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@e2718pi3142

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Malaysia Katılım Kasım 2020
62 Takip Edilen56 Takipçiler
Azri
Azri@e2718pi3142·
@mathwithvan You should be grateful that you can call directly to Allah to get exact date of eid.
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Azri
Azri@e2718pi3142·
@mnd233445 S(Blue)+S(Gray)=(16^2+12^2)/2= 200
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EXTRA
EXTRA@mnd233445·
Total Painted area = ?
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Stanislav Kuzmin
Stanislav Kuzmin@sealingwaxcreak·
@e2718pi3142 As ΔABO=ΔDCO by two sides and the angle between them, angles AOD and BOC are equal. As COBF is cyclic, angles AFD and COB are also equal, which makes AFOB cyclic. Then, CB is the O-Simson line of ΔAFD and OE is the altitude of isosceles ΔAOD with AO=OD. Thus, AE=ED.
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Azri
Azri@e2718pi3142·
@brixkitkam How do you know that arc AB=arc CD?
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kitgum
kitgum@brixkitkam·
@e2718pi3142 arc AB = arc CD (same angles at circumference) so AB = DC r = semiperimeter - hypot = (8+4+4*sqroot3)/2 - 8 = 2*sqroot3 -2 diameter is 4*sqroot3 -6
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Azri
Azri@e2718pi3142·
@PeterPW01 Greeting from Malaysia!
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PeterPaul01
PeterPaul01@PeterPW01·
@e2718pi3142 (1/2) Hello there!, <CAD = <CBD = <ABD = <ACD ---> ∆ACD is isósceles ---> AD = DC = x . Ptolemys's theorem: ---> BD*AC = 4*x + 8*x ---> BD*AC = 12*x (1) Pythagorean theorem in the ∆BCA: ---> AC = 4√3 ---> AC in (1) ---> BD = x*√3 (2)
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Azri
Azri@e2718pi3142·
@PeterPW01 Very well done! 😊
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PeterPaul01
PeterPaul01@PeterPW01·
@e2718pi3142 (2/2) Pythagorean theorem in the ∆BCD: ---> 64 = 3* x^2 + x^2 ---> x = 4 ---> BD = 4√3 Poncelet: ---> Diameter = BD + x - BC = 4√3 + 4 - 8 ---> Diameter = 4(√3 - 1) units
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Azri
Azri@e2718pi3142·
@mathwithvan AB can be equal to DC, but not always.
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