)=10l² (l => length of the square)
But, l² = Area of the square !
So (A(square)-8)(A(square)-6)=10A(square) A(square)=12+4sqroot(6).
Yep, I did the same, but I messed up with arithmetic >_< and got the wrong answer :)
Please post your solution about more than 172 people wants see your idea to solve the problem
Same it took me a page and 10 minutes to do it
Looking at the thumbnail, I thought you could use Heron's Formula to find the area:
s = (3+4+5)/2 = 6
s-a = 6-5 = 1
s-b = 6-4 = 2
s-c = 6-3 = 3
Area = √s(s-a)(s-b)(s-c) = √6*1*2*3 = √36 = 6
But this doesn't work because the AREA's of the outer triangles are 3, 4, 5. They're not the sides of the outer triangles.
My solution also started with denoting the sides and coming up with systems that will end up being quadratic in form. I’m glad I got another ome of these with almost the same approach you did.
Problem:
Resistors 1 and 2 in parallel yields total current of 2A
Resistors 1 and 3 in parallel yields total current of 3A
Resistors 2 and 3 in parallel yields total current of 4A
The power supply is 12V
Calculate:
a. Total current when all 3 resistors are connected in parallel
b. Total current when all 3 resistors are connected in series
Same problem:
a and b can do a work in 2 hours
a and c can do the same work in 3 hours
b and c can do that work 4 hours then how long it would take if a, b and c all work together.
Ans: in 2 hours a,b completes 1
in 1 hour a,b complete 1/2 of the work
Same way in 1 hour a,c completes 1/3
b,c complete 1/4 of the work
in 1 hour a+b+a+c+b+c can do 1/2+1/3+1/4 of the work
= 13/12
in 1 hour 2a+2b+2c can do 13/12
in 1 hour a+b+c can do 13/24
=>the a+b+c completes whole work In 24/13 hours.
Good generalization. For specific case, I used the constraints and set one of short bases as a fraction of length of square. That simplified to a quadratic homogeneous equation of the ratio.
Rest of it is algebra, getting the expected result.
here is my solution to the problem with the square. Just a bunch of equations that rely on "area of a triangle" and "area of the square"-function
a = Distance from A to F
b = Distance from C to E
c = length of each edge of the square
x = area of triangle
3 = 1/2 * c * b;
4 = 1/2 * c * a;
5 = 1/2 *(c-b) * (c-a);
x = c² - 3 - 4 - 5
This easy approach does only work with the square and not with the more general rectangle
I'm happy that I managed to solve this one orally without much effort, although took 15 odd minutes to correct my earlier equation. Honestly it's a very easy problem as other lengths can be easily obtained after assuming square side length as say 'x'.
Solving the equation
x^4 - 24 (x^2) + 48 = 0,
x^2 = 12 + or - 4 √6
Obviously area sum of 3 triangles is 12.
So x^2 will be greater than 12.
Hence,
Required area
= 4 √6 square units
Elegant solution. 4:57 I did it differently and got the same answer, 9.798. I ended up with A^4 -24A^2 +48=0 (A raised to the power of 4 minus 24A raised to the power of 2 plus 48=0) and factoring for A get the length of the square as 4.668, and so the AREA square is then 21.798 and thus by subtracting the TOTAL length of the three triangles 12 the result is therefore 21.798-12=9.798.
To show how I arrived at A^4 -24A^2 +48=0
Let A = the length of the square, let B= the base of the triangle with area 3, and let C the height of a triangle with area 5.
Then (1) A x B=6 or B= A/6 since the area of triangle = 1/2 base x height
(2) (A-B)x C=10 C=10/A-B or C= 10/A-6/A or AC-AB =10
(3) (A-C) x A=8 or A^2 - AC=8
(4) substituting equation (1) into (2)= (A-6/A)x C=10 or AC-6/AC=10
Adding equations (3) and (4) = A^2-6/AC=18
(5) substituting equation (1) into equation (3) = A^2-A(A/6)= 18
........ ....
I ended up ( as said above) with A^4 -24A^2 +48=0 and solving for A resulted in 4.668
What is the length of the square? Therefore, the area is 4.668 x 4.668 or 21.798 and 12 (the sum of the three triangles) from this yields 9.798. Answer: let's plugin 4.668 in the above equation to check. 4.668^4- 24 x 21.798 +48=0
475.152-523.152+48=0 or 523.152-523.152=0
So, yes, 4.666 satisfies this equation.
1
I found it out similarly but It's much easier if you apply a trick: when you get to the equation A^4 -24A^2 +48=0 just simply introduce a new "'x" which is a^2. This way you get that x is 12+4sqrt6 but the trick is that this way x is the area of the entire square so you just deduct 3+4+5 from it and you get 4sqrt6.
@@hhgygy, Thanks for the trick I will use it next time.
This video make so much sense and meaning and quadratic equation is true I can understand so much from it. Positive thank you. Lv u.
AB=BC=a，AF=b，FD=c，DE=d，EC=e
b+c=d+e=a
1/2ab=4
1/2cd=5
1/2ae=3
五元一次方程算出a
中間的面積=aa-12
完成了
Dear Presh, thank you for the video.It is highly informative.And its nice to come across to one of my friends Mr Ince as a reference. Have a nice day.
suppose
square's lenght = a
triangle(area = 4)'s base = b
triangle(area=3)'s height = x
½.a.b = 4
ab = 8
b = 8/a
½.a.x = 3
ax = 6
x = 6/a
last triangle's base = a-b
and its height = a-x
its area = 5
½(a-x)(a-b) = 5
(a-x)(a-b) = 10
a² - ab - ax + bx = 10
a² - (8) - (6) + (8/a)(6/a) = 10
a² - 14 - 10 + 48/a² = 0
a² - 24 + 48/a² = 0
a²(a² - 24 + 48/a²) = a²(0)
a⁴ - 24a² + 48 = 0
substitute t = a²
a²(a²) - 24a² + 48 = 0
(t)(t) - 24(t) + 48 = 0
t² - 24t + 48 = 0
*here, i'm taking the positive one,*
*i mean x1 = (-b + sqrt{b²-4ac})/2a*
*, otherwise the square's area would be less than 12*
*and if it equaled to less than 12, the unknown area would be negative*
*why? Because i substitute*
t = a²
*which the square's area = a²*
t=(-(-24) + sqrt{ (-24)² - 4(1)(48) } )/2(1)
= ( 24 + sqrt{ 576-192 } )/2
= (24 + sqrt{384} )/2
= (24 + sqrt{8²×6} )/2
= (24 + 8 sqrt{6} )/2
= ( 2(12 + 4 sqrt(6) )/2
= 12 + 4 sqrt{6}
t = a²
a² = t
= 12 + 4sqrt{6}
Unknown area
= square's area - 3 - 4 - 5
= a² - 12
= (12 + 4sqrt{6}) - 12
= 4 sqrt{6}
And how did you solve the general case?
@@leif1075 i beg your pardon?
@@spiderjerusalem4009 the case of the rectangle with x y and z areas.
I tried for more than hour. Never thought of generalizing. And finding that it was a quadratic equation was genius
It took me awhile to figure out it was a quadratic equation, A^4 -24x^2 +48x=0
Very impressive, many thanks! 😊 Another way solving the problem: divide the rectangle (ab) into four rectangled parts: xy, y(b - x), x(a - y), (a - y)(b - x).
xy = ?
ax = 8 = xy + x(a - y)
by = 6 = xy + y(b - x)
(a - y)(b - x) = 10 = ab + xy - 14 = ab +(8/a)(6/b) - 14
→ 24 = ab +(48/ab)
substitute: k = ab
k² - 24k + 48 = 0
k1 = 4(3 + √6) = ab
k2= 4(3 - √6) ≠ solution, not big enough - re-substitute: ab = k
w = ab - 12 = 4√6 ≈ 9,8 🙂
You can easily see: xy = 24 - ab = 4(3 - √6)
Interesting:
xy = 12 - w (not obvious)
ab = 12 + w (obvious)
(w/ab) = (√6 - 2)
(w/xy) = (√6 + 2)
(ab/xy) = (5 + 2√6)
Really amazing Mr Presh, my regards sir.
You are a true gentleman Mr. Jalal Alarabi!
@@barabasbulcsu3102 No, you both are true gentlemen, Mr. Barabás Bulcsú and Mr. Jalal Alarabi
Simpler solution:
Define base b from left to right and height h from top to bottom
b = b1 + b2
h = h1 + h2
Rectangle areas are twice the given triangle areas
b.h1 = 2.3 = 6
b1.h = 2.4 = 8
b2.h2 = 2.5 = 10
Define areas
A = b.h
Q = b1.h1
S = (b.h1 +b1.h + b2.h2) / 2 = 12
W = A - S
Product (b.h1)(b1.h) = 6.8 = 48 = AQ
Solve for W in terms of known S and AQ.
Add the three rectangles (2x, 2y and 2z in your scheme) and notice a double counting of Q = b1.h1. Thus the key insight by inspection...
2S = A + Q
Subtract S from both sides
S = W + Q (or W = S - Q)
Add the definition of W
2W = A - Q
Subtract the squares of the 2S and 2W equations, and only the cross terms remain,
4S^2 - 4W^2 = 4AQ
Divide by 4 and rearrange
W^2 = S^2 - AQ = 12^2 - 48 = 96
W = 4 Sqrt(6)
For completeness
A = S + W = 12 + 4 Sqrt(6)
Q = S - W = 12 - 4 Sqrt(6)
FYI, these solutions also result from the quadratic equation obtained by multiplying the 2S equation above by A, and rearranging...
0 = A^2 - 2SA + AQ.
As soon as you said “multiply both sides by ab” I saw exactly where you were going and damn that was beautiful
First case:
The area we're looking for = A
Square sides: x:
one formed from x and x-a
The other from x and x-b
ax=2×3=6
bx=8
(x-a)(x-b)=10
==> x²+ab=24 means x²
0:38 Mustafa Kemal İnce'ye teşekkürler, izleyen türklerle selam olsun.
Evet duyunca bir farklı hissettirdi hocam :)
çok teşekkürler
Simply putting heron's formula will also work s=(a+b+c)/2
Area = [s(s-a)(s-b)(s-c)]^0.5
a,b,c are sides of triangle
Well..that's one way to start a morning 😄
At 0:20 use Heron's Formula :
Which states Area of a triangle = √{S(S-A)(S-B)(S-C)}, where S is the semi - perimeter of the riangle which is the sum of all all side divided by 2.
So, here S will be (3+4+5)/2=6
So, therefore area will be, √{6*(6-3)*(6-4)*(6*5)} = √6*3*2*1
Which can be written as √3*2*3*2 = (*Taking off the square root*) 3*2 = 6unit^2
Love your work sir
Love from India
Thanks from you tube channel I like and love Maths and others.
Let x be the side of the sqr be x
So the other side of triangle ( that area is 4 ) be 8/x
The other side of triangle (that area is 3) be 6/x
So the area of third triangle 1/2(x-8/x)(x-6/x)=5
=>X^2=12+-4✓6
Hence W=x^2-(3+4+5)
=12+4✓6-12=4✓6 (ngtv vlu not acceptable)
I listed a few equations and used some substitution to solve this.
PS: I have seen a question features the same graph but asks a different thing
At time 3:47 minute in this video, inside the square root , it will be 16xy instead of 4xy. Although it is very minor error. Thank you.
Thanks for pointing out! There was another comment talking about using Heron's formula which i kind agree but the final answers are different and now i know why, due to the error you have mentioned
Mustafa Kemal İnce the best mathematician. 👏👏👏
Let
s = side length of the outer square
= a + b, the side lengths
= c + d, the bottom lengths
Then rectangles
sa = 2*3 = 6
sc = 2*4 = 8
bd = 2*5 = 10
Sum and product
2S = sa + sc + bd = s^2 + ac (S = 3+4+5 = 12)
P^2 = sa*sc = s^2 * ac (= 6*8 = 48)
Desired area
W = s^2 - S
2W = s^2 - ac
W^2 = S^2 - P^2 = 12^2 - 48 = 96
W = 4 Rt(6)
For completeness
s^2 = S+W = 12 + 4 Rt(6)
ac = S-W = 12 - 4 Rt(6)
ad = sa-ac = -6 + 4 Rt(6)
bc = sc-ac = -4 + 4 Rt(6)
sb = bc+bd = 6 + 4 Rt(6)
sd = ad+bd = 4 + 4 Rt(6)
Get a, b, c, d by dividing sa, sb, sc, and sd by s.
Problem:(exists)
Mr. Math Man: Δ
تمرين جميل. وشرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين
Not gonna lie, that was some dope stuff yo!
2:08 , if you still in the square situation (a^2 and not ab) without formula, you can say that a^2 = w + x + y + z = w + 12, so you will get after the distribution and simplication : w - 12 + 48/w + 12 = 0
(w - 12)(w + 12) + 48 = 0
w^2 = 96
w = sqrt(96) = 4sqrt(6)
4:52 SUCH A SIMPLE FORMULA ?!? *laughs in equation*
Elegant solution for the general case of a rectangle. There is an easier approach for this example of a square, however. Setting X as the length of one side of the square, using the triangle formula 1/2(b)(h) and the given areas one finds X^4 - 24X^2 = -48 after some basic algebra. Complete the square and find X^2 (the total area of the square) is 12 + 4(6^1/2). Subtract the given areas 3 + 4 + 5 and this also solves w as 4(6^1/2). Thank you for the problem!
Should not assume it is a square. For versatility, set X for one side of the rectangle and Y for the perpendicular side. If X =Y then it is a square otherwise it is a rectangle. One will realise it does not matter if it is a square or rectangle, the key thing is X.Y = overall area of the square or rectangle and > (3+4+5 = 12).
That feeling when you approached it differently and get the same answer :)
There is a much simpler way without having to get to quadratic equation.
Complete parameterization:
a x b is the area of rect.
ab = 12 + z (assume area of inner triangle/inscribed wrong word as no circle)
Area of ∆ = 1/2 X base X height
∆1 = a x X = 8
∆2 = b x Y = 6
∆3 = (a-Y)(b-x)=10
Now expand ∆3 and solve collect all variables in one side,
We have:
XY + z = 10+8+6-12 »12
z = 12-XY----------------(main equation)
Now find XY
X = 8/a
Y=6/b
Therefore XY = 48/ab
But we know ab = 12+z
So XY = 48/(12+z)
Now sub this value in main equation
z = 12 - (48/(12+z))
Upon rearranging we have
z = 4√6
Area of inner triangle is 4√6 sq units
This can be extended to parallelograms with areas S1 S2 S3.
Let x, a and b be side of the square, bases of right-angled triangle-(4) and right-angled triangle-(3) respectively. My solution with simplified steps: xa = 8, xb = 6 leading to ab = 48/x^2, (x - a)(x - b) = 10; combining the latter two gives x^4 - 24x^2 + 144 = 96 or x^2 - 12 = 4√6, i.e. area of triangle BEF = 4√6.
Hay I first found the side length of square x(a)= 4.67 by solving polynomial quadratic equation x^4-24x^2+48=0 (actually simple
but also got √2.28 as an extra answer)
Именно так сторону квадрата за х, основание треугольника 4 за у, основание треугольника 3 за z, выписать для всех трёх треугольников площадь вот и система, выразить все через х и вот вам уравнение см выше....оно решается и ответ....
I found root 12
Dont ask how i found it
a=4.668828437 units,AF=c=1.713491962 units,CE=d=1.285118972 units,FD=2.955336475 units and ED=3.383709465 units if ABCD was a square.All the other numbers w,x,y and z would all be the same.Note that d/c=0.75 because if they are 3/4 in area the bases would have the same ratio since they have the same height,the side of the square ABCD.The triangles simply change their shape to suit the dimensions of the rectangle.ab always has the same value and in the case of a square a^2=ab with ab having the same value given the area of the triangles.
Check this out;
CE=6/a=1.285118972
AF=4/3*CE=1.713491962
FD=a-AF=2.955336475
ED=a-CE=3.383709465.
Area triangle FED=FD*ED/2=5.000000001
w=a^2-12.000000001=9.797958973=4*sqrt6=9.797958971 error 0.000000002
The error is caused by calculator rounding and rounding surds.
After watching the whole video 5 times : this question is so easy
Mustafa kemal ince aslanim benim
Ah yes, of course, quadratic equations! These must've slipped my mind.
wow thanks presh. Very nice. I solved it the hard way (ugly). Reduced the problem to 3 *(x^4) - 96*(x^2) +256 =0. where x = base of triangle with A=4. With other information given , The side of the square = 8/x. Setting u = x^2. we can solve for x = sqrrt ( (48 +/- 16*sqrrt(6)/3). Side of the Square = 8 * sqrrt (3 / (48+/- 16 * sqrrt(6) ) ) . Area of the square = 12 /(3 - sqrrt(6)) . We can ignore the solution for 48 + 16* sqrrt (6) as it will result in a negative area for the 'W' in your example. W = 12 * (sqrrt(6) - 2) / (3 - sqrrt(6)) ~= 9.8
It's easy to solve,but it's hard to imagine.
exactly 👌✊
Yeah
Yea
Yessir
Of course it's easy to solve for you, your name is in chinese 🤣🤣😂
Using the notation of Presh,
x=blue=3,y=green=4 and z=yellow=5
c=2y/a and d=2x/b. a=width,b= length.
Area ABCD=ab= 2x+2y+2z-cd=6+8+10-4xy/ab=24-48/ab
(the sum of the blue and green areas overlap by the area of rectangle cd hence the -cd)
Therefore, ab=24-48/ab...………...……………….(1)
Area ABCD=ab=x+y+z+w=3+4+5+w=12+w
Therefore, ab=12+w …………………………………..(2)
Substitute (2) into (1)
12+w=24- 48/(12+w)
(12+w)^2-24(12+w)+48=0
144+w^2+24w-288-24w+48=0
-144+w^2+48=0
-96+w^2=0
w^2=96=16*6
w=sqrt(16*6)=4*sqrt(6) square units as required.
To find ab,
ab=12+w from (2)
ab=24-48/ab from (1)
=24-48/(12+w) from (2)
=24-48(12-w)/{(12+w)(12-w)}=24-48(12-w)/(12-w^2)=24-48(12-w)/(144-96)
=24-(12-w)=12+w,therefore the answer checks out.
ab=12+w from (2)
=12+4sqrt(6) square units.
If you work out ab from the quadratic in (1) you get ab=12+4sqrt(6) square units.
In the case of a square, a=b and therefore, a^2=12+4sqrt(6)
Giving a=sqrt{12+4sqrt(6)} >>>>>> 4.668828437
That's really an incredible explanation...just wow.
We can look for integer triplets x,y,z giving integer w. First solutions by increasing x+y are
2,2,1 ; 2,3,2 ; 2,4,3 ; 2,5,4 ; 3,4,1 ; 3,4,6 ; 2,6,5 ...
Presh to Albert Einstein : hey Einstein can you figure this out , this problem was given to new borns in China 😂😂
Einstein : 😥😵
The hills are alive with the sound of music doh
People that allow themselves to be ruled by a communist dictatorship that harvests organs from their political opposition and uses terrorism against their own population can't be all that clever......
@@crazyt1ger08 ua-cam.com/video/46KCbYe6gEA/v-deo.html
@@ansalan7734
China is certainly into manipulating the figures.
Your list is a country's average over the population as a whole, China's 1.4 odd billion population wasn't tested overall, they would have restricted areas of the country to the tests and only allowed the tests to be done to a preselected portion of the country only in order to make it seem China's scores are higher on average than they in fact are in order to 'save face' before the world and to deceive the world into believing they have the best system.
A computer glitch here, a few thousand lost test results there.
How do I know this?
It's what they do with everything.........lie.
@@crazyt1ger08 ua-cam.com/video/AL5Hjg30b_M/v-deo.html So how can you explain it when China emits the most carbon dioxide ?
Let the side of square to be y and the short side of triangle with area of 4 to be 4x with xy=2. The area to be calculated is y^2-12:
We have (y-4x)*(y-3x)=10 => y^2 - 7xy +12x^2=10 => y^4 -24y^2 + 48 =0 => y^4 -24y^2 +144 = 96
Area = y^2-12 = 4 sqrt(6)
Captions : I am Presh to occur 😂😂
The captions always play with his name😂😂
lmao
Hello, this is pressure cooker
Lmao
There is a much simpler way.
Let the side of the square be a.
Then the square area A = a^2.
Let the short side of the 3 triangle be b
Let the short side of the 4 triangle be c
Then ac= 8 (1) and ab = 6 (2).
Finally (a-b)(a-c) = 10. (3)
Multiply out (3) and substitute (1) and (2) so we get a^2 +bc = 24 (4)
MULTIPLYING (1) and (2) gives a^2bc = 12 or bc = 12/a^2 (5)
Substituting (5) in (4) gives a^2 + 48/a^2 = 24. Or A + 48/A = 24.
Multiply this by A to get the quadratic for the square's area:
A^2 - 24A + 48 = 0.
Soving the quadratic gives A = 12 + 4√6. Subtracting the known areas (3+4+5) gives your answer.
I had the particular solution. Thanks for the general one.
Hey i wanna know. Please.
I had the same answer, but derived it differently. I solved these 4 equations:
(1) a(c+d)=8 --> a = 8/(c+d)
(2) b*c=10 --> b = 10/c
(3) d(a+b)=6 --> 6 + 5*d/c = 3*c/d (substituting a and b) and then solving for d/c using the quadratic equation.
(4) b(c+d) -4 = w --> 6 + 10*d/c = w
solving (3) gives d/c = (-6 + sqrt(96)) / 10
Substituting d/c in (4) results in w = sqrt(96)
Mustafa Kemal İnce is a math genius 👏👏👏
2:26 The area of the rectangle is larger than the area of the triangle, so we keep the positive and discard the negative.
Understood, but the negative also solves the equation. What would the triangle look like if we went with the minus sign?
Could this even be drawn?
turkler like layinda sayimizi bilelim
I liked your comment not because I have any idea what you wrote but because Turkish is cool B-)
Ben hiçbir şey anlamadım
Ehem ehem izninizle,
AS BAYRAKLARI AS AS 🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷
@@Tiqerboy he wrote " Take like Turks and let's know how many we are"
Let CE=a, AF=b, square side length = L. Then aL=3/2, bL=2, (L-a)(L-b)=10. Substitute for a, b , and letting L^2 = u produces: u^2 -24u +48 = 0. Solving the quadratic: u= 12 + or - 4sqrt(6). The required area is A = L^2-12 = u-12 = 4sqrt(6), dropping the -ve root. The more general rectangle case with sides: L, L' follows the same kind of way, only there L*L' = u in the same quadratic.
When I found that the square area was 12+- sqrt(6) I thought I was going way off...
I found it in 6 years.
You do not need to solve quadratic equation here. Starting from 2:52
If S=ab and T=x+y+z then
S^2-2ST+4xy
(S-T)^2-T^2+4xy=0
w=S-T=\sqrt{T^2-4xy}.
Is there any other way to solve this? Like by finding the side of the square and then . Side squared-(3+4+5)
Rizon yup I did that. got the same answer
Also;
5 = (1/2) (a - 8/b )(b - 6/a ) ;
simplifying,
(ab)^2 - 24(ab) + 48 = 0 ,
But, (ab) is the area of rectangle, (A). Then substitute,
A^2 - 24A + 48 = 0,
by Quadratic formula, we get
A = 21.80
Thus, the remaining area of triangle (w) is,
A = 3 + 4 + 5 + w = 21.80
w = 21.80 - 12 = 9.80 sq. unit. Ans.
I want to ask a geometric problem, which has made me confused for a long time.
In a triangle ABC, AB=AC, point D is at AB, then angle ACD=angle BCD, and AD+DC=BC.Find angle ABC
My English is not very well, so I don't know if you can understand what I mean. I have solve the problem with sin,cos but I want to know a better solution. I can tell you angle ABC equals 40°. I hope you can see this comment.
Can you mention from where did you get this question??
There's actually another solution, but since there is a formula for this kind of questions, this way will actually be quite inefficient.
Assump the area of the rectangle as x.
First add all the triangle areas, and then ×2, and we have 24. There will be an extra area A which is consisted of the two short sides of the two triangles (3 and4) that had the full length of the the rectangle's sides as one of the sides.
Then multiple the two triangles mentioned previously, which is 3×4=12. 12 actually contains x and A
Now we can write a function which is
48÷A=24-A
-> A²-24A+48=0
(24±√24²-4×1×48)2×1
(24±8√6)/2=12±4√6
Since the area of the rectangle must be bigger than 12, 12+4√6 will not be a considerable answer for A
Therefore A=12-4√6
The area of the rectangle will be 24-(12-4√4)=12+4√6
The area of triangle in the middle will be (12+4√6)-3-4-5=4√6
Solved it in 1 min
Bahut tej ho maths me
How?
We can skip the quadratic formula by rearranging the process. Let me name s=x+y+z, a=AD, a′=AF, b=CD, b′=CE.
We have
(1) ab=s+w
(2) ab'=2x, a'b=2y
(3) 2z=(a-a')(b-b')=ab-ab'-a'b+a'b'=s+w-2x-2y+a'b' , resulting in
(4) a'b'=s-w
Now we have 2 expressions for aa'bb' :
from (2) : aa'bb'=4xy and from (1) and (4) : aa'bb'=s²-w²
Hence w²=s²-4xy, and the conclusion since w>=0.
2:20 you actually factor a -2, but hey, it's still understandable :)
Indeed he does. This kind of mistake is often what prompts him to upload corrected videos. It's minor but he strives to be correct when it comes to the math.
Not in my opinion. The - is not part of the term. So there's the minus, and then there's 2 times the terms.
@@kasperjoonatan6014 True
@@kasperjoonatan6014 but if you were to factor only a "2", the term would look like +2(-x-y-z)
Array of Squer is x^2
(x-8/x)(x-6/x)=10 . After substitution a=x^2 we have x^2=12± 4 sqrt6, there Is ONLY solution x^2= 12+4 sqrt6, we have the same result 4 sqrt6 for unknow array of triangle.
Lol I solved a lot of these in a 5xxx level university geometry course. We also did Inscribed Circles in triangles. Later I realized I didn’t need to take that class since it was mostly for people who were going to teach math, not applied mathematics. I enjoyed differential geometry much more than geometry.
Very nice problem! Please note that from an algebraic point of view, one should point out that \sqrt{(x+y+z)^2-4xy}>=0. Namely, by inserting 0=-(x+y)^2+(x+y)^2 under the root, one has
\sqrt{(x+y+z)^2-4xy}=\sqrt{(x+y+z)^2-(x+y)^2+(x-y)^2} (1)
since (x-y)^2=(x+y)^2-4xy
holds. Now the right hand side of (1) is clearly nonnegative, hence we're all safe ;-).
Of course, from a geometric point of view, this problem does not occur since, as Talwalkar shows, w=\sqrt{(x+y+z)^2-4xy}, so the left hand side is nonnegative by definition.
Well im impreshed
The area of the triangle that is to be determined equals sqrt((A+B+C)^2-4AB) with A, B and C the areas of the three given triangles and C the triangle in the bottom-right corner of the rectangle.
Good problem with great solution.
Hard to think quadratic in ab...
Here's the complimentary solution using side lengths that exposes the underlying geometry.
The key is to factor (ab/2) out of everything.
From the problem:
ab=x+y+z+w, a=a_x+a_z, b=b_y+b_z
x=(1/2)a_x*b, y=(1/2)b_y*a,
z=(1/2)a_z*b_z =(1/2)(a-a_x)(b-b_y)
=ab/2+2xy/ab-x-y
Let's factor the general area of a triangle and solve for w.
x=(ab/2)(a_x/a)
y=(ab/2)(b_y/b)
z=(ab/2)(1+(a_x/a)(b_y/b)-(a_x/a)-(b_y/b))
Notice how z cancels the area contribution from x and y.
Now we can solve for w:
ab=x+y+z+w=(ab/2)(1+(a_x/a)(b_y/b)+w(2/ab))
(ab/2)(1-(a_x/a)(b_y/b))=w
I find the underlying ratios and difference between product and sums of these ratios fascinating. Thanks Presh for the great video 😁🙏🙋♂️
Oh solving the Maths is so amazing and logical 😍
Easier solution without (complete) quadratic equation:
a*b=x+y+z+w
b*d=2*x (I)
a*c=2*y (II)
(a-d)*(b-c)=2*z (III)
from the last equation one gets a*b-a*c-b*d+c*d=2*z where the product c*d is unknown.
From (I)*(II) one gets a*b*c*d=4*x*y or c*d=4*x*y/(a*b). Substituting back in (III) this and the other products:
x+y+z+w-2*y-2*x+4*x*y/(x+y+z+w)=2*z or w-x-y-z+4*x*y/(x+y+z+w)=0 then (w-x-y-z)*(x+y+z+w)+4*x*y and finally w^2-(x+y+z)^2+4*x*y=0 since the first two terms in parents simplify to a notable product. From this last equation we get the final answer without baskara...
first time saw the video is released seconds ago
i've seen him video released seconds ago like 2 times
Using area of x and y as a proportional constant
(T-2X)(T-2Y)=2ZT
T is area of bigger square
Roots 4(3+√6) ,-4(√6-3)
Required area 4√6
My classmates be like:
6
Got this one! I defined the left side as Y and the top side as X, this gave me 3 formulas using triangles (8 = ay, 6 = bx, 10 = (x-a)*(y-b)).
Plugging into the last eqn, I got: 0 = (xy)^2 - 24(xy) + 48. Since xy is the area of the square, I don't need to independently solve X and Y, I instead just solved for the total area which is 12 + 4sqrt(6) and then subtracted the three triangles to get the answer.
vay be türklerden kanalı bilen varmış ismi görünce şaşırdım :D
Solved this question in 2 min..
Was relatively easier question than what u posted earlier...
Love from India ❤🇮🇳🇮🇳
Indian and their inflated ego will never die.
@@sabyasachirimpa This is not my ego.. I solved really...
.
@@naturelover4148 May be you solved it really, but I have noticed that in other videos where Presh has given very complicated problem, so many Indians say that they have solved the problem within a minute or simply say we have done those problems when we were studying in 4th or 5th grades.
May be... But I am not one of those...
@@sabyasachirimpa
Calm down man....you are showing your ego by posting such comments....Its not about Indians, it's about individuals....I mean ego is not associated with a nation but individuals....
Writing the answer before watching the solution:
Area of triangle BEF (inside the square ABCD) is 9.8
Almost the same solution, but a bit more visual:
Flip the three triangles on the long sides. Then the complete rectangle a*b is filled one time - except the upper left rectangle c*d, which is filled two times. So ab+cd=2*(3+4+5)=24
Because of (ac)/2=4 and (bd)/2=3 we have ac*cd=ab*cd=48.
Calling p=ab and q=cd you have to solve the eqations p+q=24, p*q=48 which leads to your result.
That's not clear can you please explain again?
@@leif1075 Draw two lines into the given figure: a vertical line through the touching point of the green an yellow triangles and a horizontal line through the touching point of the yellow and blue triangle.
This way you split the given rectangle into four smaller rectangles. For example the lower right one is just the yellow triangle doubled.
If you double the green and blue triangles in the same way, the whole figure is covered, but the upper left rectangle is covered two times.
Using the notation of the video the whole figure a*b plus the upper left rectangle c*d must be twice the area of the area of the tree triangles.
So ab+cd=24.
To compute c*d multiply the areas of the green triangle ac/2=4 and the blue one bd/2=3 to get ac*bd=48 or ab*cd=48.
Rename p:=ab, q:=cd we get
p+q=24 and p*q=48.
I used to watch your videos and this was the first time I solved the problem by myself
very much happy
Staying in a square, x(x-y)=8 [Eq1]; yz=10 [Eq2]; x(x-z)=6 [Eq3]; x^2-12=A [Eq4]; Putting Eq1, Eq3 and Eq4 in Eq2 we get (A+4)*(A+6)=10A+120 leading to A^2=96=16*6 so A=4*Sqrt(6).
By the given formula the answers may vary depending upon which triangle we take as z, because (x+y+z)is getting squared but -4xy is subtracted there is no variable z. As a result answers vary accordingly
@Adam Romanov okay so while computing the values to the equation we have to keep in mind that triangle x and y should b the triangle sharing the common vertex. 🙂
If we draw a horizontal line through "E" and another vertical through "F", the original rectangle is divided into four rectangular cells that, ordered from left to right and from top to bottom, have areas of value (a), (6- a), (8-a) and (10). Since the relationship between cells in the same column or row is constant, the following equality is verified: a/(8-a)=(6-a)/10 ⇒ a²-24a+48=0 ⇒ a=2.2 ⇒ (6 -a)=3.8 and (8-a)=5.8 ⇒ The total area of rectangle ABCD is: 2.2+3.8+5.8+10=21.8 ⇒ The area of triangle FBE = 21.8-3-4-5=9.8