Відео

Drop OH perpendicular to BD. We have: OH= AD/2= 2 sqrt14 and HC= CD-BD/2=18-13=5 -> sqx=sq (2sqrt14) + sq5=81 X= 9 units

Iso. triangle. Two sides equal, base comes out to 18. Drop a line from the top apex of the total triangle down to the base. Basically set up two triangles out of the green area. The one on the right will have a base of 5. (One-half of total base [18 is 9] then less 4 in the white area. ) New base is 5. Then 13 becomes hypotenuse, 5 becomes cosine. Going Pythagorean plus a little trig, gives a sine of 12, which is the height of the total triangle, greeen plus white. Total area of green triangle is thus green base 14, times height of 12, divided by two, gives area of green triangle of 84 Short and sweet. Even I did it without too much trouble .

Angle HBD = 60 degrees. With a sine of 3 at line HD, that makes the line BH 3.4642. Thus the cosine of BH gives BD = 1.732. Unlike most of this guy's problems, this one was mostly simple phythagoreanism.

x² - x - 12 = 0 , x ∈ ℕ1 { 0!, 1!, 2!, ... } (x + 3)*(x - 4) = 0 x = -3, 4 x = 4

15 : 5 = 3 : 1 3+1=4 a*a*1/2=5 a^2=10 15+5=20=2a^2 4a*4a*1/2=8a^2=80 Green Area : 8a^2 - 2a^2 = 6a^2 = 60

in your video : ua-cam.com/video/aPcjW9TQAiQ/v-deo.html the values on your problem are wrong, if BE is parallel to OF then the radius is "√10" and AB is "√10 minus √2" and not just√2

A brute force solution would be to use Cartesian coordinates, label the vertices of the large triangle, etc, determine the vertices of the red triangle, etc, could use the fact that the center of the inscribed circle is the intersection of the angle bisectors, etc...then find the intersection of the circle with the sides, etc, treating those sides as lines, etc...then find the area using calculus, or using Heron's formula or something, lol...cumbersome, brute force, but should work...I don't remember all the properties, had to look up that but about the angle bisectors...often Cartesian coordinates can largely simplify things, lol...

I was trying to understand this, I don't quite understand why x! = x(x-1)!, is that just a given?

5*9=12*x x=45/12=15/4 (12+15/4)/2=63/8 (5+9)/2=7 12-63/8=33/8 R²=(33/8)²+7²=1089/64+3136/64=4225/64 R=65/8 2r+33/8=65/8 2r=32/8=4 r=2 Area of the yellow circle : 2*2*π=4π

Thank you… But having ‘easily’ obtained the values for BC, AB, and ED, can’t you just use 1/2absinC to get the areas of triangles ABF, BCD, and FED, given that sinA, sinC, and sinE are immediately 4/5, 3/5, and 1… Then subtract these from the 3,4,5 triangle, giving the red area as 6/5 (or 1.2)… True, you can use Pythagoras, and it’s tempting… But it’s a bit of a Riemannian approach…

gereat work and great video

Dear, thank you.

Keep getting 2.22.

An alternative solution to share with. With all your notations except B point, Area(CDF)/area(EDF)=CD/DE ->CD/DE=15/5=3 ->CD=3*DE Where triangle CDF and EDF have the same height DF. DF/AC=DE/CE=DE/(DE+CD)=1/4 ->AC=4*DF Where triangle FDE and ACE are similar. area(CFE)=1/2*CE*DF=15+5=20 -->CE*DF=40 area(ACF)=area(ACE)-area(CFE)=1/2*AC*CE-20=1/2*4*DF*CE-20=80-20=60

Well done. Finally approximating to 10/3, the exact value. Best regards.

Using all your detonations except H point, my version with trigonometry to share with is as follows: Let angle FDC be beta, tan(beta)=FC/DC=y/(2*y)=1/2 --> cos(beta)=2/sqrt(5) and sin(beta)=1/sqrt(5) 1/2*GE*EF=10 -->1/2*EF*EF=10 -> EF=2*sqrt(5) Let angle EFC be alpha, alpha + beta+45=90 -->alpha =45-beta FC=y=EF*cos(alpha)=EF*cos(45-beat)=2*sqrt(5)*(cos(45)*cos(beta)+sin(45)*sin(beta)) =2*sqrt(5)*sqrt(2)/2*3/sqrt(5)=3*sqrt(2) x=2*y=6*sqrt(2) Square area=x*x=72

<DCK is supplementary with <ACB (which is equal to 106.26°). The value shown for <DCK is incorrect. It should be 73.74° Which should give a final answer of x=10/3

52=43+9，52+6Sqrt（43）=43+2*3*Sqrt（43）+9=（Sqrt（43）+3）^2， Let A=Sqrt（43）， V=（A+3）^3+（A-3）^3=2（A^3+3A*3^2）=2A（A^2+27）=140Sqrt（43）

An alternative way to get the cubic x equation is as follows. Let F be on BC and EF is perpendicular to BC, then EF bisects BC andAC, hence AE=EC=BE=2+x BE^2=BD^2+DE^2 ->(2+x)^2=x^2+y^2 ….(1) 1/2*x*y=6 -> y=12/x ….(2) Replacing (2) to (1) and simplifying x^3+x^2-36=0 -> x=3 …….

I got the same answer that you did but I went a completely different route. I didn’t want to pull out my calculator so I went some strange directions

We can get the exact x=10/3 value without any approximation. Here’s my version to share with. Denoting angle ABC as b and connecting CE, CE is perpendicular to AB. cos b=8/(8+2)=4/5 angle DCK=angle ACK=2* b angle CDK=angle ADE=(180-b)/2=90-b/2 angle DKC=180-angle DCK-angle CDK=180-(90-b/2)-2*b=90-3/2*b Applying sine law to triangle CDK sin(90-b/2)/x=sin(90-3/2*b)/2 cos(b/2)/x=cos(3/2*b)/2 cos(b/2)/ x=(4*cos(b/2)^3-3*cos(b/2))/2 1/x=(4*cos(b/2)^2-3)/2 1/ x=(4*(cos(b)+1)/2-3)/2 1/x=(2*cos b-1)/2 1/x=(2*4/5-1)/2=3/10 x=10/3

Me immediately after watching the equation: isn't it 1?

I also obtained x=arctan (2) = 63.435 by a different method. We see L=2R. Now FC = 2R*cos (y) where y=BCF. Since FD=CD, x=90-y. Now FDC=180-2x --> FC/sin (180-2x) = 2R/sin (x) = FC/sin (2x) = 2R*cox (y) / sin (2x) = 2R*sin (x) / sin (2x) = 2R/sin (x) = 2R*sin (x)/ (2cox (x) * sin (x)) = 2R/2cos (x). Thus 2*cos(x) = sin (x) --> sin (x) / cos (x) = 2 = tan (x) --> x=63.435

The most complicated way to get the solution is labeling points instead of lines and using "IxI" for absolute values of x as lengths allways are positive. All in all: This is not the way to educate.

Let's name the tangential points: E between A-B, F between B-C, G between C-D, and H between A-D. It is given that HD=FC and thus FC=GC=HD=GD. Thus ODH=ODG=OCF=OCG=b. Thus HDG=2b=FCG. We see similarly that BF=BE and AE=AH, so OBE=OBF and OAE=OAH. Now OE bisects AB. Thus OBE=OBF=OAE=OAH=a -> EAH=EBF=2a. Now 2b+2a=180 -> b + a = 90. Now sin (a) = R/2 and sin (b) = R/3 --> (R/2) ^2 + (R/3) ^2 = 1 = [R^2]/4 + [R^2]/9 --> (9+4) *R^2 = 36 -> R^2 = 36/13 -> R = 6/√13. Now [ABCD] = (AB + CD)/2 * 2R = (AB+CD) * R. AB = 2*2*cos (a) and CD = 2*3*cos (b). Sin (a) = R/2 = 3/√13 -> cos (a) = 2/√13 and sin (b) = R/3 = 2/√13 so cos (b) = 3/√13 (Note cos (a) sin (b) and sin (a) = cos (b). Thus AB = 2*2*2√13 = 2*4√13 and CD = 2*3*3/√13. [ABCD] = (AB + CD) * R = 2* (2^2 + 3^2)/√13 * 6/√13 = 2*13*6/13 = 2*6 = 12 cm^2

OAC=arctan (1/2) = 26.565 --> ACO = 90-OAC = 63.435. Now C is the midpoint of OB and ODB = 90 --> DC=OC=BC = 5 u [BDO] = OD*BD/2 = OB^2 * cos (DOB) * sin (DOB)/2. Now DC=OC -> CDO=DOC = (180-DCO)/2 = (180-63.435)/2 --> 2*DOC = 180-63.435 = 116.565 --> [BDO] = OB^2 / 4 * (2*cos (DOC) * sin (DOC) = 10^2/4 * sin (2*DOC) = 25 * sin (116.565) = 25 * sin (63.435) = 25 * 2√5/5 = 5*2√5 = 10√5 u^2

this eq doesnt have a solution dont spread false info

I got an alternative and simple solution for this problem. C is the midpoint of the hypotenuse of the right triangle ODB, then OC=BC=CD=5 Let angle ACO=b, then tan b=AO/OC=10/5=2 from which cos b=1/sqrt(5) and sin b=2/sqrt(5) Area ODB=area OCD+area BDC=1/2*OC*CD*sin b+1/2*BC*CD*sin(180-b)=1/2*5*5*sin b+1/2*5*5* sin b=25*2/sqrt(5)=10*sqrt(5) where sin(180-b)= sin b

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Bravo, this is the right way to be a responsible UA-camr. Listening to and interacting with your viewers wherever possible. Best regards.

I got an alternative solution using only Pythagorean theorem for your reference. Triangles DAB ,BCD and BCE are all right triangles. AD^2+AB^2=BD^2. ……(1) BC^2+CD^2=BD^2. --> BC^2+(2+3)^2=BD^2 BC^2+25=BD^2. ……..(2) Equating (1) and (2) AD^2+AB^2=BC^2+25. ……(3) BC^2+CE^2=BE^2 -->BC^2+3^2=AB^2. -> BC^2+9=AB^2 ….(4) where CE=2 and AB=BE Substitutg (4) to (3) AD^2+AB^2=AB^2-9+25. -->AD^2=16 -->AD=4

BCD=90 is sustained by BD -> BD is the diameter -> BAD=90 also. Let y=EB=AB, then AD^2 + y^2 = 5^2 + BC^2. Now 3^2 + BC^2 = y^2 -> BC^2 = y^2 - 3^2 --> AD^2 + y^2 = 5^2 + y^2 - 3^2 = y^2 + 5^3 - 3^2 = y^2 + 16 --> AD^2 = 16 --> AD = 4 cm

This is a challenging and interesting question and we can get the exact result without calculating any angles. Draw a line DH parallel to AC intersecting FJ at K and H is on FB, then KD=4 ,JC=4 ,BJ=2 and triangle FHK is similar to triangle FBJ. HK/BJ=FK/FJ=FD/FC=4/10=HK/2 --> HK=4/5 HD=4+4/5=24/5 triangle HDG is similar to BAG AG/DG =AB/HD AG/DG=8/(24/5)=5/3 AGE+DG=AD DG=3/8*AD AD=sqrt(AC^2+CD^2)=sqrt((8+6)^2+6^2))=2*sqrt(58) DG=3/8*AD=3/4*sqrt(58). …..(1) FD=4*sqrt(2). ….(2) Denoting angle ADC as a Area(FDG)=1/2*FD*DG*sin(135-a). …..(3) sin(a)=(8+6)/(2*sqrt(58) )=7/sqrt(58) cos(a)=6/(2*sqrt(58))=3/sqrt(58) sin(135-a)=sin 135*cos(a)-cos 135*sin(a)=sqrt(2)/2*3/sqrt(58)+sqrt(2)/2*7/sqrt(58)=5*sqrt(2)/sqrt(58). …..(4) Replacing (1) (2) and (4) to (3) Area(FDG)=1/2*4*sqrt(2)*3/4*sqrt(58)*5*sqrt(2)/sqrt(58)=15

[FDG] = Yellow Area = [FEBC] - [FED] - ([DCA] - [ABG]). Now [FEBC] = (6+4) * (4+6)/2 = 10^2/2 = 50; [FED] = 4^2/2 = 8; and [DCA]=14*6/2 = 42. For [ABG] we need its height: y=6x/14 = 3x/7 and y=10/2*(x-8) = 5x-40. Now 5x-40 = 3x/7 -> (5-3/7) *x = 40 = 32x/7 -> x= 40*7/32 = 5*7/4 = 35/4 -> y = 3/7 * 35/4 = 3*5/4 = 15/4 --> [ABG] = 8*15/4/2 = 15. Then [FDG] = 50 -8-42+15 = 15 u^2

There is a simple way to solve it. Let x/12=y then x/3=4*y sin y+cos y=2/3. …..(1) Squaring (1) on both sides sin(y)^2+2*sin(y)*cos(y)+cos(y)^2=4/9 1+sin(2*y)=4/9. -->sin(2*y)=-5/9 cos(x/3)=cos(4*y)=1-2*sin(2*y)^2=1-2*25/81=31/81=0.38271

In ∆QGB: Cos B=23/27 because In ∆ACB: 10^2=(12^2)+(18^2)-(2×12×18×cosB). , ok 100=144+324-432cosB; -368=-432cosB"; CosB=23/27 In ∆QGB :: Tan(B/2)=r/12 Too tan(B/2)=sqrt[(1-cos(B))/(1+cos(B))] Tan²(B/2)=(r^2)/(12^2) (1-(23/27))/(1+(23/27))=(r^2)/144 4/50=(r^2)/144 r^2=(144×4)/50 r^2=(144×4×2)/2×50 r^2=1152/100 r^2=11.52 thus S of mini circle is 11.52π Good luck

solved for x using angle bisector theorem. In ADC, AE is a bisector, therefore AD/AC = 4/x. Construct a triangle ABD' next to ABD such that these two are congruent. In AD'C, AD is bisector, therefore AD'/AC = 6 / 4 + x, since AD' = AD 4/x = 6/4+x and solving you get x = 8

I got an alternative solution using only the law of cosine twice. Here is my version. Let angle CBD =2*a and angle CED=b, then b=180-2*a cos(2*a)=(16^2+10^2-14^2)/(2*10*16)=1/2 -->2*a=60 --> a=30 and b=120 cos(b)=(DE^2+EC^2-14^2)/(2*DE*EC) DE=EC since they both extend to 30 degree. 3*DE^2=196 hence DE^2=196/3 that is the area of the square containing DE as the side.

(3)^2O/A/Tano°+ (2)^2O/A/BCCoso° =;9O/A/ATano°+4O/A/BCCoso°}= 13O/A/ATano°BCCoso° 120°ac+60°a}= 180°aca^2 180°aca^2/13O/A/ATano°BCCoso°= 10.50Oaca^2/A/ATano°BCCoso° 2^5.5^10 2^1.1^2^5 1^1.1^2^1 2^1( O/A/ATano°BCCoso°aca^2 ➖ 2O/A/ATano°BCCoso°aca^2+1)

The result -1+ln 2=-0.30685. Double check it, will you? Thanks.

I have a suggestion for you. In this video, if you choose a hexagon instead of a heptagon, the concept you want to convey will be the same. But the tedious and cumbersome calculation can be avoided and the presentation will be more interesting and challenging, too. My best regards.

Thank you for your quick response and correction video. But my answer 12/5*sqrt(2) is about 3.39411 while yours is about 3.38289. They are not within two decimal matching . Your reasoning that AT is perpendicular to TC is not justified to me. Can you double check it again. My best regards.

I got r=12/5*sqrt(2). Can you double check your result? Thanks. I was wondering why DO must pass through Q, the center of the small circle. Moreover, DO is not equal to R apparently. Here is my version leading to r =12/5*sqrt(2) for your reference. Let angle ABC=b by law of cosine cos b=(18^2+12^2-10^2)/(2*18*12)=23/27 cos b=2*cos(b/2)^2-1=23/27. --> cos(b/2)=5/(3*sqrt(3)) cos b=1-2*sin(b/2)^2-1=23/27. --> sin(b/2)=sqrt(2)/(3*sqrt(3)) tan(b/2)=sqrt(2)/5 tan(b/2)=OG/BG=r/12 hence r=12/5*sqrt(2)

Nice! φ = 30° → sin⁡(φ) = 1/2 → cos⁡(φ) = √3/2 → sin⁡(3φ) = 1 → cos⁡(3φ) = 0 δ = 24°; ∆ ADB → AB = y; AD = b; BD = c; BDA = δ; ABD = φ + δ; DAB = 3φ + δ/2 ∆ BCD → BD = c; BC = b; CD = y; DBC = x = ? BCD = 6φ - 2δ; CDB = 2δ - x ∆ ADB → sin⁡(δ)/y = sin⁡(3φ + δ/2)/c = sin⁡(φ + δ)/b → sin⁡(3φ + δ/2) = sin⁡(3φ)cos⁡(δ/2) + sin⁡(δ/2)cos⁡(3φ) = cos⁡(δ/2) = √((1/2)(1 + cos⁡(δ))) ∶= k√2/2 → y = csin(δ)√2)/k ∆ BCD → sin⁡(x)/y = sin⁡(6φ - 2δ)/c = sin⁡(2δ)/c = 2sin⁡(δ)cos⁡(δ)/c → y = csin(x)/sin⁡(2δ) = csin(δ)√2/k → sin⁡(x) = sin⁡(δ)sin⁡(2δ)(√2/k) = (√2/2k)(cos⁡(δ - 2δ) - cos⁡(δ + 2δ)) = (√2/2k)(cos⁡(δ) - cos⁡(3δ)) = 0,309016994 → x = π/10 = 18°; k ∶= √(1 + cos⁡(δ))

Na figura os Ys não podem ser iguais.

We don't know that BE=BQ. Why does it say BE=BQ? I see nothing that indicates this. We don't know that is a square or the lines are parallel.

2r + (πd)/4

Using the Intersecting Chords Theorem is more straightforward, and means you don't have to solve a quadratic. Call the side length of the square 2x. The ICT then gives you x² = 1·(2R - 1) = 2R - 1 It can also be seen from inspection that 2x = 2R - 1 Hence 2x = x² and so x = 2 So diameter 2R = 2x + 1 = 5 P = 4·2x = 16

Great to see such a detailed exploration.

Great vid