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.

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

S=32(3-2√2)π≈17,097

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

I understood the mechanics and thought process completely. Does that mean that I should be able to do this?

R+R/√2+2=6...R(1+1/√2)=4...R=8-4√2...mah,troppo semplice?!?

Fantastic aproach. I used the Pytagoras Theorem to find the radius, and the aproximately value to the circle area, was 5,5 . PI 👍👍

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

Can you solve IOQM maths paper 2023, question 19.

(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.

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

Nice! r(√2 + 1) + 2√2 = 6√2 → r = (4√2)(√2 - 1) → πr^2 = 32π(3 - 2√2)

(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)

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🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉

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 ,

Did it in me heed lol