Can you find the area of the circle inside a quarter circle | A Nice Geometry Problem
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r + r√2 = 2, so r = 2(√2 - 1) and A = πr² = π(12 - 8√2).
You can calculate the segment 2-x easier by knowing that a disgonal of a square is its side multiplied by square root of two. So radius of quarter circle 2=x+x.sqrt2.
Thank you for the problem and the solution. ❤ I managed eventually to solve it.
This took me a while to see. Blindly obvious now.
D is the point where the circle touches OA
OD = radius of circle = r Drawing diameter PF across the circle, from O
r.r = OP.OF OF is given , it is 2 or R is given as 2
OP = 2 - diameter = 2- 2r It is also (sqrt(2) -1)r
In general , for next time R- 2r = (sqrt(2) -1) r
r (sqrt(2) -1 +2) =R r = R /(sqrt(2) +1) = R 😌 r = R (2^(1/2) -1)
radius = 8^(1/2) - 2 or 2 (sqrt(2) -1) r = approx 0.828
So the area of the circle
pi (4)(3 - 2sqrt(2)) = pi (12- 8sqrt(2)) approx 2.156
The following was redundant work. That comes from tryng and trying on the wrong path, (again 🤔😐😑).
F where the circle touches the arc AB, E where it touches OB, D where it touches OA and C at the centre,
gives OD = OE = DC = EC = FC = radius, r
Angles at C are 90, 135, 135 degrees with ODCE being a square OC = (sqrt(2)).r
Also extending OA to G and OB to H in the way that GFH is the only common tangent to both these curves,
gives triangle OGH is rightangled and isosceles. OF divides this triangle, the mentioned square, the given qudrant and the square
each into two congruent halves
In this problem OF = 2
GH = 4 OG = OH = 2sqrt(2)
GD = OG - OD = 2sqrt(2) - r
Considering CG, there are two Pythagorean ways to write its square which might be equated but they are congruent
In triangle CDG CG.CG = DC.DC + DG.DG = r.r + (2sqrt(2) - r) (2sqrt(2) -- r) = 2.r.r + 8 +r.r - 4 sqrt(2).r
In triangle CFG CG.CG = FC.FC + FG.FG = the same also there is triangle CGO , with cosine rule
In triangle CGO CG.CG = OG.OG + OC.OC - 2. OG.OC cos (COG) = 8 + 2r.r. - 2. (2sqrt(2)).(sqrt(2)).r (1/sqrt(2) the same
ALTHOUGH OG/OC = 2/r so OG: OC: CG
There is no need to set up a quadratic equation.
By drawing the perpendicular "QT" from Q to OB, it is sufficient to set the proportion:
OQ : QT = OP : PN
2 : √2 = (2-r) : r
that is,
r = 2√2 - 2
The rest is simple
R = 2cm = r + r√2 = r(1+√2)
r = 0,82843 cm
A = πr² = 2,156 cm²
2-r=r√2
r=2√2-2
A=π*(2√2-2)²
A=(12-8√2)π
I used trigonometry and got to the same answer:
Call the point of intersection of OQ with the lower-left of the circle K.
Make two triangles:
OMQ and KMQ
In triangle OMQ we have (all numbers degrees):
angle at O = 45
angle at Q = 22.5 (45/2 by opposite angle to a 45 degree arc).
angle at M = 180 - 45 - 22.5 = 112.5
In triangle KMQ we have:
angle at M = 90 (inscribed angle in a semicircle)
angle at Q = 22.5 (as before)
angle at K = 180 - 90 - 22.5 = 67.5
We can find the radius by repeating the law of sines on these two triangles:
Side QM = (sin 45 / sin 112.5) * 2 (from triangle OMQ)
diameter of red circle = KQ = (sin 90 / sin 67.5) * (side QM)
-or-
diameter of red circle = (sin 90 / sin 67.5) * (sin 45 / sin 112.5) * 2
The area of red circle is therefore:
pi * [(sin 90 / sin 67.5) * (sin 45 / sin 112.5)]^2
Numerically this is the same as the video's answer.
The circle is symmetric with respect to the quarter circle's angular bisector, so its center C and the point D, where the circle touches the arc AB, must be located on the angular bisector. By symmetry, C has distance r from OA and OB, respectively. Also, the distance CD is r as well. So the distance OC is r√2, as OC is the diagonal of a square. From this we have r√2 + r = r(√2 + 1) = 2 and therefore r = 2 / (√2 + 1) = 2(√2 - 1) . Then we have the circle area = 4(2 - 2√2 + 1) π = 4(3 - 2√2) π.
赤の半径をrとすると、
r+r√2=2
r=2/(1+√2)=2(√2-1)
π2^2(√2-1)^2=4π(3-2√2)
=(12-8√2)π
D = point of contact between the quarter circle and the circle,
C = point of contact between circle and the line OB,
M = middle point of the circle,
r = radius of the circle,
R = radius of the quarter circle = 2.
Pythagoras for the right isosceles triangle OCM:
OC²+CM² = OM² ⟹
r²+r² = (R-r)² ⟹
r²+r² = (2-r)² ⟹
r²+r² = 4-4r+r² |-r²+4r-4 ⟹
r²+4r-4 = 0 |p-q-formula ⟹
r1/2 = -2±√(4+4) = -2±√8 ⟹ r1 = -2+√8 and r2 = -2-√8 < 0 [not in geometry]
Area of circle = π*r² = π*(-2+√8)² = π*(4-4*√8+8) = π*(12-4*√8) ≈ 2,1560
1/4円に接する円の半径をr、中心をkとする。またOBとの接点をC、弧ABとの接点をDとする。
OC₌r、OK₌√2r OD₌√2r₊r₌(√2₊1)r₌2よりr₌2/(√2₊1)₌2(√2₋1)
to find radius of circle I use other way
let Q is point where circle and quarter circle meet so line OPQ is a straight line that also radius of quarter circle = 2
from point P we create a line that is perpendicular to line AO (point M) and BO (point N) so OMPN is a square with side = radius of circle, say r
diagonal of square OP = r √2
we get PQ is also radius of circle, so we get OQ = OP + PQ = r √2 + r = r (√2 +1)
OQ = 2 = r (√2 +1) => r = 2/(√2 +1) and if we multiply numerator and denominator with (√2 -1) we get r = 2√2 - 2
S=4π(3-2√2)≈2,156
По теореме о касательной и секущей r²=Rx=2x, откуда х=r²/2. Но мы также знаем, что R=2r+x или 2r+r²/2=2. Избавляемся от знаменателя: 4r+r²=4. Переносим, получаем приведённое: r²+4r-4=0. D=16+16=32. r=(-4±4√2)/2=-2±2√2. берём положительный, получаем r=2√2-2=2(√2-1). S=πr²=4π(√2-1)²=4π(2-2√2+1)=4π(3-2√2).
Brilliant !
X+X√2=2, X(1+√2)=2, X=2/1+√2, A=π4/3+2√2
OP=√2x
OQ=OP+PQ=√2x+x=(√2+1)x=10
x=2(√2-1）
x^2=4(3-2√2）
area=π*x^2=4(3-2√2)*π
4*Pi*(3-2*sqrt(2)) ... a cappella, messieurs dames !
(2)^2=4 90°ABO/4=22.2ABO 2^11.2 1^11^1.2 1^1^1.2 1.2(ABO ➖ 2ABO+1).
Radius = R
R * sqrt(2) + R = 2
R = 2 * (sqrt(2) - 1) ~ 0,82843
Area = (2*(sqrt(2) - 1))^2 * Pi
Area ~ 2,16
Exact Area = (2sqrt(2) - 2)^2 * Pi = (12 - 8*sqrt(2)) * Pi
Let r is the radius of the circle
r(2^1/2)+r=2 => r=2/(1+(2^1/2))
Area=π(r^2)=π(2(2^1/2)-2)^2=π(12-8(2^1/2)
Area=π(12-8(2^1/2)) units squared
This Problem can be for grade 9 and 10
According to the picture you made,
2 - x = x√2
x = 2√2 - 2
Why you make harder what is simple and elegant?
which is a better writing, 4(3 - 2√2) or 12 - 8√2? I think, no, I insist on 12 - 8√2. Math Booster, please, improve your contents.
You complicate it ..keep it simple
No need for so much!! Skip known basics, i.e. simple algebra steps, and quadratic formula.
should be < 4 minute video.
Dai....sforzati un po' di piu
2=r√2++r
r=2/(√2+1)
❤️=π4/(3+2√2)
❤️=(4π)(-2√2+3)
💛🧡💜
Hamas=Résistance💚💚💚💚❤️