Відео

Can you share more folding questions

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Name the circle tangent points: P on AB, T on AD and Q on BE. BP = 12 - r = BQ (by tangents to a circle theorem). So QE = BE - BQ = 15 - (12 - r) = 3 + r = TE (by the same theorem as above). So AT + TE = 9 (as given) r + 3 + r = 3 + 2r = 9 So 2r = 6 and therefore r = 6/2 = 3 So [circle] = 3²·π = 9π

[DEFC] = [△DEF] + [△DCF] Radius of circle A = R and radius of circle B = r DE = R√2 and EF = r√2 and easy to show that ∠DEF = 90° so [△DEF] = (R√2)(r√2)/2 = Rr [△DCF] = (R + r)(R - r)/2 = 8(R - r)/2 = 4(R - r) and as r = 8 - R so [△DCF] = 8R - 32 So R(8 - R) + (8R - 32) = 79/4 After rearranging this gives 4R² - 64R + 207 = 0 which solves as R = 11.5 or 4.5 As R < 8 then R = 4.5

Might have been a nice touch to spell the name of the formula correctly.

👍👍👍👍 Interesante problema

Using the inscribed angle theorem (and calling the centre of the circle O), it can be shown that ∠DOC = π/3, hence △DOC is equilateral, and so DC = R Law of cosines gives BC = √7 as per the video, and the angle bisector theorem splits BC into BD:DC = 3:1 as per the video.

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Didn't spot the cyclic quad. I instead used the law of cosines on △PCB as ∠PCB = arctan(2) - 45°. Did need a calculator unfortunately, to get to (16√10)cos(arctan(2) - 45°) = 48

FALSE , Completely FALSE !!!! In contradiction with BC/sin(BAC) #AC/sin(ABC) C'est faux 2/sin44 no egal à 3/sin/66 Désolé mais c'est n'importe quoi

Dear, what is the radius of outer big circle?

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Get the correct answer in a few seconds. ------The given equation is a "quadratic equation" of the form, ax^2 + bx + c = 0, with a = 1, and (at 0:17) 18 and 10 are the values of the two numerators of x, when the "quadratic formula" is applied to solve for x; this is so, so nice; therefore, x = 18/(2a) = 18/2 = 9, or x = 10/(2a) = 10/2 = 5.

WADR, the thumbnail shows that the shaded area to solve for is not that of a triangle but instead a closed figure bounded by two straight lines (CE & DE) and the smaller arc of the circle between points C & D, and whose solution would have been even more interesting.

Like it!!!

Excellent. Very simple approch !!!🖖

0:12 error: Points A and B >> Points A and C

SATs seem a bit harder judging from this problem compared to when I took them many years ago. Did solve it, basically the solution given, though after trying a wrong approach. Enjoyed it, thanks.

Am I wrong for not using coordinate geometry?

Turn the drawing, and consider a quater circle with radius 1, then the unknown area is (6^2) times the integral from cos(Pi/3) = 1/2 to cos(Pi/6) = sqrt(3)/2 of sqrt(1 - x^2) dx. We use x = sin(t), dx = cos(t).dt, we obtain 36 times the integral from Pi/6 to Pi/3 of (cos(t))^2 dt As (cos(t))^2 = (1/2).(1 + cos(2.t)) we obtain 18.[t + ((1/2).sin(2.t)] between Pi/6 and Pi/3 and finally get 3.Pi.

Dear, what is the radius ?

r = OA = 6 a = EC = OF = r/2 = 6/2 = 3 b = EF = OE - OF = EC√3 - OF = a√3 - a = 3√3 - 3 c = CD/2 = b√2/2 = (3√3 - 3)√2/2 d = c√3 + 2c = [(3√3 - 3)√2/2]√3 + 2[(3√3 - 3)√2/2] Area = ab + (1/2)b² + (30°/360°)πr² - 2[(1/2)cd] = 3π

3PI

Yea total area 4.06

Good one too a while to solve. I took similar triangles ACE and BDE and used sine rule in triangle CDE since sine and cos of angle CDE are known.

hello sir method 2 in right triangle AB+AE=BE+2R 12+9=15+2r r=3 ur videos r interesting as usual sir thank u

R=4,5

method 2 extend BK and OA to meet at point m BOM is right triangle 30 60 90 MO+OB=MB+2R 4( root3)+4=8+2r thank u sir that was nice video

área=(79/4)+área ADE+ área BEF ; 8X=(79/4)+(X²/2)+((x-8)²/2)

جميل وممتع حل هذ السؤال🌺👏👏👏

Guess before the watching the video 36+6pi

Deae, excellent. GOD BLESS YOU AND YOUR FAMILY

This is definitely not on the SAT

All six is 18𝛑(-4√3+7)

The link for the original video under the logo and MathrootLTP

Enjoyed. Go on.

From △OAP, we get that cos(∠OPA)=AP/OP=10/(5√5)=2/√5 But ∠OPA=x/2 i.e., cos(x/2)=2/√5 From cos(x/2), we can find the value of cos(x) using the double-angle formula, namely: cos(2θ)=2cos²(θ)-1 Thus, cos(2*x/2)=2cos²(x/2)-1=2(2/√5)²-1=8/5-1=3/5 i.e., cos(x)=3/5

The link for the original video under the logo and MathrootLTP

Which program do you use?

The link for the original video under the logo and MathrootLTP

The link for the original video under the logo and MathrootLTP

Can it be applied to any quadrilateral?

The link for the original video under the logo and MathrootLTP

Height of the rectangle is (-2a+20) Multiplying by the width (a) gives (-2a+20)a The first derivative with respect to a is -4a+20. When set to 0 to find the maximum, we get a=5 We get the height by plugging a back into -2a+20 to get 10. 10*2+5*2=30 I have no clue what is going on in the explaination in the video. It hurts to look at.

The link for the original video under the logo and MathrootLTP

The link for the original video under the logo and MathrootLTP

70 degree?

The link for the original video under the logo and MathrootLTP

The link for the original video under the logo and MathrootLTP

The link for the original video under the logo and MathrootLTP