Find the shaded area in the circle | 2 Different Methods

[@DocumentingBTC]

great content , in which program do you create the thumbnails?

[@jimlocke9320]

At 10:20 we have 13² = h² + x² and 15² = h² + (14 - x)². The second equation expands to 15² = h² + 14² - 28x + x² or 15² = h² + x² + 14² - 28x. Replace h² + x² by 13²: 15² = 13² + 14² - 28x. Solve for x and find x = 5. From 13² = h² + x², we find h = 12. So, tan(

[@0ZkYtEKE]

I’m a big fan of the second method because it doesn’t rely on remembering complicated formulas.

[@Ramkabharosa]

A copper atom, a silver atom, and a gold atom walk into a night-club. The club owner says, “Eh you, get out of here.” Question #1: Who had to leave the club? After the first atom left, a second one followed it. Question #2: Who was the second atom - and what did it say when it was leaving? The club owner says, “ ‘E Ji, you can stay.” Question #3: Who was the third atom? -------------------- This is a Chemistry joke. " 'E you " sounds like "Au" and Au is the Chemical symbol for gold in Chemistry ( 'Aurum' means gold in Latin). So the first atom was the Gold atom. Also, the Copper atom said "See you" and left with the Gold atom because Cu sounds like 'See you' ('Cuprum' means copper in Latin.) So the second atom was the Copper atom The Chemical symbol for silver is "Ag" ( 'Argentum' means silver in Latin ) & Ag sounds like " 'E ji " - that is why silver is always treated with respect. So finally, the third atom was the Silver atom. .

[@RobertHering-tq7bn]

Hi, following my way to solve the problem and again using lines in (x,y) with D being the point on line through A and B that is on the line through C and being perpendicular to the line through A and B. Furthermore, F is the point in the middle of AC. First we easily see with Heron's theorem that [ABC] = 84 . Next we see that h = CD is 12 as we have (1/2)*14*h = 84 . Now we look for the coordinates of important points... A(0,0), D(dx,0), B(14,0), F(fx,fy) . Based on what we have already, we see dx² = 13² - 12² = (13+12)(13-12) = 25 ==> dx = 5 . So, the line through A and C has a gradient value of 12/5. We also know now that fx = 5/2 and fy = 6. With O(ox,oy) being the center point of the circle, we have in addition a gradient value of -5/12 for the line through F and O, as it is perpendicular to AC (as always for secants). For the same reason we know that ox = 7 as AB is 14. For the line through F and and O we have... (y-6)/(x-(5/2)) = -5/12 y = (-5/12)x + 25/24 + 6 = -(5/12)x + (25+144)/24 = -(5/12)x + 169/24 . This means for oy... oy = -(5/12)*ox + 169/24 = -70/24 + 169/24 = 99/24 = 33/8 . For the radius r of the circle this results in r² = 7² + 33²/8² = (49*64 + 1089)/8² = (3136+1089)/8² = 4225/8² = 65²/8² ==> r = 65/8 . And the desired result is... pi*r² - [ABC] = (4225/64)*pi - 84 .

[@marioalb9726]

A = ½bh = ½*14*12= 84 cm² R= a.b.c/4A = 13*14*15/(4*84) R = 8,125 cm A = πR² - 84 = 123,39 cm² ( Solved √)

[@oscarcastaneda5310]

Heron's Formula gives the area of the triangle as 84. Since 14 = 2Rsin(A) and 84 = (1/2)(13)(15)/(14/2R) we have that R = 65/8. This gives the desired area A = (pi)(65/8)^2 - 84.

[@santiagoarosam430]

Perímetro ABC=13+14+15=42---> Semiperimetro =21---> Área ABC =√(21*8*7*6)=84---> Altura AD respecto a BC =2*84/14=12---> BD=√(13²-12²)=5---> 5*(14-5)=12s---> s=15/4---> (2r)²=[(15/4)+12]²+(14-5-5)²= (63/4)²+4²---> r²=4225/64---> Área sombreada tres segmentos circulares ---> (4225π/64)-84 ~123,3942 ud². Gracias y saludos

[@五十嵐特許事務所]

If ∠BAC＝θ, then by the law of cosines, cosθ＝33/65. ∴sinθ＝56/65. By the law of sines, BC/sinθ=2R (R: radius), so 2R=14/(56/65). ∴R=65/8.

[@holyshit922]

Cosine law for cosine of angle Pythagorean theorem for sine Sine law for radius Area is the difference between circle area and triangle area pi/4*(15/sin(B))^2 - 91*sin(B) where sin(B) = 12/13

[@michaeldoerr5810]

The answer is (4225/64)pi-84 units square. I have noticed that the two different methods always require a different circle theorem. Is it possible that ALL circle theorems result in making use of HL similarity? I am just asking.

[@ناصريناصر-س4ب]

The semi perimeter of triangle ABC is (13+14+15)/2=21, and from this the area of the triangle is A=√(21(21-13)(21-14)(21-15))=84. The height of the triangle related to side BC is H=2A/14=12. We have AB*AC=H*2R, and from this R=(13*15)/24=65/8, and from this the area of the shaded region is π(65/8)²-84=(4225/64)π-84.

[@jean-claudedouvry1636]

Dans la 2nde mèthode, la hauteur issue de A est ègale à 2 x surface du triangle / longueur du côté BC, soit 2 x 84 / 14 = 12.

[@ManojkantSamal]

Asked area =123.46 square unit....... May be Explanation ^=read as to the power *=read as square root r=radios of the circle S=semi perimeter of the triangle As per question the sides of the triangle a, b, c are 13,14,15 respectively. So, S=(a+b+c)/2=(13+14+15)/2 =41/2=21 S-a=21-13=8 S-b=21-14=7 S-c=21-15=6 So, S(S-a)(S-b)(S-c)=21×8×7×6 =7×3×8×7×2×3 =(7×7)×(3×3)×(8×2) We know area of the triangle = *{S(S-a)(S-b)(S-c)} So, area of the given triangle =*{(7×7)×(3×3)×(8×2)} =7×3×4=84........eqn1 According to Heron r=(a×b×c)/(4×area of triangle ) So here, r=(13×14×15)/(4×84) =(13×15)/(4×6)=(13×5)/(4×2) =65/8 So, r^2=(65/8)^2=4225/64=66.01 So, the area of given circle = π(r^2)=(22/7)×(66.01) =(22×66.01)/7 =1452.22/7=207.46...... Eqn2 Hence, Asked area= eqn2 -eqn1 =207.46- 84=123.46 (There are numerous methods to get the solution )

[@2012tulio]

R= 8.11 Area of the circle = 207 Area of the triangle = 83.76 Wanted area = 207- 83.76= 123.23

[@murdock5537]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = 13; BC = BM + CM = k + (14 - k) → sin⁡(BMA) = 1 AM = h; AC = 15 → h^2 = 169 - k^2 = 225 - (14 - k)^2 → k = 5 → 14 - k = 9 → h = 12 AO = BO = CO = QO = r; AQ = AM + QM; QM = m → 5(9) = 12m → m = 15/4 → h + m = 63/4 = AQ → AQ/2 = 63/8 = AN = QN → sin⁡(ONQ) = 1; BC = 14 = BS + CS = BC/2 + BC/2 = 7 + 7 → NO = 7 - k = 2 → ∆ NQO → QO = r → r^2 = 4 + (63/8)^2 = (65/8)^2 → red area = πr^2 - 7h

[@ina-j2p]

s=(13+14+15)/2=21 s-13=8 s-14=7 s-15=6 △ABC=√(21・8・7・6) =7・3・4 =84 14/sinA=2R R^2=49/sin^2A cosA=(13^2+15^2-14^2)/(2・13・15) =(169+225-196)/390 =198/390 =33/65 sin^2A=1-33^2/65^2 =(4225-1089)/65^2 =3136/65^2 =(56/65)^2 sinA=56/65 R^2=49・65^2/56^2 =4225/64 ∴πR^2-△ABC=4225π/64-84

[@kateknowles8055]

Thank you for this. The centre of the circle, where OA = OB = OC, will be on the perpendicular bisectors of each side of the triangle . With r = radius , the square of the area of the triangle is the sum of three smaller triangles: [OBC][OBC] = 49(r.r-49] [OAC][OAC]= 225/4 (r.r- 225/4) [OAB][OAB] =169/4(r.r- 169/4) = (half base)^2 . (height)^2 The square of the area of ABC is s(s-13)(s-14)(s-15) also and s is (13+14+15)/2 . it is the semiperimeter for use with Heron's formula. So 21.8.7.6 is this square of the area of ABC It is 84^2. 49r.r - 2401 + 225/4r.r - 50625/16 + 169/4 r.r - (170-1)(170-1)/16 is also 7056 r.r (49 + 225/4 +169/4) = 7056 +2401 + 50625/16 +(28901-340)/16 (reaching for a calculator) (295/2)r.r = 9457 + 79186/16= 115249/8 r.r =97.67approx Area of circle = 306.835 approx ( No it is pi times 4225/ 64 see the video ) The total of the shaded red areas is approximately 307❌ - 84☑ Approximate answer 223 square units. I shall check now and enjoy the video and the comments. My answer was wrong More practice with the calculator or) just remember to use angles that stand in a circle on equal length chords !! I will find out where the height of twelve came from.

[@RealQinnMalloryu4]

(13)^2 (14)^2.(15)^2={169+196+225}=580/360=1.220 1.110 15^5 2^3^2^3 2^1^1^3 23 (CD ➖ 3CD+2).

[@marioalb9726]

sinθ= 12/13= 15/2R --> R= 8,125 cm A = πR² - ½bh A = π*8,125² - ½*14*12 A = 207.39 - 84 = 123,39 cm² (Solved √)

[@marioalb9726]

13-14-15 is a special triangle, that can be divided into two Pitágorean triplet triangles, 5-12-13 and 9-12-15. all sides integer numbers and both right triangles. Hence, the height of the original inscribed triangle, is h=12. perpendicular to side 14