Thank you so much! My first attempt was the same as yours so it is more fun to find an alternative one.😊 1/ angle ABE= angle ABD-angle EBD= 45-30= 15 degrees With the value of tan 45=1 and tan 30=1/sqrt3 We have tan 15 degrees= 2-sqrt3 So 2/x = tan 15 = -2- sqrt3-> x=2(2+sqrt3) R=sqrt2. (2+ sqrt3) Area=2pi.sq(2+sqrt3) =87.51 sq units
a simple way to have the value of tan 15 degrees using pure geometry: Just build a 30-90-60 triangle of which the short leg=a so the hypotenuse is=2a and long leg= a.sqrt3. Extend the long leg a segment such that the value= 2a. We have a new isosceles of which the base angle= 15 degrees. tan 15 degrees= a/(asqrt3 +2a)= 1/(2+sqrt3)=2-sqrt3
AB=2tan 75, OA=ABsqrt(2)=tan( 30+45) sqrt(2)=sqrt(2)(1/sqrt(3)+1)/(1-1/sqrt(3))=sqrt(2)(sqrt(3)+1)/(sqrt(3)-1), therefore the answer is pi 2(sqrt(3)+1)^2/(sqrt(3)-1)^2=pi 2(2+sqrt(3))/(2-sqrt(3))=pi2(2+sqrt(3))^2=pi2(7+4sqrt(3))=87.5 approximately. 😊
AEB and BFC are congruent for simmetry reasons. So the side S of the square is: AB=2*tan 75 (Its value is known) and radius is half the diagonal of the square
... Happy weekend to you, An equilateral triangle has angle measures of 60 deg. angle(EBF) = 60 deg. ... I AE I = I FC I = 2 ... triangles (ABE) & (CBF) are congruent ... angles (ABE) & (CBF) are equal and ( 90 - 60 ) /2 = 15 deg. ... now finding I AB I by applying .... TAN(15 deg.) = I AE I / I AB I = 2 / I AB I ... I AB I = 2 / TAN(15 deg.) ... first finding TAN( 15 deg. ) = TAN( 60 - 45 ) = TAN(60) - TAN(45) / (1 + TAN(60) * TAN(45) ) = ( SQRT(3) - 1 ) / ( SQRT(3) + 1 ) ... now I AB I = 2 *(SQRT(3) + 1) / (SQRT(3) - 1)) = (SQRT(3) + 1)^2 = I BC I ... I AC I = Diameter (green circle) = SQRT( 2 * I AB I^2 ) = SQRT(2) * I AB I = SQRT(2) * ( SQRT(3) + 1 )^2 ... radius of green circle .... R = I AC I / 2 = SQRT(2)/2 * ( 1 + SQRT(3) )^2 .... finally Area ( green circle ) = pi * R^2 = (1/2) * pi * ( 1 + SQRT(3) )^4 = pi * (28 + 16 * SQRT(3) ) / 2 = 2pi * ( 7 + 4 * SQRT(3) ) u^2 ... same outcome ... thank you for your clear presentation ... Jan-W
Angle DEF = 60 degrees Angle DEF = 45 degrees Angle AEB = 75 degrees 60 + 45 + 75 = 180 degrees As Angle AEB = 75 degrees the Angle ABE = 15 degrees tan (15) = 2 / AB AB = 2 / tan(15) ~ 7,464 lu Diagonal of Square [ABCD] = AB * sqrt(2) ~ 10,556 Radius = 5,278 lu Green Circle Area ~ 87,5 su Answer: Green Circle Area = (5,278)^2 * Pi ~ 87,5 su
Posto l lato del quadrato e a lato del triangolo equilatero,x=DF, risulta a=√(l^2+4)=√((l-2)^2+x^2)=√(l-x)^2+l^2...quindi si ottengono l,a,x...in particolare risulta una equazione in l...(l^2+4)(l^2-8l+4)=0...cioè l=4+√12=4+2√3...quindi r=√(l^2/4+l^2/4)=l/√2=2√2+√6
Let's do it: . .. ... .... ..... Let s and t be the side lengths of the square and the triangle, respectively. Since we have a couple of right triangles, we can apply the Pythagorean theorem a few times in a row: Triangle ABE: BE² = AB² + AE² t² = s² + 2² t² = s² + 4 Triangle BCF: BF² = BC² + CF² t² = s² + CF² s² + 4 = s² + CF² CF = 2 Triangle DEF: EF² = DE² + DF² EF² = (AD − AE)² + (CD − CF)² t² = (s − 2)² + (s − 2)² s² + 4 = 2*(s² − 4s + 4) s² + 4 = 2s² − 8s + 8 s² − 8s + 4 = 0 s = 4 ± √(4² − 4) = 4 ± √(16 − 4) = 4 ± √12 = 4 ± 2√3 Since 4−2√3 is less than two, the only solution is s=4+2√3. Since the diagonal of the square corresponds to the diameter of the circle, we have: d = 2r = √2s = 4√2 + 2√6 ⇒ r = 2√2 + √6 = √2(2 + √3) A(circle) = πr² = π*2(4 + 4√3 + 3) = (14 + 8√3)π ≈ 87.51
Angle DEF is 45 degrees because triangle DEF is isosceles. Angle FEB is 60 degrees because triangle FEB is equilateral. Therefore, Angle AEB is 75 degrees. tan75 = AB/2. AB = 7.464. Diagonal DB = 7.46 * SQRT(2) = 10.555. Radius = 5.278. Area of circle = 87.511. No need for quadratic equation or Pythagorean theorum that way.
Let t be a side of the equilateral triangle ∆EBF, s be a side of the square ABCD, and r be the radius of the green circle. By observation, as ∆EBF is equilateral with a shared vertex at B and further vertices E and F on AD and DC respectively, ∆EAB and ∆FCB are congruent and FC = AE = 2. ∆EAB: a² + b² = c² AE² + AB² = EB² 2² + s² = t² t² = s² + 4 ∆EDF: ED² + DF² = EF² (s-2)² + (s-2)² = t² t² = 2 (s-2)² = 2s² - 8s + 8 EB = EF so EB² = EF²: s² + 4 = 2s² - 8s + 8 s² - 8s + 4 = 0 s = {-(-8) ± √[(-8)²-4(1)4]}/2(1) s = [8 ± √(64 - 16)]/2 s = (8 ± √48)/2 = 4 ± 2√3 4 - 2√3 < 2 so s = 4 + 2√3 Let BD be both a diagonal of ABCD and the diameter (and thus 2r) of green circle, passing through center point O. ∆DAB: DA² + AB² = BD² s² + s³ = (2r)² 2 (4 + 2√3)² = 4r² 4r² = 2(16 + 16√3 + 12) = 32 + 32√3 + 24 r² = 8 + 8√3 + 6 = 14 + 8√3 Area of green circle: A = πr² = (14+8√3)π ≈ 87.51
Similar to other postings. ABE = (90 - 60) / 2 = 15 degrees. Tan 15 = 2 / AB. AB = 2 / tan15 = 7.4641. Triangle ABO. ABO = 45 degrees. Sin 45 = AO /AB. AO = sin 45 x AB. AO = sin 45 x 7.4641. .AO = 0.7071 x 7.4641 = 5.2779. Area = Pi x 5.2779 x 5.2779 = 87.51.
Yay! I solved the problem. I calculated angles, and then used a trig function instead of the quadratic formula in the video. Tan(15) = 2/x. Looking online, I found that Tan(15) = (√3 + 1) / (√3 -1), so (√3 + 1) / (√3 -1) = 2/x. Simplifying and multiplying by the denominator conjugate gives x = 4 + (2)(√3), as in the video. After I solved for x, my solution was then the same as given in the video.
Thank you. I went a little different. I called the square's sides x+2. This made the triangle sides x*sqrt(2). However, under this method, the triangle sides could also be the square root of ((x+2)^2 + 2^2) . It all got a bit unwieldy and I employed a calculator part way through, but I got there eventually.
Everybody had my idea, to figure out the angles and use the Trigonometric Tangent function to get x, the square side then multiply by the square root of 2 to get the diagonal/ diameter and divide by 2 to get the radiuto use to find the area of the Circle. I find that much easier than working with the quadratic equations and identities. And really fewer steps.
AE = 2 and ∠AEB = 75° and ∠ABE = 15°⇒ x = 2 sin 75° / sin 15° ⇒ x = 2 [(1+√3)/(2√2)] / [(√3-1)/(2√2)] = (1 + √3)² Therefore, the radius of the circle is (1 + √3)² / √2 = √2 (2 + √3) = 2√2 + √6 and its area is 2π (7 + 4√3) square units
I see that quite a few others used the tan(15°) (= 0.267949) function to find side of □. Divide 2 BY that and we get 7.464102. Then the diagonal as you say, divided by 2 to get radius. Then π𝒓² to get 87.5135. All in all, a good exercise in trig and geometry.
