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Я решила правильно.

Here's a much easier way: (I am almost sure that this works) The 64 square has height of 8, the 49 square has a width of 7, and the 81 square has a height of 9, so if you connect the top right corner of the 64 square to the top left corner of the 81 square then the Pythagorean theorem says that this new line has a length equal to the square root of 50. From this we can construct a square with area 50, and this new square fits perfectly inside the yellow circle (I am not sure how to prove that it fits perfectly, and this might not always be the case, but I can just somehow tell that it fits perfectly as far as this particular problem goes). This means that the yellow circle's radius is equal to half the new square's diagonal. And we know that the diagonal of any square is the side length multiplied by the square root of 2. So multiply the square root of 50 by the square root of 2 to get 10, and then cut it in half to get 5.

I spent countless hours solving it by trial and error, and once I started using 1 as y, it was just an iterative process to get to the right answer. I didn't know about the trigonometric identity tan( x +y ) = [ tan(x) + tan(y) ] / [ 1 - tan(x) * tan(y) ]. Even then, it's a lot of manipulation to get to the value of y. Nice challenge!

tanA= h/a tanB= h/b a + b = 7 atanA = h btanB = h atanA = btanB 25 = a² + h² 36 = b² + h² 11 = b² - a² = (b+a)*(b-a) 11 = 7*(b-a) b-a = 11/7 b+a = 7 2b = 60/7 b = 30/7 a = 19/7 tanA/tanB = b/a = 30/19

شكرا😊

Nice!

4a = 180 degrees Divide each side by 4 a=45 2a=90 However 4^2+4^2 does not equal 6^2. So im wrong....

Orthodiagonal quadrilateral: a²+c²=b²+d² The table formula! 😁

2:22-3:28 <FCD=<GCH=θ ΔFCD: tanθ=FD/CD=x/2x=½ ΔGCH: CH=GH/tanθ=2/½=4 x-2=4; x=6 Without similarity!

m≈2023.0005 😁

Thanks to the Algorithm, I guess. Interesting to see such 'heavy' algebra in what feels like it should be a geometric problem.

My kids have just learnt about simultaneous equations and the heavy algebra that goes along with it. This will make such an awesome "try and get the same answer" exercise! The problem seems so "elegant" at first, then the Algebra creeps in...

2:42 [Red]=½(3•4)10+½(3•4)3=60+18=78

When the hypotenuses of a quadrilateral intersect at 90 degrees, the total of squares of opposite sides are equal. Because when you add the squares of opposite sides (there are two sets of opposite sides), in either case, the total is equal to x^2 + y^2 + z^2 + t^2.

You can easily solve If drew line from E parallel DA intersect AB at k Then angle Tan DAE =1/3 that equal to tan KEA…then Tan KEB =tan (45-AEK)=(1-1/3)/(1+1/3)=1/2 ,then KB segment =1/2*6=3--> then area of red triangle =1/2*3*6=9

Lets call lower angel of red triangle b …then tan a =tan(45-b)=tan45-tanb/(1+tan45*tanb)=3/15-->Tanb=2/3 then ..we calculate the base of red which equal 3 squre root 26…the hight 2square root 26--> area=78

... 2021 no math, just a W.A.G. (wild assed guess) 😉

Ok understood,thanks

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What is the problem? Two different pictures have been shown

We draw a perpendicular CE on BD. Take 🔺 ACE. AC=7 It is a 30-60-90 triangle. CE=1/2 AC=7/2 BE=√(49-49/4=√(147)/2 BD=49√3 (easy and time saving method to know BD)

It is a very simple problem. You have a triangle with sides 11 13 7sqrt 3. Use the area formula with three sides and subtract the area of BCD which is (sqrt 3)*49/4.

I Think i have a quicker Solution though Maths is only a Hobby: Triangel under the 120 degree angle isosceles, both angles 30 degrees, length of base 12,12 and now Heron of Alexandria: area of Both Triangels 61,90 then subtract 6,06*3,5=21,21 and result 40,69. Elegant!

Why didn't you simplify the answer? Both 94 and 4 are divisible by 2. 😉

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6/4 = sin 2α/sin α = 2 cos α, so cos α = 3/4. So, AB/4 = [ sin (180 - α - 2α) = sin 3α = 3 sin α - 4 sin^3 α ] / sin α = 4 cos^2 α - 1 = 4 x 9/16 - 1 = 5/4. So, *AB = 5*

Muito bom, parabéns!

Bela aplicação do teorema da bissetriz interna e das propriedades das proporções!

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I found it easier using sin(C) = sin(180 - 3a) = sin(3a) = 3sin(a) - 4sin^3(a) and then calculating sin(a) using the sine rule with sin(2a)/6 = sin(a)/4 = sin(180 - 3a)/|AB|, giving cos(a) = 3/4, sin(a) = sqrt(7)/4, and |AB| = 5.

Method using sine and cosine laws: 1. By sine law: 4/sinA = 6/sin2A cosA = 3/4 (as sin2A = 2sinAcosA) 2. Let AB be x. By cosine law: x^2 + 6^2 - (2)(6)(x)cosA = 4^2 3. Solve the quadratic equation x^2 - 9x + 20 = 0. x = 4 or 5 x = 4 is rejected as this makes angle BAC = 90 (isosceles triangle with base angles = 45) then by Pythagoras theorem 4^2 + 4^2 = 32 not 6x6 (as given).

An alternative method without using construction: 1. By sine law: 4/sinA = 6/sin2A cosA = 3/4 (as sin2A = 2sinAcosA) 2. Let AB be x. By cosine law: x^2 + 6^2 - (2)(6)(x)cosA = 4^2 3. Solve the quadratic equation x^2 - 9x + 20 = 0. x = 4 or 5 x = 4 is rejected as this makes angle BAC = 90 (isosceles triangle with base angles = 45) then by Pythagoras theorem 4^2 + 4^2 = 32 not 6x6 (as given).

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WAY too elaborate!! I mean: immediately note that it's a 3-4-5 triangle thus the hypotenusa is 5x. (3 seconds of work). Then A==P. 6x^2==12x. x^2-x=0. (x-1)^2 = 1. x-1= +/- 1. x = 1+1=2 or 1-1=0. ALL could have been done in way under 60s.

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Rotate the green triangle 90° CW around its right node , and you see that its base is sqrt(2) and its height is 2*sqrt(2), so its area is 2.

Very nice

elementary problem in Korea

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5x

The Pythagorean theorem does not apply to triangles without a right angle. In a triangle without a right angle, you need to use other methods, such as the laws of cosines or sines, to solve problems related to side lengths and angles.

Triangles ABC and IEF are similar then AC=5k, AB=3k. Triangle ABC is also similar to triangle ICH, that gives 3k=8/5 and 5k=8/3 thus red area=32/15 squ

My way of solution would be➡ [CE] is equal to R R= 4+4 R= 8 length units ⇒ [CE]= 8 [CB]= 4 according to the Pythagorean theorem: CE²= CB² + BE² 8²= 4²+BE² BE= √48 BE= 4√3 length units consider the ΔCEB, ∠CEB= α tan(α)= BC/BE tan(α)= 4/4√3 α= artan(1/√3) α= 30° this is also equal to ∠ECD ! if we subtract from the quarter circle, the small quarter circle and the triangle, plus the 30° circle segment, we get the green area ! ⇒ A(ACD)= π*R²/4 A(ACD)= π*8²/4 A(ACD)= 16π A(ΔCEB)= 4*4√3/2 A(ΔCEB)= 8√3 square units A(ECD)= π*R²*(30°/360°) A(ECD)= π*8²*(30°/360°) A(ECD)= π*64*(30/360°) A(ECD)= 16π/3 square units A(ABF)= π*r²/4 A(ABF)= π*4²/4 A(ABF)= 4π square units Agreen= A(ACD) - [A(ΔCEB) + A(ECD) + A(ABF) ] Agreen= 16π - [8√3 + 16π/3 + 4π ] Agreen= 20π/3 - 8√3 Agreen ≈ 7,088 square units

Gostei muito dessa questão. Traçar CE foi o pulo do gato!

Shouldnt it be: the whole structure minus (ABF+BCE+CDE)? That would be [8²π/(4)] -[4²π/(4*3)] -8*(3^0.5) -[8²π/(4*3)] Green area ~~ 15.46 or 28π/3-[8*(3^0.5)]

I solved it in my head. (Just kidding) Nice problem!

Questão ótima, e difícil.

Nice

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You can also move the white and the green circle to the center, it may get the answer more directly.😀