Very difficult for most students

[@MarieAnne.]

Here's a faster way to calculate CQ and CR: Extend line DC to point Q on circle. Extend line BC to point R on circle. Since AD and BC are parallel, then so are chords AD and BR Let M be midpoint of AD and N be midpoint of BR Perpendicular bisector of a chord passes through center of circle. Perpendicular bisector through M is perpendicular to AD and perpendicular bisector through N is perpendicular to BR. But since AD and BR are parallel, then so are their perpendicular bisectors. But both perpendicular bisectors must pass through center of circle, so these two perpendicular bisectors are actually the same (otherwise they would have no point in common), and this singular bisector of AD and BR passes through M and N and is perpendicular to AD and BR. So DMNC is a rectangle, with DM = CN Since M is midpoint of AD, then DM = 4/2 = 2 Therefore, CN = 2, BN = BC - CN = 8-2 = 6 Since N is midpoint of BR, then RN = BN = 6, and CR = RN - CN = 6 - 2 = 4 Now chords BE and DQ intersect at C with BC = 8, CE = 4, DC = 6, CQ = ? By intersecting chord theorem we get: BC * CE = DC * CQ 8 * 4 = 6 * CQ CQ = 32/6 = 16/3 Triangle ADQ has right angle at D, so AQ must be diameter of circle (since angle subtended by arc AQ at center of circle = twice angle subtended by arc AQ on circle = 2 * 90 = 180). Using Pythagorean theorem, we get: AQ^2 = (2r)^2 = AD^2 + DQ^2 = 4^2 + (6 + 16/3)^2 4r^2 = 4^2 + (34/3)^2 = 16 + 1156/9 = 1300/9 r^2 = (1300/9)/4 = 325/9 Area of circle = πr^2 = 325π/9

[@santiagoarosam430]

Prolongamos BC y DC hasta que corten a la circunferencia en E y F respectivamente → Por simetría respecto al diámetro horizontal: CE=CB-DA=4 → Potencia de C respecto a la circunferencia: DCxCF=ECxCB → 6CF=4x8 → CF=16/3 → DF=6+(16/3)=34/3 → Diámetro circunferencia = AF → 4²+(34/3)²=AF² =1300/9 → Área círculo =Pi[(1300/(9x4)] =Pi(325/9) =113.4464 Gracias y un saludo.

[@mathswan1607]

BD=10 and AB=2sqrt(13) By sine Formula 2R=10/sin DAB=10/(6/2sqrt(13))=10sqrt(13)/3 Hence R=5sqrt(13)/3 AreA of circle =325 pi/ 9

[@AajanmaHindu]

That means, use of Trigonometry is involved, he is doing it using only property of circles in plane geometry.

[@sarantis40kalaitzis48]

CONGRATULATIONS,but first you must prove that angle DAB + angle ABC= 180 degrees,from parallels DA and CB so sin(DAB) = sin(ABC). Then create right triangle AEB , where AE is perpenticular to CB and angle (ABC) is THE SAME angle (ABE) . So sin(ABC) = sin (ABE) = AE/AB =6/sqrt(52) =6/(2*sqrt 13)= 3/sqrt(13) . THEN 10/ ( 3/sqrt13) = 2R ,LAW OF SINES, and R=5/ (3/sqrt 13 )= 5* sqrt13/3. Finaly R^2 = (5^2)* (13/3^2) = 25*13/9 = 325 / 9 and AREA of circle = 325/9 * π.

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين

[@andrescarlosrodriguezsanch4401]

I find it easier to see it in the x y plane. Lets say A=(0,0), B=(6,-4) and D=(0,4). Then replace those points in the circumference equation and get the radious of the circle

[@amit1508]

i did the same way. midpoint of ab is (3,-2) eqn of perp bisector of chord AB is y=1.5x-6.5 eqn of perp bisector of chord AD is y= 2 so centre of the circle is (17/3, 2) and radius**2 = (17/3)**2 + 4 = 325/9

[@AajanmaHindu]

That means use of coordinate Geometry is involved, he is doing it using only property of circles in plane geometry.

[@leif1075]

@@AajanmaHindu Why draw those 2 triangles like he does at all though

[@Doonburn]

I agree - the quickest and easiest way to solve the problem is to use coördinate geometry.

[@sarantis40kalaitzis48]

@@amit1508 Very clever method. Congratulations amit.

[@Ni999]

By symmetry, CR = 4 therefore 8*4 = 6 CQ or CQ = 16/3 By Thales, diameter is √( 4² + ( 6 + 16/3 )² ) Radius is half of that and the rest is arithmetic.

[@07Pietruszka1957]

Math is great. I find much simpler resolution. Two rectangles 12 on x and (6-x) on 4 gives two equations: (12-x)^2+4^2=D^2 and 12^2+x^2=D^2 hence x=2/3 and D^2=(144+4/9)

[@isg9792003]

Extend BC to E. CE would be 4 by symmetry. Draw a perpendicular on AD and extend it to BC. Let the pointe be F and G. Join and E with Center of the circle O. Let OF be X and hence OG would be 6-X. By pythagorous theorem, 2^2+X^2= 6^2+(6-X)^2. X=17/3. Using theorem again we get 2^2+17/3^2=r^2. r^2=289+36/9=325/9. Area = Pyex325/9

[@sfratini]

Yet another approach is to use analytic geometry. To get some zeros in the equations, place the segment of length 4 along the y-axis, and the bottom endpoint of the segment of length 8 on the x-axis. This gives you three equations with three unknowns. You will also determine the circle's center in the process.

[@juanalfaro7522]

Here's a faster solution: Extend DC to point Q and let AS be a chord parallel to DQ for S on the right circumference of the circle and passing through G on the BC chord. Since CB=8 and AD=4, then BG = 8-4=4 and CR = BG = 4. Then DC*CQ = BC*CR --> 6*CQ = 8*4=32 --> CR = 32/6 = 16/3. Then we draw AQ which is the diameter of the circle. AQ^2 = AD^2 + DQ^2 --> AQ^2 = 4^2 + (6+16/3)^2 = 4^2 + (34/3)^2 --> to get R we divide everything in the squares by 4 --> R^2 = 2^2 + (17/3)^2 = 4 + 289/9 = (36+289)/9 = 325/9 --> A = Pi*R^2 = Pi*325/9 (so R = 5*sqrt(13)/3)

[@ybodoN]

Notice that AD is parallel to BC. So CR is CB - AD then CQ is CB × CR ÷ CD. Radius of the circle is ½ √(AD² + DQ²). Area of the circle is π ¼ (AD² + DQ²). If AD = a, CD = b and BC = c, then the area A = π ¼ ((b + c (c - a) / b)² + a²). Amazingly, when CD is 9 and BC is 7, we get exactly the same final result.

[@cb_q]

circle circumscribed around a triangle ABD, so we can use formula R = a * b * c / 4 * Striangle. Striangle = 1/2 * a * h, a = AD = 4, h = CD = 6, because ADC is 90 degrees and point B belongs to the line BC. a = AD = 4, b = BD = sqrt(CD^2+BC^2) = sqrt(6^2+8^2) = 10, AB = sqrt(6^2 + (8-4)^2) = sqrt(36 + 16) = sqrt(52) = 2 * sqrt(13) R = 4 * 10 * 2 * sqrt(13) / (4 * 1/2 * 4 * 6) = 80 * sqrt(13) / 48 = 5 * sqrt(13) / 3 Scircle = pi * R^2 = (5 * sqrt(13) / 3)^2 * pi = 25 * 13 / 9 * pi = 325 * pi/9.

[@dainiusb1114]

BC extend to circle line. CE=CB-DA(by symetry). BC*CE=DC*(32/6). (2R)^2=(DA)^2+(DC+32/6)^2

[@dimuthdarshaka7985]

Bisect DA Line center of the circle lies on bisect line say "O" With two Eqn. Can be solved

[@tontonbeber4555]

I hate geometry, so just wrote equations. We know coordinates of 3 points on the circle, so ... (x-a)^2 + (y-b)^2 = r^2 (0,0) a^2 + b^2 = r^2 (0,4) a^2 + (4-b)^2 = r^2 => b=2 (well, expected ^^) (6,-4) (6-a)^2 + (-4-b)^2 = r^2 a^2 + 4 = r^2 a^2 -12a + 36 + 36 = r^2 -12a + 72 = 4 -12a = -68 a = 68/12 = 17/3 289/9 + 4 = 325/9 = r^2 Area = pi 325/9

[@Waldlaeufer70]

The 4 units on the left show that the 8 units must be extended upwards by 4 units to form a chord. The chord theorem says that x * 6 = 8 * 4 must apply. => x = 32/6 = 16/3 (x = extension of 6 to form a chord) According to Pythagoras, the diameter is d² = 4² + (6 + 16/3)² = 16 + 36 + 64 + 256/9 = (144 + 324 + 576 + 256) / 9 = 1300/9 r² = 1300/9 / 4 = 1300 / 36 = 325/9 A(circle) = r² π = 325/9 π

[@jayeshkumar3861]

Can u explain how 4 units on Left implies 8 unit line can be extended by exactly 4 to form a chord. Please

[@jayeshkumar3861]

You meant 8 - 4 = 4 below side => 4 above side too

[@Waldlaeufer70]

@@jayeshkumar3861 Exactly. Due to the given right angles and the symmetry we therefore have, we know that a difference of 4 to the bottom of the circle implies the same distance to the top of the circle as well.

[@jayeshkumar3861]

@@Waldlaeufer70 thank u

[@krishnamoyghosh6047]

It is very easy by applying cosine law related to sides and angle subtended at centre is twice the angle on the same cord.

[@claudeabraham2347]

Very good! I love it!

[@LongZhao-bz9te]

BD²=CD²+BC² BD=10 AB²=CD²+(BC-AD)² AB=2*13^1/2 CosA=AD²+AB²-BD²/2AD*AB SinA=(1-Cos²A)^1/2 BD=2R*sinA Then R=BD/2sinA Area circle=pi*(BD/2sinA)² No calculations

[@thiagaopiaui]

Há um atalho. quando descobriu CQ, vamos ao triangulo ADQ. O segmento AQ correponde a um angulo reto; entao ele é um diâmetro. Calcula-se seu comprimento por pitágoras; o raio é a metade; com o valor do raio, calcula-se o diametro.

[@ThomasLoganRitchie]

Ans: (325/9) × π. Proof: Call A the lower vertex of line 4, B the upper vertex of line 4, C the upper vertex of line 8, D the lower vertex of line 8. Make A=(0,0) B=(0,4) C=(6,4) D=(6,-4) Call K=(x,2) the center of the circle determined by A, B and D, R the radius of this circle, M=(3,-2) the medium point of AD and E=(6,0). Observe that (1) ∆ABC ~ ∆AED, (2) slope of AD = -2/3, (3) slope of MK = 3/2. Hence, 4/(x-3)=3/2 --> x= 17/3 --> R²= (17/3)² + 2² = 325/9. Q.E.D. For a different solution, see the link kzbin.info/www/bejne/hGOto6qBir2Nhrs math, geometry.

[@sarantis40kalaitzis48]

POST SCRIPT. I saw other comments and MarieAnne and i solve it almost the same !!!!!! But you can see my comment to 温海濤 , who has done great job. If you bring perpendicular from the midle M of DA it passes from the midle N of the parallel chord RB.From rectangle DMNC CN=DM=2. So NB = CB - CN = 8 - 2 = 6. RN = NB = 6 and RC = RN - CN = 6 -2 = 4. From intersecting chords theorem we have DC*CQ = RC* CB so 6* CQ = 4*8 = 32 or CQ= 32/6 or CQ =16/3. The inscribed angle ADQ = 90 degrees and the chord AD must be diameter of the circle, so AQ = 2R. PYTHAGORAS THEOREM to triangle ADQ gives AQ^2 = AD^2 + DQ^2 or (2R)^2 = 4^2 +(6+(16)/3)^2 = 16 + (34/3)^2 = 16+ (1156/9) = (144/9)+ (1156/9) = 1300/9 so 4* R^2 = 1300/9 or R^2= 1300/36 (or R = sqrt (1300) / 6 ) , but we need R^2 for the area of circle. FINALY AREA of circle = π * R^2 = π * 1300 /36 = (325/9) * π , (simplyfing 1300 and 36 by 4 each one).

[@blackranger8775]

Interesting! But not really difficult, i can just find the circumradius by creating ABD triangle, and then just simply find every side of the triangle, and then find the circumradius, square it and multiply it with pi

[@thatweakpowerlifter2515]

Learnt some new today, thanks bro.

[@aaradhanagoodnewsavm4881]

From Where do you getting this problems sir

[@yakupbuyankara5903]

(325/9)×(3,14)

[@daddykhalil909]

18:42 where from you got the formula for 4R^2??

[@MathBooster]

Watch this video kzbin.info/www/bejne/foXceaJjq5eZl9k

[@daddykhalil909]

@@MathBooster noted but not acceptable at all. You cannot base the solution of a new problem on the solution of an old one considering this method as taken for granted. Sorry

[@MathBooster]

in that problem I have proved the formula. Otherwise you need to remember the formula because it is very basic formula of circle. [Basic formulas are supposed to be used directly without proving, for example formula for (a+b)³]

[@manjunathkulkarni2828]

Nice👌👌🙏

[@MathBooster]

Thank you 🙂

[@rogerphelps9939]

Your answer is excruciatingly complicated and long winded.. Use the facts that the perpndicular through the mid point of a chord goes through the centre of the circle, the line between the ends of two orthogonal chords that meet at a point is a diameter of the circle and the rectangle property of a circle.

[@uiopazerty9471]

There is easier.