Thank you

Thank you

Fantastic solution! 👏👏👏👏👏

Sir why do we use there thales theorem.... Though you make a new question by marking circle from the side,if the question is just like before where height and Base given it must be 120 only??is it right?if not please correct me🥲

NICE

Won’t the sums of the blue and yellow triangles alway be 1/2 of the square? 126/2 is 63. 63 minus 35 is 28.

sorry sir, but exercising my old brain, i solve this problem using area between curves and integrals and i found an area of 4.3636 then i divide the areas in triangles finding the same result, please can you check where is the mistake, thanks a lot

I did a coordinate system solve. Point A set to (0,0) so 2 triangle points are C(0,3) and F(2,5). y=mx+b for each sloped line FD and CB: y = -3/7x + 3 and y = -5x+15. I then set them equal to each other to find the intersection, Point E coordinate = (21/8,15/8) Then used the coordinate formula: A = (x1y2 + x2y3 + x3y1 - x1y3 - x2y1 - x3y2)/2 Area = 15/4 = 3.75

I could only find a Geometrical Solution with Analytic Geometry. 1) Two Equations: a) Straight Line DF : y = - 5x + 15 b) Straight Line Bc : y = - (3x/7) + 3 Solutions : x = 21/8 and y = 15/8 And those are the Coordinates of the Point of Intersection E The Perpendiclar Distance from this Point E (21/8 ; 15/8) to Line CF (y = x + 3) is equal to (15*sqrt(2))/8 Base = 2*sqrt(2) : Length of Line CF. Height = 15*sqrt(2)/8 Area = (2*sqrt(2)) * ((15*sqrt(2))/8)/2 Area = (30 * 2)/16 Area = 60/16 ; Area = 30/8 ; Area = 15/4 ANSWER : Area of Blue Triangle is equal to 15/4 Square Units or 3,75 Square Units.

Allow me to present the stupidest possible approach, which also works! I'm very sorry I went down this rabbit hole. I intended to find the location of point E, using X,Y coordinates as if this were a graph. With this information, I could calculate the lengths of the three triangle sides, then find the area with Heron's formula. What could go wrong? 1) I make equations for CB and FD in x,y coordinates: y = 15 - 5x for FD and y = 3 - 3x/7 for CB. 2) Use the two equations to solve for x: 15 - 5x = 3 - 3x/7. Multiply all by 7: 105 - 35x = 21 - 3x. 84 = 32x or x = 21/8. Not too ugly, right? 3) Plugging x into the first equation: y = 15 - 5 * 21/ 8. y = (120 - 105/8. y = 15/8. 4) Use pythagorus to find CE and EF lengths. For CE, delta x = 21/8 and delta y = 9/8. Sum of squares is 441/64 + 81/64. Result: sqrt(522)/8. For EF, delta x = 5/8, delta y = 25/8. Sum of squares = 25/64 + 625/64. Result is sqrt(650)/8. 5) We already know CF is 2 * sqrt(2). So now plug this mess into Heron's formula. None of the square roots are reducible. I had to give up and just use Excel to solve it! The result is indeed 3.75. But why? There is no benefit and lots of computational complexity in this way.

Do we assume that DC is parallel to AB ?

Your explanations are very clear. Thank you so much!

CEF = ADFP - CFP - ABC + DBE = 12,5 - 2 - 10,5 + 15/4 = 15/4. Only DBE was a bit more tricky to calculate, the rest are very simple. I set point A as an origin in a coordinate system and calculated the coordinates of point E where two lines cross. That gives the height of DBE and the rest is easy. Fun fact, CEF = DBE.

Hello sir..I m from India..and watch your videos...I really like it and encourage others to watch

If you put the value and find the angle of DCE Then it will be much easier

I got lost on this explanation halfway thru the video, so Im glad I didnt miss a simpler solution. I used coordinate geometry and found the coordinates of: C (0,0) F (2,2) and E (21/8,-9/8) It took 3/4 of a page of calculations, but I got there. What a chore this one was. Thanks for the mental gymnastics!

Just one more solution: 1. Let us drop a perpendicular from point F to [AB]. Let its intersection with CB be K and M - with AB 2. Area of the Blue triangle = KF*2/2 + KF*h/2 (1) 3. Let A(0;0) 4. Equation of CB y =-3/(3+4)*x +3 = -3/7*x+3; (2) equation of FD y = -(3+2)/1*x +15 = -5*x+15.; (3) 5. Let us put x=2 into (2) : MK = y(2) = 15/7; 6. KF = MF - MK = (3+2) - 15/7 = 20/7; 7. x coordinate of E {(2),(3)} => -3/7*x+3 = -5*x +15 => x = 21/8; 8. h = 21/8 - 2 = 5/8; 9. Let us "fill in the blanks in" (1): A(CFE) = KF*(2/2+ (5/8)/2) = 20/7 * (1+5/16) = 15/4 = 3.5 sq.u.

Drop a perpendicular from E to AC and label the intersection G. Let EG = x. Note that ΔABC and GEC are similar. From ratio of sides, CG = 3x/7. Because AG + CG = AC = 3, AG = 3 - 3x/7. Drop a perpendicular from F to AD and label the intersection H, and drop a perpendicular from E to AD and label the intersection J. Note that ΔFHD and EJD are similar. FH = 5 and HD = 1. EJ = AG = 3 - 3x/7, so DJ = (1/5)(3 - 3x/7). AJ = EG = x, so x + (1/5)(3 - 3x/7) = 3. Doing the algebra, x = 21/8. Construct a line through C parallel to AB. Lebel the intersection with FH as M and FD as N. Note that ΔFMN and FHD are similar and we find MN = 2/5, so CN = 2 + 2/5 = 2.4. ΔFCN has area (1/2)(2.4)(2) = 2.4. ΔCEN has area (1/2)(2.4)(3x/7) = (1.2)(9/8) = 1.35. Blue area = sum of areas of ΔFCN and ΔCEN = 2.4 + 1.35 = 3.75, as PreMath also found.

Very nice problem I tried to solve with one formula: The answer is: ■ green area: (338·(π/2) - 338)/2 + 338/2 = 265.46 cm² ~~~~~~~~~~~~~~~~~~~~~~~~~~ Explanations (step by step): • 338·(π/2) is the circle area (*) • 338·(π/2) - 338 is the area of the 4 circular segments • (338·(π/2) - 338)/2 is the area of 2 circular segments • 338/2 is the half of the square area (**) • (338·(π/2) - 338)/2 + 338/2 is the green area (final answer) ■ green area = 265.46 cm² ----- notes ----- (*) property: • the area of a circle is π/2 (≈ 1,570796) multiplied by the area of its inscribed square (**) property: • the area of ​​a triangle inscribed in a square, having as base one of the sides of the square, is equal to half the area of ​​the square 🙂

Thanks. It can be solved another way. Area of Trapezoid = area of triangle + area of Parallelogram.

Second method: 1. CH ⟂ AP, FK ⟂ CH, M(CH ∩ FK). 2. FM/MH = FK/KD = FK/(AD - PF). 2/MH = 5/1, MH = 2/5, CH = 12/5. 3. CHBD - trapezoid. CH/BD = CE/BE, BE = (5/8)BD. 4. S(ABE) = (5/8)S(ABC), S(DBE) = (4/7)S(ABE). 5. CFBD - trapezoid. S(CFE) = S(DBE) = (21/2) × (5/8) × (4/7) = 15/4 = 3,75.

1/ AP and DF intersect at point I and draw a parallel line from C with AD intersecting FD at point M. Label IP= a and CM=x We have a/(a+5)= 2/3--> a= 10 2/ CM/AD = IC/IA-->12/15= 4/5 x/3= 4/5--> x= 12/5 So the area of triangle FCM = 1/2 . 2.12/5= 12/5 (1) 3/ The 2 triangles CEM and BED are similar. Just name the heights from E to CM and AD as h and h’ We have h/ h’ = x/4= (12/5)/4= 12/20=3/5--> h/3=h’/5 = (h+h’)/(3+5) = 3/8--> h=9/8 Area of the triangle CME= 1/2 . 12/5 .9/8= 27/20. (2) Area of triangle FCE= (1) + (2)=12/5 + 27/20= 15/4 sq units

Very nice❤❤

It is difficult. 😢

coord geometry

PC=2 CA=3 CG=2+2/5=12/5 CG : DB = 12/5 : 4 = 12 : 20 = 3 : 5 3*3/8=9/8 area of the Blue triangle : △CFG+△CEG = 12/5*2*1/2 + 12/5 * 9/8*1/2 = 12/5 + 108/80 = 2.4 + 1.35 = 3.75

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Let's find the area: . .. ... .... ..... According to the sketch we can assume the following coordinates: A: ( 0 ; 0 ) B: ( 7 ; 0 ) C: ( 0 ; 3 ) D: ( 3 ; 0 ) E: ( xE ; yE ) F: ( 2 ; 5 ) P: ( 0 ; 5 ) The lines CB and FD can be represented by the following functions: CB: y = (yB − yC)/(xB − xC)*(x − xB) = (0 − 3)/(7 − 0)*(x − 7) = (−3/7)*(x − 7) DF: y = (yD − yF)/(xD − xF)*(x − xD) = (0 − 5)/(3 − 2)*(x − 3) = −5*(x − 3) E is the point of intersect of the lines CB and FD: (−3/7)*(xE − 7) = −5*(xE − 3) 3*(xE − 7) = 35*(xE − 3) 3*xE − 21 = 35*xE − 105 84 = 32*xE ⇒ xE = 84/32 = 21/8 yE = (−3/7)*(xE − 7) = (−3/7)*(21/8 − 7) = (−3/7)*(21/8 − 56/8) = (−3/7)*(−35/8) = 15/8 yE = −5*(xE − 3) = −5*(21/8 − 3) = −5*(21/8 − 24/8) = −5*(−3/8) = 15/8 ✓ Now we are able to calculate the size of the blue area: v(CE) = ( xE − xC ; yE − yC ) = ( 21/8 − 0 ; 15/8 − 3 ) = ( 21/8 ; −9/8 ) v(CF) = ( xF − xC ; yF − yC ) = ( 2 − 0 ; 5 − 3 ) = ( 2 ; 2 ) A(CEF) = (1/2) * |v(CE) x v(CF)| = (1/2) * |(21/8)*2 − 2*(−9/8)| = (1/2) * |21/4 + 9/4| = (1/2) * (30/4) = 15/4 Best regards from Germany

Спасибо за интересные задачи!

Please give heart 😊😊😊

First comment and first like, can you pin it?

21..?

Thanks ! What is the app you use?

but this method doesn't work when all coefficients are coprime.

Thanks sir, I wanted to ask if there were any limitations to the formula you showed

What value is the variable X for?

why 8x and 72x?

(7X-5)(X+2)=X(5X+1) ; X=1 ; Longitud de la cuerda vertical =2v; v=(2+3)/2=5/2; su distancia al centro es b=[(1+6)/2]-1=5/2=v; Radio =r =(5/2)√2=5√2/2. Gracias y un saludo cordial.

I solved this problem in 3 seconds. Who's faster? 😄

Ooooòoooooooyoooooooooo

Thanks alot sir 💪😬❤

Thanks alot sir 💪😬

Method 2 is faster

This was helpful. Thank you

By Intersecting chords theorem, AE•EB = CE•ED. AE•EB = CE•ED x(5x+1) = (x+2)(7x-5) 5x² + x = 7x² + 14x - 5x - 10 2x² + 8x - 10 = 0 x² + 4x - 5 = 0 (x+5)(x-1) = 0 x = -5 ❌ | x = 1 AE = 1 EB = 5(1) + 1 = 6 CE = 7(1) - 5 = 2 ED = (1) + 2 = 3 As AB and CD are chords, any radii that perpendicularly intersect the chords will do so at their midpoints. Draw radii OM and ON perpendicular to AB and CD respectively. The intersection points will be P and Q respectively. As AB = 1+6 = 7, AP = PB = 7/2. As CD = 2+3 = 5, CQ = QD = 5/2. As AE = 1, EP = OQ = 7/2-1 = 5/2, and as CE = 2, EQ = OP = 5/2-2 = 1/2. Triangle ∆OPB: OP² + PB² = OB² (1/2)² + (7/2)² = r² r² = 1/4 + 49/4 = 50/4 = 25/2 r = √25/2 = 5/√2 = (5√2)/2

tan x = 4/3 tan y=7 tan (x+y) = (tanx + tany)/1-tan x*tan y tan (x+y) = (4/3+7)/1-7*4/3 tan (x+y) = (25/3)/-25/3 tan (x+y) = -1 x+y = arctan (-1) since x and y are acute angles so x+y<180 x+y=135º

The hardest part of solving geometry problem is to find a point like P.😂😂😂

The general formula for any trapezoid is 2s(s-x)(s-y)(s-B+b)÷(B²-2abB+b²)*B Where s=(x+y+B-b)÷2 and x,y are the 2 nonparalel lines and B is the biger base and b is the smaller one

(7x-5)*(x+2)=x(x+1), x^2+4x-5=0, x=1. Letёs draw perpendicular OF to CD, OF=AB/2-x =3,5-1 = 2,5. CF=CD/2=2,5, OC = r = \/2,5^2+2,5^2=2,5\/2=3,53553390592...!