Excellent explanation 👍

@PreMath

10 ай бұрын

Glad you liked it! Thanks for your feedback! Cheers! 😀 You are awesome, Ramani. Keep it up 👍

Using the Intersecting Chord Theorem at P, 1*1 = {r -(2+1/√3)}*{r +(2+1/√3)} 1 = r² -4 -1/3 - 4/√3 r = 2√{4+√3)/3}

@PreMath

10 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍

I'm in my 50's and love watching your teachings with my son. Its always first class. We found an alternative method without using trig and only using Pythagorean Theorem: Since CF=√3, then OF= √((√3-OF)² - 1²) = 1/√3 = 0.577 OP = 2 + 1/√3 = 2.577 r² = OP² + PD² r² = 2.577² + 1² r² = 6.6409 + 1 r² = 7.64 r = √7.64 = 2.76

@PreMath

10 ай бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 You are a great role model for your son 😀

There is another justification for the length OF to be (√3/3)cm: In an equilateral triangle every height is given by (√3/2) times the side length, so every height in the triangle ABC is (√3)cm. In an equilateral triangle every height also corresponds to a median line and O, which is the point of intersection of the median lines and the heights as well, does divide the median lines with a 2:1 ratio. So, OF is the height, divided by 3. Best regards from Germany

@PreMath

10 ай бұрын

Great! Thanks for sharing! Cheers! You are awesome. Keep rocking 👍 Love and prayers from the USA! 😀

@SkinnerRobot

10 ай бұрын

I've found this proposition to be very useful in many geometry puzzles.

Thanks for video.Good luck sir!!!!!!!!!!!

@PreMath

10 ай бұрын

Thank you too, dear Love and prayers from the USA! 😀

Great solution!❤🥂😀

@PreMath

10 ай бұрын

Glad you think so! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

Let's connect the points O and A, O and E. Then we can find that OA=1/cos 30⁰= 2√3/3. Using the law of cosines in the triangle OAE, we'll find OE

@user-ru7pl4tx5v

9 ай бұрын

Αυτή είναι και η δική μου λύση.

When I realized the diagonal was a radius, ... I realized that I realized how to solve it!

@PreMath

10 ай бұрын

Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

That interesting🤩

@PreMath

10 ай бұрын

Glad you think so! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

Thank you

@PreMath

10 ай бұрын

You are very welcome! Thank you! Cheers! 😀 You are awesome. Keep it up 👍

Very biutiful, this one i did not be alble to solve. Very biutiful! Kind regards from Portugal!

@PreMath

10 ай бұрын

Excellent! Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍

Radius=2•√((4+√3)/3)≈2,78

Solved it with Heron's formula: The area of the triangle ABC: A = √(s(s-a)(s-b)(s-c)) = √((s(s-a)^3) , with s being the semiperimeter of the triangle and a, b, c its sides. =√ ((3a/2)*(((3a/2)-a)^3)) = √ ((6/2)*((6/2)-2)^3)= √((3 * (1)^3) = √3 The area of the triangle ABC is equal to 3* the area of the triangle ABO: A = 3* (1/2)* 2*h= 3h , which is equal to √3 h= (√3 )/3 And from the Pythagorean theorem: r² = OP² + PE² = (a+h)² +(1/2)a)²= (((√3 )/3)+2)² + ((1/2)a)² = (((√3 )/3)+2)² + ((1/2)*2)² = 2.577² + 1² = 7.64 r = √7.64 = 2.76

No peeking, but starting out with a more algebraic approach: By Pythagoras, altitude of triangle ABC = sqrt(3); Let OF = x; then OC = OA = sqrt(3) -- x. Since triangle AOF is 30-60-90, OF = AO/2; so x = (sqrt(3) -- x)/2 = sqrt(3)/2 -- x/2; 3x/2 = sqrt(3)/2; 3x = sqrt(3); and x = sqrt(3)/3. From there we proceed the same as you did: OE^2 = 1^2 + OP^2 = 1+ (2+sqrt(3)/3)^2, etc. to get OP = r = 2.7645, approx. Cheers, 🤠

PreMath went in the downward direction but I went upward, and we got the same answer. Here's my solution method: Going clockwise around the square with vertices B and C, label the next vertex as point G. Drop perpendicular from G to the vertical red line and label the intersection as point R. Note that

Side of square = 2 cm Half side of square = 1 cm a = Apothem of eq. triangle tan 30° = a / 1 a = 0,577 cm tan α = 1 / (2 + 0.577) α = 21,206° sin α = 1 / R R = 1 / sin α R = 2,7645 cm (Solved √ )

One could try filling the circle so that the center area shape is triangle (as here), then square, pentagon, etc and express the resulted series.

First each square is 2x2, especially BD=2, OB is 1/cos 30=2/root3, angle is 120, therefore by cosine rule the radius is the squdare root of 4+4/3+(2)(2)(2/root3)(1/2)=root(16/3+4/root 3)=2.76455approximately. 😊

@PreMath

10 ай бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍

تمرين جيد. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين

@PreMath

10 ай бұрын

Thank dear🙏

Let OP be a perpendicular bisector of ED. Let W be the point on BA that intersects with OP. By observation, side length of all squares is √4 = 2 cm and ∆ABC is equilateral, so all its internal angles are 60°. ∠OAW is 30°, and ∠AWO is 90°, so ∠WOA is 60° and ∆AWO is a 30-60-90 triangle and OA = (2)WO. Let the length of WO = x. Triangle ∆AWO: a² + b² = c² x² + 1² = (2x)² = 4x² 3x² = 1 x² = 1/3 x = 1/√3 Triangle ∆EPO: a² + b² = c² 1² + (2+(1/√3))² = r² r² = 1 + 2² + 2(2)/√3 + 1/3 r² = 1 + 4 + (4√3)/3 + 1/3 = 16/3 + (4√3)/3 r² = 4(4+√3)/3 r = 2√((4+√3)/3) ≈ 2.76 cm²

how do you conclude the angle of OAF is half of FAC which is 60 degree, is it by the eyes or any mathemtical reason for that?

@j.r.1210

10 ай бұрын

Line AO bisects the angle. He kind of glossed over that point quickly, but it could be proved separately if necessary.

You don't need trig, if you know the fact that on a median the distance from the vertex to the centroid if 2/3 of the entire median This mean that CO = CF * (2/3) and OF = CF * (1/3). Using the pythagorean theorem on CBF gives CF = sqrt(3), and therefore OF = sqrt(3) * (1/3). Same result as with the trig, just using a different piece of knowledge

I enjoy doing these problems in Excel : =SQRT(1+(2+TAN(PI()/6))^2)

After solving this problem, I was a little surprised that the resulting expression, which includes the sqrt of a sqrt, couldn't be cleaned up further. I thought, "Surely the polynomial with sqrt3 will turn out to be the perfect square of another easy polynomial" -- but no! I couldn't find it. In that respect, I guess this problem is more "realistic," because everything is not exactly configured in advance.

I got r² = [16+4sqrt(3)]/3 = 7.64 and therefore r = 2.76. I presumed the center triangle to be equilateral and that perpendiculars from the center to the sides of the triangle would cut the sides in half.

The first time I solved the problem I did it by tracing a line from the center of the circle to the circumference through A but found I was wrong. It is because the line from A to the border does not equal 2 as the square border. It is hard to tell the length of the segment from A to the border when it is part of the line from O through A. I believe your method is the best way to tackle the problem, if not the only way. Any other way would involve an exhausting analysis.

@juanalfaro7522

5 ай бұрын

Actually, using the Intersecting Chord Theorem at P works too. And tracing the radius through A to the border works as well, if we apply the Intersecting Chord Theorem at the midpoint of the segment between E and the border point of the square at the left. If I call that left square border point G and the midpoint of the EG segment as H, then EG = 2*sqrt (3) so EH = sqrt (3), AH = 2 cos (120/2) = 1, and OA = 2/ sqrt (3). Then OH = 1 + 2/sqrt (3) = (2*sqrt (3) +3)/3 and, by the Intersecting Chord Theorem at H, (R - [(2*sqrt (3) +3)/3]) * (R + [(2*sqrt (3) +3)/3]) = [sqrt (3)] ^2 = 3 ---> R^2 - [(7+ 4*sqrt (3))/3] = 3 ---> R^2 = [(16 * 4*sqrt (3))/3] and we take the sqrt of the expression.

I went wrong somewhere. I got to the initial h being 1/(sqrt(3) due to 30-60-90. I then took a different path. I went OA is, therefore, 2/(sqrt(3). I think that is probably correct. My next part was to extend OA outward to the circle and make a triangle of AE and the circumference. I called this another 30-60-90 with 2 (AE) as the hypotenuse so the short side which is the extension of OA would be 1. This gives a total radius of 1+(2/sqrt(3)) which comes to 2.155(3dp). I suspect my error is in calling the second triangle a 30-60-90. The angle at A would be 60 but I suspect the other two are NOT 30 and 90. Your explanation makes complete sense, so thank you for that.

Generalized: the radius of the circle is _√(S ⅓ (4 + √3))_ where _S_ is the area of one yellow square.

Solution: 4[cm²] = area of a square ⟹ 2[cm] = side of a square = s, height of the isosceles triangle = √[s²-(s/2)²] = √[s²-s²/4] = √[3/4*s²] = s*√3/2 = √3[cm] = h. It is h/3 = √3/3[cm] from O to the next side of the square, and to the oposite side it is √3/3+2[cm]. Pythagoras: r = √[(√3/3+2)²+1²] = √[1/3+4/3*√3+4+1] = √[16/3+4/3*√3] = 2*√[(4+√3)/3] ≈ 2,7645

2.764 cm

It is √3/2

asnwer= 240cm isit my math matter isit

1+13=14