วีดีโอ

Nice explanation. Thanks

Merry Christmas

I have find your videos by chance I get found of it thanks alot for your simplicity the math is my hoppy

Thanks from Morocco

شكرا األك

❤❤❤ Thank you thank you thank you so much 😊❤

You are the best teacher I have ever seen ❤️🥰 lots of love 😘

I'm trying to study for my final and have always struggled with this concept, but this is the easiest-to-understand video I've ever seen. Thank you so much!

[180-[[(180-25)/2]-25]]/2=63,75°

Oh fux, I started trying to "asking Pythagoras", but got no answer! Thanks for a less complex response.

Can you find answer for the value of radius equal to 7 cm😢

Thank you

Awesome! Thanks for the simple explanation!

Simple explanation and nicely put 🤔💭🧮

so there is a equation to find radius of incircle, Area of triangle/half of the perimeter of triangle

They're obviously all 345 triangles so all the angles are given. Move on...

Wow great explanations, I thought even and odd was hard until i saw this video, this help me a lot thank you

This was interesting, thanks for teaching!

Lado del cuadrado =c---> 3=(5c/4)+(3c/5)---> c=60/37. Gracias y saludos

Really awesome explanation, thank you!!

Awesome! Thanks for another great math video!

I used cosine rule to calculate the angle ACB = 71.8 degree. Let a = distance of C to the point P, the tangent of line BC to the circle. b = distance of B to the point Q, the tangent of line BC to the circle, c = distance of A to the point R, the tangent of line AC to the circle. Given that the line CO (center of the circle) bisects the angle of ACB, BO bisects angle CBA and AO bisects angle CAB, I got a = 3.5, b = 5.5 and c = 4.5. Angle OCP = 71.8 / 2 = 35.9 degrees. The radius r = a x tan 35.9 = 3.5 x 0.7237 = 2.53 units

I study for te exams, this is a big help, thank you

Method 3 and 4 are easy to apply

Great proof, but why is this important?

thank you for this! i have been struggling about understanding transformations and you explained this very well. i hope this channel gain more viewers, your videos are very helpful.

Iam very glad... best teacher

Maestra, gracias por su ayuda, soy un pobre latino indefenso en busca del sueño americano, pero el álgebra me impide alcanzar mi sueño de ser alguien en la vida, confío en que gracias a su tutorial, pueda salir de este gran problema que me está carcomiendo el alma, y logre grandes cosas al igual que usted, es un honor ver este vídeo sin morir en el intento, gracias una vez más.

40^2-36^2+4^2=72x-8x=64x X=(4*76+16)/64=(76+4)/16=80/16=5 Height =X+4=5+4=9

Great illustrations, thanks for explaining!

Brilliant

You are a great teacher. Thank you so much.

Let's solve this in an entirely general way. We can describe the problem as follows. (1) Impose a standard x/y coordinate system on the problem (2) Rectangle dimensions: width w, height h (3) Rectangle corners: (0, 0), (w, 0), (w, h), (0, h) (4) Target triangle corners: (u, 0), (w, h), (0, v) (5) Area of triangle (0, 0), (u, 0), (0, v): X (6) Area of triangle (0, v), (w, h), (0, v): Y (7) Area of triangle (u, 0), (w, 0), (w, h): Z (8) Area of triangle (u, 0), (w, h), (0, v): a (this is the variable we seek) (9) Define S = X + Y + Z Note that we have used uppercase variables for quantities that are "given" and lower case for quantities that we do not know. So our ultimate goal is to get an expression for a that is in terms of uppercase variables only. We can immediately write the following four relationships: (10) w*h = a + S (11) u*v = 2*X (12) (w-u)*h = 2*Y (13) w*(h-v) = 2*Z This is simple, but as a linear system it has five unknowns: w*h, a, u*v, u*h, and w*v, but there are only four equations. This tells us immediately that no amount of linear algebra will get us to the solution we seek. We have to do something nonlinear. First, expand equations (12) and (13): (14) w*h - u*h = 2*Y (15) w*h - w*v = 2*Z Now form the sum and product (14)*(15) and substitute (10). The product is our nonlinear action: (16) 2*w*h - (w*v + u*h) = 2*Y + 2*Z (17) -(w*v + u*h) = 2*Y + 2*Z - 2*(S + a) (18) (w*h)*(w*h) - (w*h)*(w*v) - (u*h)*(w*h) + (u*h)*(w*v) = (2*Y)*(2*Z) Substitute in (10), (11), and (17): (19) (a + S)^2 - (a+S)*(w*v) - (a+S)*(u*h) + (a+S)*(2*X) = (2*Y)*(2*Z) (20) (a + S)^2 + (2*X - (w*v + u*h))*(a+S) = 4*Y*Z (21) (a+S)^2 + (2*X + 2*Y + 2*Z - 2*S - 2*a)*(a+S) = 4*Y*Z (22) (a + S)^2 - 2*a*(a + S) = 4*Y*Z Finally just expand and simplify: (23) a^2 + 2*S*a + S^2 - 2*a^2 - 2*S*a = 4*Y*Z (24) -a^2 + S^2 = 4*Y*Z (25) a^2 = S^2 - 4*Y*Z (26) a = sqrt(S^2 - 4*Y*Z) This is our final answer. One might ask why Y and Z appear explicitly while X does not, but if you look at the original diagram you see that X is "special" in the sense that it is the only triangle with a corner at the origin. Note that we can always rotate any problem of this type to bring it into this form - to ensure that only one of the four triangles has a corner at the origin. I'm sure it could be solved in any orientation, but as I've played with this a few times I've felt like the algebra is simplified by putting it into this "standard orientation." Finally, we note that this does give the right answer. For our given problem we have S = 54, Y = 15, Z = 27. If you plug those in you do get 36.

Molto bello❤

Heron e A = a b.c/4R

This is very helpful. I've saved this off.

Thanks for the great explanation 🤩

Excellent performance of resolving the problem

Yes y=x^3 at 0

The formula for the shaded triangle area is A = sqrt( (a+b+c)^2 - 4bc), with a=12, b=15 and c=27. You’re welcome.

Wonderful.❤❤

How was this 'Think outside the box'?

rararomama

I worked out the answer in my head, just by looking at the thumbnail. Here's how: the top triangle has an area of 27 units. Assume that the area of a triangle is length x height instead of length x height / 2. This is just a scale factor and we'll correct for that factor of 2 later. Ignoring this scale factor, the sides of the top triangle are probably 3 x 9 if they're whole numbers, since this equals 27. Similarly, the sides of the other two triangles are 2 x 6 and 3 x 5 respectively. Looking at how these could fit together, it all works if the top of the rectangle has length 9 and the bottom 3 + 6 = 9 as well. The right side of the rectangle will have length 5 and the left side 2 + 3 = 5 as expected. So the area of the rectangle is 9 x 5 = 45. Because of this scale factor of 2, the real linear dimensions will all be √2 times as much as the above figures and the area 2 times as much. So now we need to correct for that missing factor of 2. Therefore the area of the rectangle after correction is actually 2 x 45 = 90 square units. The area of the blue shaded section is therefore 90 - 15 - 27 - 12 = 36 square units.

I didn't know that formula, so once I'd calculated x, I constructed isosceles triangles from the circle centre (O): ∠OCA = 36.5° ∠OCD = 71.5° Named the vertex containing θ as P, so therefore ∠PCD = 180 - (36.5 + 71.5)° = 72° (supplementary angles) Likewise ∠ODB = 53° and ∠ODC = 71.5° so ∠PDC = 55.5° So in △PCD: θ = 180 - (72 + 55.5) = 52.5°

Theta is 1/2(105°). By golly I am glad to have learned od a new circle theorem!!!

Thanks for helping me

arctg[(12,5-1,8-5,1)/(9,6-6,4+4,7)]=arctg(5,6/7,9)=arctg(0,70886..) =35,3313..º=θ Gracias y saludos.

Nice simple solution. At first glance it looks like there are too many variables and too few equations.