The lines from the top right corner to the points of contact with the circle are both tangent and 5 cm long (-> square), and therefore, the 90 degree angle of that corner gets split into the 30° and two identical angles α & α, which must be 30° each (α + α + 30 = 90). Taking the triangle betwen the top right corner, the point of contact of the circle with the top, and the centre, we get that the radius must be 5 tan(30). That's it.

The law of cosines works ; but , it requires extra work , that is not really required , to solve this problem . Once we come up with the 30,60, 90 triangle BQO with long leg QB = 5 and short leg QO =r , we can get r =5tan30 = 5/sqrt3 . We then have the area of the circle is A=pir^2= 25pi/3 .

Sir you don’t need trigonometry, triangle OBQ is a 30,60,90 degree right triangle in which we know the major cathetus =5

1. AP = AQ (by tangents from a point.) 2. QB = 5 (since ABCD is a square so that AB = AD.) 3. Label the center of the circle O, and construct right triangle △OQB (by perpendicular to the tangent point.) 4. Observe that ∠ABT is 60° (since ∠TBC is 30° and ∠ABC is 90°.) 5. Further, observe that OB bisects ∠ABT (inscribed circle) resulting in ∠OBQ = 30°. 6. Therefore, △OQB is a 30-60-90 right triangle, with QB (5) as the longer leg, and OQ (the radius of the circle, R) the shorter leg. 7. In a 30-60-90 right triangle the length if the shorter leg is equal to the length of the longer leg divided by √3. 8. Therefore, R = 5/√3. 9. And, finally, the area of the circle is: πR² => π(5/√3)² => (25/3)π So, area = (25/3)π Now to watch the video to see if I got it.

@skwest

2 ай бұрын

Got it! Thanks for the challenge.

|AP| = |AQ| and |AD| = |AB| => |BQ| = |DP| = 5 |angle OBQ| = 1/2 x 60 (degrees) = 30 (degrees) OBQ rectangular triangle => |OQ|/|BQ| = tg(30 degrees) => r/5 = sqrt(3)/3 => r = 5/3 x sqrt(3) Therefore P = Pi x r^2 = Pi x (5/3 x sqrt(3))^2 = 25/3 x Pi

Note: You missed the obvious at 6:20. Triangle(QOB) ~= Triangle(BOS) implies Angle(QBO) ~= Angle(OBS) = 30 degrees implies r/5 = 1/sqrt(3) implies r = 5 /sqrt(3) implies A(circle) = 25/3 * pi.

O centro del cerchio...si può scrivere OT^2 in due modi ....5^2+((r+5)-(r+5)tg30-r)^2 e r^2+((r+5)/cos30-5)^2..eguaglio le due espressioni e risulta r=5/√3

AB=AD cause ABCD is square => AQ+QB = AP+PD => QB = PD => QB =5 (AP=AQ Theorem - The lengths of tangents drawn from an external point to a circle are equal) In orthogonal triangle BOQ, QB=2•OQ. Pythagoras theorem in Δ BOQ : OQ²+QB²=0B² => R²+5²=(2R)² => R²=25/3 So area of the circle = πR²=25π/3

@Ray-qb7tk

Ай бұрын

Simple.Meanwhile, "His Excellency blew a fuse after one hour of abstruse reasoning and was quartered in the University hos"..

AD=r+5=AQ+QB. ; QB=5=BS. Si O es centro del círculo; los ángulos QBO=OBS=(90-30)/2=30° ; QO√3=QB ; r√3=5 ; r=5/√3 Área del círculo=25π/3. Saludos.

Я так и не понял. для чего провели ВТ. Для чего эти лишние построения, если можно построить треугольник POD с углом PDO =30 градусов, из которого видно, что OP=R, a OD=2R, откуда 3R^2=25, a R^2=25/3. Area =pi*23/3!

BQ=5, so BS=5(tangents from same point), in triangle BOS, angle OBS=(90-30)/2=30, angle BOS=60°, tan60°=BS/r=5/r=√3 , r=5/√3, area circle π×25/3 sq.units

Menggunakan trigonometri lebih mudah. *Karena ABCD adalah persegi maka AB=BC=CD=AD dan AP=AQ=OP=OQ=r maka PD=BQ=5 *karena sudut ABC 90° dan sudut TBC= 30° maka sudut OBQ=sudut OBS=sudut TBC=30°. *perhatikan segitiga OBQ dengan siku-siku di Q dan sudut OBQ=30°. *dengan trigonometri; tangen 30°=OQ/BQ ==> √3/3=r/5 ==> r=5√3/3. *luas lingkaran =26,17

Si ∆OSB ~∆OQB and

BQ=PD=5. Let's draw OB, triangles QOB and OSB are congruent with 30-60-90 degrees angles, BQ=√3*r -> r=5/√3

Join OB Then angle qbo=obs=30 r = 5 * tan 30 = 5 / sqrt(3)

Let O be the center of the circle. Look at the triangle formed by SBO. One can derive sin(alphabar/2) = r /sqrt(H^2 +r^2) where H =5, alphabar = pi/2 - pi/6. Then r = H tan(alphabar/2); So pi r^2 = 25 pi/3

@jimlocke9320

2 ай бұрын

Yes, I did basically the same with congruent ΔBQO. It is a special 30°-60°-90° right triangle. Side BQ is √3 times as long as OQ, so r = 5/√3. Area of circle equals π r², or 25 π/3.

BS, radius is 5/sqrt(3) > 2.5, so the circle will expand outside of the square. In fact, radius is 5/(1+sqrt(3)).

@Grizzly01-vr4pn

2 ай бұрын

DP = 5 AD ≠ 5 AD = r + 5 Don't worry, I made the same mistake when I first tried the problem.

BQ=5, BO=2r, so r^2=25/3, it is much much easier, right?

Что за чушь, радиус получился 5/sqrt(3) > 2.5, так что окружностьвылезет за пределы квадрата. На самом деле, как легко посчитать куда более простым способом, радиус равен 5/(1+sqrt(3)).

This is impossible, r = 2,89, but square is 5×5. R = 1,82

@Grizzly01-vr4pn

2 ай бұрын

Yes, I made the same mistake as you initially by also not reading the question properly, and took the side length of the square ABCD as = 5, when it obviously isn't. However, if you actually do take AD = 5 (even though it isn't), you end up with r = (5√3 - 5)/2 ≈ 1.83 which is in agreement with the decimal approximations you give.

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