Construct the similar kites DOPC and EOPA. Recognise the second is half the size of the former thus EA is 1/4 the side length of the square. Calculate the side length of the square using 96 = 0.5 * a * 0.75a then calculate the area of the semicircle
@petrdub865011 ай бұрын
Great solution, much more intuitive, however it is quite difficult to spot the similarity.
@jimlocke9320 Жыл бұрын
Solution by tangent double angle formula. Let radius of semicircle = r. Construct OC and OP. Note that
@kimchee94112 Жыл бұрын
The math is simple but logic is head scratching until it's explained in this video as how to apply the two tangent theorem. Thank you, another clear solution.
@User-jr7vf10 ай бұрын
Yea, the two tangent theorem is very helpful in this case.
@marcelowanderleycorreia8876 Жыл бұрын
Beautiful question!
@PreMath Жыл бұрын
Thanks❤️
@georgerobartes2008 Жыл бұрын
I just recognised the area as the result of a multiple of 12 in a 3 ,4 ,5 triangle giving me 16 for the side . Therefore the radius is 8cm and obtained the result using pi r 2 ÷ 2 . Quick and simple
@CrackerJackRed Жыл бұрын
How did you do that? I’ve looked at the diagram, again and again, and can’t still figure out how you can tell the area is a multiple of 13 in a 3,4,5 triangle.
@georgerobartes2008 Жыл бұрын
@@CrackerJackRed Multiple of 12 you mean , I guess a typo there . 8 x 12 = 96 , so for a right angled triangle ( side lengths multiples of 3,4 and 5 for the hypotenuse - a geometric rule ) the side must be 16 as we have the short side of 12 , the radius 8 and the area pi r2 ÷ 2 from mental arithmetic is 100 cm 2 to the nearest makes no difference . I knew the radius was 8 as soon as I looked at the diagram because of the area before confirming it using the basic geometric rule in the 3,4, 5 triangle as I saw the triangle as 12 x 16 ÷ 2 . I used to find economic ways of solving other mathematical problems at school , particularly in "modern math" like modular arithmetic that gave the teachers " ureka " moments , often being unable to explain how I had arrived at the correct answer without showing the method which resulted in failed tests .
@brianscott5008 Жыл бұрын
Thinking out loud, the relationship of the semi-circle area is always pi / 3 of the triangle area and the side of the square is always 1/3 of the triangle area. Never noticed that before. Very well laid out and explained, as your videos always are, thank you.
@PreMath Жыл бұрын
You are very welcome! Thanks❤️
@Stuv017 Жыл бұрын
Side of square is 2x Circle rad is x DC = CP = 2x L(DCA) + L(EAC) = 180 L(DCA) + L(DOP) =180 L(DOP) = L(EAP) DC/DO = OE/EA = 2 EA = PA = x/2 Hyp of triangle is 2x + x/2 (2.5x)^2 = (2x)^2 + (1.5x)^2 6.25x^2 = 4x^2 +2.25x^2 96 = (2x)(1.5x)/2 3x^2 = 96x2 X^2 = 64 X = 8 Area semi- circle is A = pi x^2 /2 = 32pi A= 100.531 sq units
@PreMath Жыл бұрын
Thanks❤️
@raya.pawley3563 Жыл бұрын
Thank you
@alster724 Жыл бұрын
Wow, that's amazing! After seeing the area of the square in the Pythagorean Theorem alongside the diameter being the square's side measure, I knew where this was going.
@Stereomoo Жыл бұрын
I had angle DCA = CAB (Z angles), triangle OPC = ODC (side, side, side). So let θ = angle OCD, angle DCA = 2θ = CAB, angle EAP = 180 - CAB = 180 - 2θ. Triangle EOA = POA so angle EAO = (180-2θ)/2 = 90-θ. Hence angle EOA is θ and triangle ODC is exactly twice AEO (same angles, OE = half DC). By similar triangles, EA is half DO, and thus a quarter of the square's side. So AB = 3/4 L, BC = L, triangle formula L*3/4*L*1/2 = 3/8 L^2 = 96. ∴ L = 16. Making the circle area 8^2 π / 2 = 32π Your way's definitely tidier on the back half, same similar triangles to get into it though.
@MrPaulc222 Жыл бұрын
Great stuff. I made harder work of that than necessary through using values and r rather than your a and b variables. I had AB as 96/r and AE and AP as 2r - 96/r. It made very hard work of it all, so thank you for enlightening me :)
@wackojacko3962 Жыл бұрын
This problem is cool! And too me Fundamentally Binary. 🙂
@PreMath Жыл бұрын
Thanks ❤️
@marcgriselhubert3915 Жыл бұрын
Let's use an adapted orthonormal. C(0;0) D(a;0) E(a;a) B(0;a) O(a;a/2), R beeing the diameter of the circle. The equation of the circle is: (x-a)^2 + (y-a/2)^2 = (a/2)^2, or: x^2 + y^2 -2ax -ay +a^2 = 0 when developped. The straight line (CPA) has an equation of the type y = mx, with m un unknown parameter at present. At the intersection of this line with the circle we have: x^2 + (mx)^2 -2ax -amx + a^2 = 0, or (1 + m^2).x^2 - (2a+am).x + a^2 = 0. At the point P (tangency), this second degree equation has a double solution, so its delta is equal to 0. delta = (2a +am)^2 -4.a^2.(1 +m^2) = 4.a^2.m -3.a^2.m^2, which is equal to 0 when m = 0 or m = 4/3. m=0 is corresponding to point D and m = 4/3 to point P. So the equation of the straight line (CPA) is y = (4/3).x Its intersection with (BE) (which equation is y = a) is point A ((3/4).a; a), and so the length AB is (3/4).a We finish. The area of triangle ABC is then ((3/4).a).a. (1/2) = (3/8).a^2 = 96, then a^2 = (96). (8/3) = 256, and finally a = 16. The area of the green semi disk is then (1/2).Pi. (a/2)^2 = (1/2).(8^2).Pi = 32.Pi
@allanflippin2453 Жыл бұрын
Again, I have a stupid approach, but it works! I'll use your names: square side = a, triangle base is b. Also, ab = 192. Next I prove that triangle OEA is similar to CDO. If DC/DO = 2, then OE/EA = 2. DC = a, DO = a/2. OE is also a/2 so EA = a/4. b = a - EA so b = 3a/4. Then I can solve for a = 16, r = 8 and circle area = 32pi. To prove OEA is similar to CDO: 1) Let alpha = the angle of DCP 2) The quadrilateral ODCP angles must total 360. alpha + 90 + 90 + DOP = 360. DOP = (180 - alpha). 3) Per straight angle theorem, POE = 180 - DOP. Simplifying, POE = alpha. 4) The angle DCP = POE, therefore the triangle pair CDO, CPO are similar to OEA, OPA.
@FlatEarthMath Жыл бұрын
Fantastic, using the "two tangents" principle. I went a completely different way, using r for radius, and constructing OP = r, then constructing a segment (with length 2r) rightward from O. This creates multiple similar triangles. :-)
@ybodoN Жыл бұрын
DC is twice DO and EO is twice EA as △OEA ~ △CDO ⇒ AB = ¾ BC ⇒ ABC is a 3:4:5 Pythagorean triangle. 96 cm² is 4² times the area of a 3, 4, 5 cm triangle, so BC = 16 and the area of the semicircle is 32 π cm².
@phungpham1725 Жыл бұрын
A nice and elegant solution!
@soniamariadasilveira7003 Жыл бұрын
Obrigada querido professor por seus ensinamentos! Desejo ao senhor um Ano Novo maravilhoso, cheio de bênçãos e realizações! Viva 2024!!!
@PreMath Жыл бұрын
You are very welcome! Thanks❤️🌹
@flash24g Жыл бұрын
I first found myself constructing the circle with OC as diameter, in order to make use of Thales' theorem, and algebraically finding the intersection of this and the given semicircle. Once I got to the point where you introduce the two-tangent theorem, I saw that (since a circle tangent is perpendicular to the radius to the point of tangency) APO and OPC are similar triangles, and used the ratios AP : OP and OP : CP to determine b in terms of a.
@phungpham1725 Жыл бұрын
1/ 1/2 a.b=96 so a.b= 192 2/ Let the angle DCA = 2 alpha so the angle DCO= alpha and tan alpha = 1/2 ------> tan 2 alpha= 1 /(1-1/4)= 4/3 3/ the angle CAB= 2 alpha so a/b= 4/3 ------> b =3a/4 -----> a.b= 3a/4 . a= 3/4 sqa= 192------> a=16 Area of the semicircle= 1/2 pi . 64= 32 pi =100.53 sq cm
@phungpham1725 Жыл бұрын
My second attempt: The angle EOA= the angle ECO= alpha so the 2 triangles EOA and DCO are similar ------> EA/OE=OD/DC=1/2-------> EA= a/4------> AB=3a/4 -------> a.b= 3a/4 .a= 3sqa/4 =192----> sq a= 4 . 192/3 ------> a=16
@prossvay8744 Жыл бұрын
Let r is radius of semicircle So site of the square= 2r AE=AP=x BC=2r; AB=2r-x AC=2r+x Area of the right triangle ABC A=1/2(2r)(2r-x)=96 r(2r-x)=96 (2r)^2+(2r-x)^2=(2r+x)^2 4r^2+4r^2-4rx+x^2=4r^2+4rx+x^2 4r^2-8rx=0 4r(r-2x)=0 x=r/2 r(2r-r/2)=96 r(3r)=192 r^2=64 So: r=8cm Area of the green semicircle=1/2(π)(8)^2=32πcm^2=100.53cm^2. ❤❤❤ thanks
@PreMath Жыл бұрын
You are very welcome! Thanks❤️
@LuisdeBritoCamacho Жыл бұрын
Since BC > AB we have (Integer Solutions): Let BC = x and AB = y ; x > y 96 = x * y Factors of 96 = 2 * 2 * 2 * 2 * 2 * 3 = 2^5 * 3 Only Integer Solutions, when x > y: Dividers of 96 = {1 ; 2 ; 3; 4 ; 6 ; 8 ; 12 ; 16; 24; 32; 48; 96} (96 ; 1) ; (48 ; 2) ; (32 ; 3) ; (24 ; 4) ; (16 ; 6) ; (12 ; 8) Excluding (98 ; 1) and (48 ; 2) ; (24 ; 4) ; (32 ; 3) and (12 ; 8) because the Area of the Square is too big: 98^2 = 9.604 cm^2 ; 48^2 = 2.304 cm^2 ; 24^2 = 576 cm^2 ; 32^2 = 1.024 cm^2 and the Area of the Square is too small: 12^2 = 144 cm^2 Remaining Area of the Square : 9.604 - 96 = 9.508 cm^2 ; 2.304 - 96 = 2.208 cm^2 ; 1.024 - 96 = 928 cm^2 ; 576 - 96 = 480 cm^2 and 144 - 96 = 48 cm^2 In my opinion the only Possible and Elegant Integer Solution is (16 ; 6). Size of the Square = 256 cm^2 Side of the Square = 16 cm Radius of the Circle = 8 cm Green Area = (8^2*Pi) / 2 cm^2 = (64*Pi) / 2 = 32*Pi cm^2 There are too many non Integer Solutions within a certain Interval, where the Remaining Area of the Square is bigger than the Rectangle Area. P.S. - Please let me know if I'm wrong.
@PreMath Жыл бұрын
Thanks❤️
@Ben-wd2in Жыл бұрын
Quickest soln is to recognise that with an area of 96 sqcm, the sides are most likely comprised of two integers. One can then skip most of the geometry and proceed directly to the answer 😂
@Skandalos Жыл бұрын
Didnt see the two-tangent thing, so I got punished by having to do it th hard way. First off I created functions for the circle and for a general tangent line at x0. Then I proved that the slope of the tangent in order to go through the point (2r;2r) has to be 4/3. All the time I somehow knew that there had to be a much quicker and more elegant way.
@giuseppemalaguti435 Жыл бұрын
(2r-192/2r+2r)^2=(2r)^2+(192/2r)^2..r=8...Ag=32π
@PreMath Жыл бұрын
Thanks❤️
@yakupbuyankara5903 Жыл бұрын
32×(3,14).
@gibbogle Жыл бұрын
Neat. I barked up several wrong trees.
@devondevon4366 Жыл бұрын
32 pi
@PreMath Жыл бұрын
Thanks❤️
@xalkxalk3037 Жыл бұрын
I dont know how to say it in english. This is why i ll use my native language. Greek.: φανταστειτε οτι το σημειο Ρ ειναι κινουμενο. κ πως ξεκιναει απο το σημειο Ε. στην αρχη λοιπον το εμβαδον του κιτρινου τριγωνου ειναι 0. Οσο ομως κινειται το Ρ επανω στη περιφερεια του ημικυκλιου, υπαρχει μονο ενα σημειο οπου το κιτρινο τριγωνο θα εχει το μεγαλυτερο εμβαδο κ η μια καθετη πλευρα του τριγωνου θα ειναι "α"! Αυτη ειναι η αρχικη θεση του Ρ. Ειμαι σιγουρος πως μπορει να βρεθει λυση με τη μεθοδο της "αναλυσης" ή αλγεβρικη μεθοδος σε συνδυασμο με τριγωνομετρια αφου η πλευρα ΕΒ ειναι η εφαπτομενη στο σημειο Ε... Δυστυχως περασαν 36 χρονια απο τοτε που εφυγα απο το σχολειο. Υπεροχο προβλημα! By the way it is a wondefull problem