วีดีโอ

Bhaiya pls videos uplaod karte raho , me aapki saari videos mains ke exam ke baad dekhunga

OC^2 = CG^2+OG^2 (a/2+r)^2/ = (a/2)^2/+(a-r)^2 (a/2)^2/ +2*(a/2)*r+r^2 = (a/2)^2+a^2-2*a*r+r^2 a*r = a^2-2*a*r 3*a*r = a^2 r = a^2/3*a = a/3 = 6/3 = 2 ○O_area = (2*2*π) = 4π AREA = □ABFH - ○O = (6*3)-(2*2*π) =18-4π

u sound young, exceptional of you for solving an integral problem.

Very helpful channel!

I think D

Which whiteboard you use?

Keep doing

Alternative solution for the last part cos(pi/7)+cos(3*pi/7)+cos(5*pi/7)=1/2 can be much simpler as follows. Multiplying numerator and denominator by 2*sin(pi/7) 2*sin(pi/7)*cos(pi/7)+2*sin(pi/7)*cos(3*pi/7)+2*sin(pi/7)*cos(5*pi/7)/(2*sin(pi/7))= sin(2*pi/7)+sin(4*pi/7)-sin(2*pi/7)+sin(6*pi/7)-sin(4*pi/7)/(2*sin(pi/7))= sin(6*pi/7)/(2*sin(pi/7))=sin(pi/7)/(2*sin(pi/7))=1/2 Where sin(6*pi/7)=sin(pi-pi/7)=sin(pi/7)

Great job

I am watching all your videos.. They are very much helpful for my competitve exams... Thank you so much sir❤

Multiply will be more and more easier than this..

Very good. Thanks 👍

Bonito ejercicio, te ejercita en el algebra. 👍🏻

By the way, how did you conclude that DQ is perpendicular to AB? Which of the given information did you use to reach that conclusion?

I got another solution for your reference without any construction lines. Let angle GCF = theta sin(theta)=a/CF and cos(theta)=(a+2)/CF area(ACF)=1/2*AC*CF*sin(45-theta)=1/2*2*sqrt(2)*CF*(sin 45*cos(theta)-cos45*sin(theta)) =1/2*2*sqrt(2)*CF*sqrt(2)/2*((2+a)-a)/CF=2

Hi, nice channel keep going 👍🏼 I am doing the same in Turkish and English to promote math and geometry.

I just used synthetic division to factor the numerator and the denominator, then cancelled factors that appeared in both.

We can simplier solve this exercise : a=(2*x)*sin(36°) (you found it) b=(2*x)*cos(18°) because angle(FAE)=108°=90°+18° (just make an orthogonal projection of A on (EC) to have a rectangular triangle) Then a/b=sin(36°)/cos(18°)=(2*sin(18°)*cos(18°))/cos(18°)=2*sin(18°) Answer : a/b=2*sin(18°)=2*((sqrt(5)-1)/4)=(sqrt(5)-1)/2=0,618 on average (more precise that a/b=1175/1903) You know that sin(18°)=(sqrt(5)-1)/4 because you said at 1:40 that cos(108°)=cos(90°+18°)=-sin(18°)=(1-sqrt(5))/4

Since AM=CM and CN=BN and O is a centroid, [ABN]=[ACN] = [ABC]/2 and [ABM]=[CBM] = [ABC]/2. Also, AO = 2*ON and BO = 2*OM --> [AOB] = 2/3*[ABN] = [ABC]/3 and [BON] = [ABN]/3 = [ABC]/6. Since [CBM] = [ABC]/2 and [BON] = [ABC]/6, then [CMON] = [CBM] - [BON] = [ABC}*(1/2 - 1/6) = [ABC]/3. Another method: Trace line MN and it divides CMON into CMN and MON. [CMN] = [ABC]?2^2 = [ABC]/4 since each segment is 1/2 the corresponding parallel segment of ABC. Since MN = AB/2, MO = BO/2, and ON = AO/2, then [MON] = [AOB]/4 = [ABC]/4/4 = [ABC]/12. Then [CMON] = [CMN] + [MON] = [ABC]*(1/4 + 1/12) = [ABC]/3. Now [ABC] = 40*sqrt(3) by Heron's formula ---> [CMON] = 40*sqrt(3)/3

Error at the End : Blue = 48-4*Pi-Pi-9*Pi=48-14*Pi

Thank you sl much ❤❤❤❤❤

5

nice

अच्छा वीडियो, मेरा ऋण यह है कि आप कैसे कह सकते हैं कि त्रिभुज ANB = ACN