Vv nice presentation

@PreMath

Жыл бұрын

Glad you think so! Thanks for your continued love and support! You are awesome, Niru. Keep smiling👍 Love and prayers from the USA! 😀

@nirupamasingh2948

Жыл бұрын

@@PreMath 🙏🙏

Using trigonometry: sin α = r/3r = 1/3 ⇒ tan α = √(2)/4 h = 2r⋅tan α = 2r⋅√(2)/4 = r√(2)/2 A = ½⋅4r⋅h = 2r⋅r√(2)/2 = r²√(2)

Using trigonometry and my calculator I arrive at 22.627 square units for the area. via cos(4/12) you get the angle at the origin one of the circles (B) of the right triangle BQE, which via sin leads to the length of the opposite (QE), which via tan (r/QE) leads to the angle at E, which via tan (FC/CE) leads to FC, which is the height of one blue triangle.

I prefer the use of trigonometric functions to solve this problem. Working in construction related engineering my entire career, I needed answers in terms of US standard units to an acceptable degree of accuracy. My boss would never accept an answer in terms of square roots. Therefore, I used trigonometric functions to determine the lengths of triangle sides. I kept the decima places out as far as they were displayed on my calculator until I got to my final answer. Only then did I round off to the required degree of accuracy for what I was trying to determine.

By Pythagorean theorem , AP^2 = 12•12-4•4 =128 AP = 8•sqroot2 Triangles ACF and APD are similar CF/PD =AC/AP CF/4=8/8•sqroot2 CF =4/sqroot2 Area of blue triangle =1/2•16•4/sqroot2 =16•sqroot2

@PreMath

Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Used the fact that FC=FP as tangent lines connecting point F outside circle are equal and after first solving for AP created equation X^2-2•128^1/2+128=8^2+X^2 and solved for X (h)

QE^2=BE^2 - BQ^2 》QE=8sqrt(2) 》BQE y FCE son triángulos con razón de semejanza =sqrt(2) 》CF sqrt(2)=QB 》CF =2 sqrt(2)》Área azul =CF×CE =16 sqrt(2) Gracias y saludos.

Another example of mathematical structure...very cool!...and inspiring! Think I'll map the the position of a moon of mars this afternoon. 🙂

@PreMath

Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

Thanks for video.Good luck sir!!!!!!!!!!

I calculated the first step : AP. Then I didn't know how to go further. I knew you have to calculate the height of the triangle, but how ? That was my problem. Nice explanation

How ( and why)common tangent at C of both circles will pass through the intersection point F of AP&EQ

Love these problems :)

Very very nice, had to watch it, didn't pickup on the like triangles, ... Note to self... Pay more attention.... Thanks again 👍🏻🤓

@PreMath

Жыл бұрын

Thank you! Cheers! You are awesome, Ian. Keep smiling👍 Love and prayers from the USA! 😀

I'm getting good and can solve in my mind before you finish.

I remember we learned something about "tan", which is litterally to find tangent and tangent angles, wouldn't that make life thousand times easier?!!

Well done, thanks!

@PreMath

Жыл бұрын

You are very welcome! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

well explained

Thanks premath

Let a be the side of the blue triangle We have the altitude of AFE is the symetry axis so ACF is congruent to FCE so AFE is isoceles law of sines sin E/a=sin F/16 let angle E be @ so angle F = 180-(@+@)=180-2@ sin @/a=sin (180-2@)/16 16 sin@=a sin 2@ 8*2sin@= a*2cos@sin@ 8=a × cos@ a=8/cos @ a= 8*sec @ in QBE sin @ = 4/12=1/3 (1) cos @ = sqrt(1-sin^2@) cos @ = sqrt (1-(1/3)^2) cos @= sqrt (1-1/9) cos @ = sqrt (8/9) @

Belíssimo questão de geometria. Obrigado.

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

as always excellent.....there is one more way as well..... th=QF=FP next ECxEA=QE square. therefore QE=8square root of 2.. now EF=8square root of 2-h next, by applying pythagorean theorem in CFE right triangle we can easily find the h=4/ square root of 2...now to find the area is a piece of cake......thx

very good

Point A outside circunference: AP^2 = AE.CE = 16*8 = 128

Another approach could be as follows. Prolong DP beyond P and CF beyond F and let M be the intersection of these two lines. Right riangle MCD is equal to right triangle APD because CD = PD and PDC is a common angle. So MD=AD=12; AP=MC=√AD²-PD²=8√2. Triangles MCD and MFP are similar with similarity ratio r=MP:AP=8/(8√2)=√2/2. FP=CD·r=4·√2/2=2√2. FC=FP=2√2 --> Area=(1/2)AE·FC=(1/2)·4·4·2√2=16√2.

Another approach could be as follows. Prolong EQ beyond Q and draw the parallel to BQ passing through A and let G be the intersection of these two straight lines. Triangles AGE and BQE are similiar with a ratio AE:BE=4/3. As a consequence AG=(4/3)BQ=16/3. QE^2²=BE²-QB²=12²-4²=4²(3²-1)=4²(3-1)(3+1)=2·4³=2·2⁶ so QE=√2·2³=8√2. GE:QE=AG:BQ --> GE=8√2·(16/3)/4=(32/3)√2. FE is the hypotenuse of the right triangle FCE, similar to the right triangle AGE having the angle at E in common. Their ratio is CE:GE=FE:AE --> FE=CE·AE/GE=8·16/((32/3)√2)=4·3/√2=12/√2. The area of the blue area is (1/2)FE·AG=(1/2)(12/√2)·(16/3)=32/√2=16√2.

Great رائع

@PreMath

Жыл бұрын

Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

👍 super sir

@PreMath

11 ай бұрын

Thank you! Cheers! 😀

After AP = 8√2, AP = AF + FP = AF + FC = 8√2 AF² - FC² = AC² = 64 Divide this equation by the prev, AF - FC = 64 / 8√2 = 4√2 Solving, FC = 2√2 Area is, ½ x 2√2 x 16 = 16√2

Triangle ∆EQB: a² + b² = c² 4² + EQ² = 12² EQ² = 144 - 16 = 128 EQ = √128 = 8√2 As they share angle ∠CEF and sides CE and EF, ∆FCE and ∆EQB are similar. Triangle ∆FCE: FC/CE = QB/EQ FC/8 = 4/8√2 = 1/2√2 FC = 8/2√2 = (2√2)²/2√2 = 2√2 Blue Triangle ∆EFA: A = bh/2 = 16(2√2)/2 = 16√2

φ = 30°; ∆ AEF → AE = 4a = 16 = AC + CE = 8 + 8; sin⁡(ACF) = 1; AD = 3a; a = PD = BQ; AF = EF = k → PAD = δ → sin⁡(δ) = 1/3 → cos⁡(δ) = 2√2/3 = k/2a → k = 6√2 → area ∆ AEF = sin⁡(δ)2ak = 16√2 btw: PD = PQ + DQ = n + (4 - n) ↔ FE = FQ + EQ → ∆ FPQ → PFQ = 2δ; QPF = 3φ → FQP = 3φ - 2δ sin⁡(2δ) = 2sin⁡(δ)cos⁡(δ) = 4√2/9 → cos⁡(2δ) = √(1 - sin^2(2δ)) = 7/9 → tan⁡(2δ) = sin⁡(2δ)/cos⁡(2δ) = 4√2/7 = n/m = n/2√2 → n = 16/7 → FQ = 18√2/7 → EQ = k - FQ = 24√2/7

QE is 4root 8, so CF is 8x4/(4 root 8)=8/root 8=root 8, therefore the area is 8xroot 8=22.6 approximately. 🙂

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

This can be solved a little bit faster with Trigonometry functions. a = PD = QB = 4 c = ABCD = BCDE = 12 base = ABCDE = 16 theta = ArcSin(a/c) = ArcSin(1/3) = 19.47122063449° adjacent = ABC = 8 height = CF = adjacent * Tan(theta) = 8*Tan(19.47122063449) = 2√2 Area of triangle= ½(base)(height) A = ½(16)(2√2) = 16√2 ≈ 22.62741699797 units squared

Line A B is 28m.m.form point B is tilted 150° at C, distance from AC is 34m.m what's the length of BC? Can you solve it please

오늘은 참 재밌었다.

@PreMath

Жыл бұрын

반가워요! 의견을 보내주셔서 감사합니다! 건배! 당신은 굉장합니다. 계속 웃으세요👍 미국에서 온 사랑과 기도! 😀

@mohamadtaufik5770

Жыл бұрын

PreMath can speak Korea, awesome

@user-kq5nn6oc6t

Жыл бұрын

@@mohamadtaufik5770 나도 놀래는중입니다.awsome

Taking the appropriate right triangle: Angle of isosceles triangle: sin α = R / 3R sin α = 4 / 12 α = 19,47° Height of triangle: tan α = h / 2R h = 2R tan α h = 2 . 4 . tan19,47° h = 2,828 cm Area of triangle: Area = ½ b . h Area = ½ 4.R . 2,828 Area = 22,63 cm² ( Solved √ )

@Irishfan

6 ай бұрын

I got 22.63 Sq ft.

@marioalb9726

6 ай бұрын

@@Irishfan Excellent !!!! Much better specifying some unit in some international standard system, than nothing, like some teachers wrongly use to say, for example "units" or "square units" where not even is given if is length units or what (could be weight unit, mass unit, angle unit, etc) See in this video, teacher says "degrees" and never says "angle units or units". Same thing has to be done with length units Always is better to specify some length unit or surface unit !!!!

A(blue) =16*2sqrt2/2=16sqrt2...laltezza del triangolo risulta dalla similitudine dei triangoli rettangoli

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Καλημέρα σας. Ωραίο, απλό σχολικό πρόβλημα. Και η δική μου λύση είναι ίδια με την δική σας. Έχω όμως μία ένσταση. Όταν λέτε ότι "FC κάθετη στην ακτίνα CD ως εφαπτομένη" αν κατάλαβα καλά δεν το δικαιολογήσατε. Από πουθενά δεν προκύπτει προφανώς αυτό. Εγώ, σύγκρινα και απέδειξα ότι τα ορθογώνια τρίγωνα APD, BQE είναι ίσα. Άρα οι γωνίες Α, Ε είναι ίσες, άρα το τρίγωνο AFE είναι ισοσκελές και επειδή FC διάμεσός του προς τη βάση, θα είναι και ύψος (και διχοτόμος) του. Άρα... Ευχαριστώ.

I think there may be two quick ways of doing this. Spoiler alert. The most direct way is to construct a right triangle, BQE. The area of this right triangle will be the same as the blue area. By Pythagoras: BE² − BQ² = EQ²; 12² − 4² = EQ²; EQ² = 144 − 16 = 128; EQ = √128 = 8√2. Area of triangle = (BQ ∙ EQ) / 2 = (4 ∙ 8√2) / 2 = 16√2 square units. A slightly less direct way is to begin with triangle BQE, calculate EQ as before, and then observe that triangles FCA and FCE are similar to triangle BQE and congruent to each other, so that the blue area is the same as that of a rectangle with sides FC and AC. BQ = 4; EQ = 8√2; AC = 8. FC / AC = BQ / EQ; FC = AC ∙ BQ / EQ; FC = 8 ∙ 4 / (8√2); FC = 4 / √2 = 4√2 / 2 = 2√2. Blue area = FC ∙ AC = 8 ∙ 2√2 = 16√2 square units.

@ashieshsharmah1326

Жыл бұрын

Nice 🤩 but can u please tell me why area of blue equal to area of ∆BQE.

@AnonimityAssured

Жыл бұрын

​@@ashieshsharmah1326 Ah, I wish I could. Alas, I'm no mathematician; I'm just observant. Perhaps someone here can provide a proof. Sorry to disappoint you. Actually, I _can_ prove it, although I can't provide any explanation of _why_ it is true. Let's prove it for the general case, for two touching circles with radius r, keeping the same labels for indicated points. Line segment BQ is a radius. Line segment QE is a tangent, where Q is the point of tangency. By the Circle Theorem, the two line segments are mutually perpendicular. Therefore, triangle BQE is a right triangle. By examination, the hypotenuse of that triangle has length 3r and its shortest side has length r. By the Pythagorean Theorem: (BQ)² + (QE)² = (BE)² (QE)² = (BE)² − (BQ)² (BQ)² + (QE)² = (BE)² (QE)² = (3 ∙ r)² − r² (QE)² = 8 ∙ r² QE = √(8 ∙ r²) = √8 ∙ r = 2 ∙ √2 ∙ r Area, A₁, of triangle: A₁ = BQ ∙ QE / 2 = r ∙ 2 ∙ √2 ∙ r / 2 = √2 ∙ r² By symmetry, the blue area is twice the area of triangle FCA. By similar triangles, length of FC = length of QP times length of BQ divided by length of QE: FC = QP ∙ CA / QE FC = 2 ∙ r ∙ r ∙ / (2√2 ∙ r) = r / √2. (Rationalizing the denominator is not necessary at this point.) Blue area, A₂: A₂ = FC ∙ CA = (r / √2) ∙ 2r = (2 ∙ √2) ∙ r² = √2 ∙ r² Hence, A₁ = A₂. Q.E.D.

I had some ideas to complet the questions.(b) FA=FE=? and (c) FB=FD=?. SOLUTION. b) FB=FP=FC= 2*(sqroot2), so FA=AP--FP=8*sqrt2--2*sqrt2= =6*sqrt2. also FE=FA=6*sqrt2. c) FB^2 = FC^2 + BC^2=(2*sqrt2)^2+(4)^2=8+16=24. Hence FB= sqrt(24)= 2*sqrt6. Also FD=FB=2*sqrt6. ****(d) Calculate QP . Try to solve. ANSWER 16/3.

شكرا لكم يمكن استعمال S=1\2AF^2sin2EAF

I also decided that

Отрезок AP можно легко найти.

22.627

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

CAD solves this in a hurry.... yes, I know, what if I have no computer or program to solve things like this... if my life depends on it... I'm gonna die!

My solution is A = 22.627