A Nice Geometry Problem | You should be able to solve this!
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- Опубліковано 4 лип 2024
- A Nice Geometry Problem | You should be able to solve this!
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Make a point vertically down from O and horizontally across from Q. Call it Z.
Make a right triangle OQZ.
r (4-r),(4-r)
(4-r)^2 + (4-r)^2 = r^2
2(16-8r+r^2) = r^2
32-16r+2r^2=r^2
32-16r+r^2=0
r^2-16r+32=0
(16+or-sqrt(256-4*1*32))/2=r
(16+or-sqrt(128))/2=r
(16+or-8*sqrt(2))/2=r
8+or-4*sqrt(2)=r
8+4*sqrt(2) is greater than the square's side length, so use 8-4*sqrt(2).
r = 8-4*sqrt(2)
Square it for 64-64*sqrt(2)+32
96-64*sqrt(2) = r^2
(96-64*sqrt(2))*pi = area of circle.+
Approx 5.49pi so approx 17.248 un^2
I think you overcomplicated this one, but thank you.
2*sqrt (2) + R + R*sqrt (2) = 6*sqrt (2) --> R * (1+sqrt (2)) = (6-2) * sqrt (2) --> R = 4*sqrt (2) / (1+sqrt (2) = 4*sqrt (2) * [sqrt (2) - 1) = 8 - 4sqrt (2). Now Area = Pi*R^2 = Pi* (64+32-64*sqrt (2)) = Pi*(96-64*sqrt (2)) = 17.2484
S=32(3-2√2)π≈17,097
Let be O thé center of the circle. D Q O and B are aligned on the diagonal of the big square.
Let M be the projection of O onto the line DC.
Use of Thales théorème White DQ parallel to OM
So :
OM/QS = DO/DQ
(6- R) /2 = (2 * square root (2) + R)/2 * square root (2)
And R = 8 - 4 * square root (2)
And Area of thé cercle = PI * R square
(6-2)√2=4√2=r+r√2=r(1+√2) → r=4√2/(1+√2)→ r²=32/(3+2√2)=32(3-2√2)→ Área del círculo =32π(3-2√2).
Gracias y un saludo.
Fantastic aproach. I used the Pytagoras Theorem to find the radius, and the aproximately value to the circle area, was 5,5 . PI 👍👍
consider the diagonals of the 2 internal squares (of sides 4 and 2)
diagonal of the sqr of side 4 = r√2+r
r√2 + r = (6-2)√2 => r= 8-4√2
Nice! r(√2 + 1) + 2√2 = 6√2 → r = (4√2)(√2 - 1) → πr^2 = 32π(3 - 2√2)
R+R/√2+2=6...R(1+1/√2)=4...R=8-4√2...mah,troppo semplice?!?
Let O be the center of the circle, and M and N be the points of tangency between circle O and AB and BC respectively. Let R be the radius of circle O.
Draw radii OM and ON. As AB and BC are tangent to circle O at M and N respectively, ∠BNO = ∠OMB = 90°. As ∠MBN = 90° as well, then ∠NOM = 360°-3(90°) = 90°. As all four internal angles of OMBN are 90° and adjacent sides OM and ON are both of equal length r, OMBN is a square with side length r.
Draw BD. As OMBN and DPQS are both squares and thus each have all sides of equal length, Q and O are collinear with BD.
By Pythagoras, it can be shown that the diagonal of a square (let's use OMBN for an example) is equal to √2 times the side length:
OM² + MB² = OB²
r² + r² = OB²
OB² = 2r²
OB = √(2r²) = √2r
Therefore as BD = 6√2 and QD = 2√2, QB = 6√2-2√2 = 4√2. QB also equals OB+OQ = √2r+r = r(√2+1)
r(√2+1) = 4√2
r = 4√2/(√2+1)
r = 4√2(√2-1)/(√2+1)(√2-1)
r = 4√2(√2-1)/(2-1) = 4√2(√2-1)
Circle O:
Aₒ = πr² = π(4√2(√2-1))²
Aₒ = π(32(2-2√2+1)
Aₒ = 32π(3-2√2) ≈ 17.25 sq units
Very complicated😅
I understood the mechanics and thought process completely. Does that mean that I should be able to do this?
❤❤❤❤🎉🎉❤❤😊😊
(2)^2.H/A/DpQSDino°(6)^2 H/A/MBONCoso° 4H/A/DPQS,Sino°+36H/A/MBON} ={40H/DPQS/Sino°Cos°MBON 360°/Tano°}= 90°H/A/DPQSSino°MBONCoso° 3^30 3^5^6 3^5^3^2 1^1^3^2 3^2 (H/A/DPQSSino°MBONCoso°Tano° ➖ 3H/A/BPQSSino°MBONCoso°Tano°+2)
Did it in me heed lol
sqrt(2*2^2)+R+sqrt(2*R^2)=sqrt(2*36^2) , sqrt(2*R^2)=sqrt(72)-sqrt(8)-R , / sqrt(72)-sqrt(8)=4*sqrt(2) , / ,
2*R^2=32-R*8*sqrt2+R^2 , R^2+R*sqrt128-32=0 , / R>0 / , R=(-sqrt128+sqrt256)/2 , R=(-11.3137+16)/2 , R=2.34315 ,
To=R^2*pi , To=~ 17.2484 area unit ,
Thanks !
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