What is the area of the Circle? | (Step-by-step explanation) |

[@bigm383]

Thanks, Professor!❤😁👌

[@PreMath]

You are welcome! ❤️🌹 Thanks for your continued love and support!

[@Okkk517]

Using analytic geometry, take point D(0,0) then applying the intersecting chord theorem 9y=12*6 we find y=8 where y is the length of the segment between point C and the point of the intersection of line BC with the circle. Now we can easily deduce that point O(9,1/2) from which we can find R=5*sqrt(13)/2.

[@predator1702]

Fantastic explanation 👍, thank you teacher 🙏.

[@Uber308]

Take triangle ABD. AB=6 DB=15 AD= Sqrt(117) S(ADB)= 27 . r = (AB * DB * AD)/ 4 S

[@mdhassibulhassansshanto9448]

yes,,, applying the formula of radius of Circumcircle.

[@ybodoN]

Great solution using only the given labeled points 🤩

[@alexniklas8777]

did the same

[@rey-dq3nx]

We can also use the formula 4r²= (√a²+b²+c²+d²) after obtaining the values of a,b,c,d where a=9 b=8 c=12 d=6 4r²=325

[@alster724]

Wow, that's amazing! Cheers from The Philippines 🇵🇭

[@ybodoN]

Extend BC to E on the circle. Extend DC to F on the circle. Since ∠ABC is 90°, AE is a diameter (d) of the circle. Since ∠BCD is also 90°, DC is parallel to the chord AB. And the perpendicular that bisect AB is also a diameter. So CF = DC − AB = 6. By the chord theorem CE = DC ⋅ CF / BC = 8 ⇒ BE = 17. By Pythagoras AE² = AB² + BE² = 325. Since the area of a circle is ¼ π d², the final answer is ¼ π 325 square units or approximately 255.25 square units.

[@Waldlaeufer70]

x = C to South 9x = 6 * 12 = 72 x = 72 / 9 = 8 Therefore, O is 0.5 north of DC and 3 west of BC r² = (9 - 0.5)² + 3² r² = 8.5² + 3² r² = 72.25 + 9 r² = 81.25 A(circle) = r² π = 81.25 π ≈ 255.3 square units

[@alexbidiak2057]

the same, and the square will be 325π

[@tombufford136]

At a glance, Rotate The blue area about O, 90 degrees then DC = BC. Then DO^2 - 81 - 9 = 0 and DO^2= R^2 then R^2 = 72 and the area of the circle = PI * 72.

[@jnhrtmn]

I went by chord length DB=15 and drew that triangle from the mid point, 7.5, to the center O with one horizontal side =3. The normal angle to DB is the ∠DBC=53.13. The vertical component length from the mid point of DB to center O is just trig from angle 53.13 and side 3. I got about 4 and added that to the 4.5 (1/2 BC) Then by ∠BEO, 3^2 + 8.5^2 = r^2

[@johnbrennan3372]

Chord AB and the diameter that cuts it at right angles have the product of their parts equal. So (r-9+x)(9-x+r)= 3.3=9. Then [r-(9-x)][r+(9-x)]=9. So r^2-(81-18x+x^2)=9. But r^2=x^2+81 so x^2 +81-81+18x-x^2=9 giving x=1/2. It follows that r^2= 81.25. Then area of disc is 81.25pi.

[@DDX01]

Thanks sir I have solved it by intersecting chordes theorem😊

[@PreMath]

Great 👍 ❤️

[@mathbynisharsir5586]

Very Very nice video sir 👍

[@MrPaulc222]

I hit a couple of dead ends but ended up with 3^2 + (17/2)^2 = r^2. 9 + 289/4 = r^2 36/4 + 289/4 325/4 = r^2 (325pi)/4 = 81.25pi Approx 255.255 un^2 5*sqrt(13)/2 = r = 9.014 (rounded).

[@murdock5537]

Nice! sin⁡(β) = BC/BD = 3/5 → cos⁡(β) = 4/5 → sin⁡(2β) = 24/25 → cos⁡(2β) = 7/25 CD = 12 = 6 + 6 = DF + CF → DR = DC + CR = 12 + 6 → BR = 3√13 → 117 = 2r^2(1 - cos⁡(2β)) → r = 5√13/2 → πr^2 = 325π/4

[@Copernicusfreud]

Yay! I solved the problem.

[@soli9mana-soli4953]

If you like doing calculations there is also the solution on the Cartesian plane. Find the equation of the circle passing through the points O (0,0), B(12,9) A(6,9) knowing that r=√(a²/4+b²/4-c) 🙃

[@StephenRayWesley]

180°-141)= 0√49.7^7 (x+7x-7) (360°141) =√ 219. √ 3√^7 3^3 √1^√1. (x+3x-3)

[@giuseppemalaguti435]

L'equazione risolutiva è...3=rsin(arctg12/9-arccos√(9^2+12^2)/2r)=rsin(arctg4/3-arccos7,5/r)...r^2=25+7,5^2=81,25...A=π81,25

[@brianmcguigan4785]

Why not use coordinate geometry and the formula for a circle and solve it algebraically?

[@michaelgarrow3239]

9.014 radius.

[@ybodoN]

Bonus question: what is the area of the blue shaded region?

[@leaDR356]

area of whole circle - the yellow part

[@ybodoN]

@@leaDR356 the best answer so far! 😃

[@ybodoN]

First, calculate the area of the trapezoid ABCD: 9 (6 + 12) / 2 = 81. Calculate the chord (AD) then use the radius to find the apothem. By bisecting AOD, we got a pair of 3:4:5 triangles. So, sin θ = 24/25. So the area of the circular segment is 325/8 (sin⁻¹ (24/25)-24/25). Add and simplify. Total area: ⅛ (336 + 325 sin⁻¹ (24/25)) ≈ 94.2845.

[@Marcus-y1m]

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[@MOHAMMED-ox5yw]

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[@PreMath]

Our message to the whole world is loud and clear: peace and love! Peace on Earth 🙏

[@adept7474]

No need to complicate things. S(ABCD) = 81. S(BCD) = 54. S(ABD) = S(ABCD) - S(BCD) = 27 = (AB × BD × AD)/4r = 6 × 15 × √117. r = 5/6 × √117. r² = 325/4.

[@JSSTyger]

A = 81.25π

[@himadrikhanra7463]

( 9 - x )^2 + x^2 = r ^2 15.... 6..... 9..... Wait...have to draw digram but don't have pen & paper right now!