Such an easy math😊🥰like for premath👍

@PreMath

2 ай бұрын

Thanks for liking❤️

All too often, math teachers will skip steps, losing the interest and understanding of their students. I am also an educator and I enjoy your videos very much. You have shown me the importance of stating and restating each and every step in a solution. In this way, you are not only teaching students, you are also teaching teachers. I commend you, sir. Thank you for your videos.

1/Draw the full circle. We have r= 2 ( it's easy). 2/ By using Pythagorean theorem we have AC=sqrt5 and because the triangle ACE is an isosceles one so, CE=CA= sqrt5 3/By using the chord theorem: BCxCE= DCx (2r-DC)---> BCxsqrt5= 1x(4-1)= 3---> BC=3/sqrt5 4/By using Pythagorean theorem again: sq AB=sqAC-sqBC--->AB=4/sqrt5 Area of the triangle=1/2 x3/sqrt5 x4/sqrt5= 6/5 sq units

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Once we have created the mirrored quarter circle and extended the line BC using Thales theorem like you did, we can see that the angle ACB is 180 -2*angle OCA. Given that cos(OCA) is 1/sqrt(5) we can use the double angle formula to determine cos(2*OCA) = -3/5. As cos(180 -x) = -cos(x) then we conclude that cos(ACB) = 3/5 and triangle ABC is in fact a 3-4-5 triangle and from there we can figure out the base BC and the height AB. Keep up the good work.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

@phungpham1725

2 ай бұрын

Very nice! I think we can use the double angle formula to find the tan of the angle ACB as well (=4/3)

Radius of circle = 2 as the quarter circle area is π. Extend BC through C so it meets the extension of AO through O. The intersection point P is on the circle, due to Thales. Also triangles ABP and OCP are similar as they are both right-angled and share another common angle, OPC. Hence, OC/CP = AB/AP => 1/√5 = AB/4 => AB = 4/√5. BP = 2*AB = 8/√5. BC = BP - CP = 3/√5. Blue triangle area = 1/2 *AB * BC = 1/2*4/√5*3/√5 = 6/5 sq units.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

It is more fun to have alternate solution! 😃 1/ Once having AC = CE= sqrt5. Notice that AC= sqrt5 = 5x (sqrt5/5) 2/ The quadrilateral AOCE is cyclic so, EOxEA = ECxEB---> 2x4 = sqrt5 x (BC+sqrt5) ----> BCxsqrt5 + 5=8---->BC= 3/sqrt5= 3 x (sqrt5/5)---> the triangle ABC is a 3-4-5 triple . Therefore AB= 4x(sqrt5/5)----> Area of ABC= 1/2 x 3 (sqrt5)/5) x 4 (sqrt5/5)= 6/5 sq units😀

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

I really appreciate you Man. I watched it 3 times to understand

Gota have my morning (pre) math!!! 😎😎

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Very good question! Think outside of the box!!!

@PreMath

2 ай бұрын

Excellent! Glad to hear that! Thanks ❤️

Looks Complicated^, Mine: ∆ ABC, BC = a, AB = c a" + c" = 5 --- (1) ∆ ABE, BE = a + √5, AB = c _ still (a + √5)" + c" = 4" a" + 2a√5 + 5 + c" = 16, insert (1) 2a√5 = 6 ---> a = 3/√5 Phytagoras : b = 4/√5 Blue ∆ = (1/2)•a•b = (1/2)•(3/√5)•(4/√5) = 6/5 = 1,2 Much-much Easier & Straight. Good Maths Problem, Thanks👍 29/03/24_7am_North Sumatra

You're very clever Mr Premath, I'm learning a lot from you. I have to keep rewatching to remind myself though. 😊

Pythagorean theorem used multiple times: 1. r=2; 2. AC^2 = r^2+(r/2)^2; => AC = sqrt(5); 3. likewise CE = sqrt(5); 4. Using chords theorem BC*CE= (r/2)*(r+r/2) => BC*sqrt(5)=1*3 => BC = 3/sqrt(5); 5. AC^2=AB^2+BC^2 => 5 = AB^2 + 9/5; => AB = 4/sqrt(5); 6. Ablue = AB*BC/2 = (4/sqrt(5))*(3/sqrt(5))/2 = 6/5= 1.2 sq units.

@PreMath

2 ай бұрын

6/5=1.2 Thanks ❤️

@phungpham1725

2 ай бұрын

Thank you! I did it the same way but I posted my solution a little bit later! 😃

CE = rt5, so once you have BE = 8/rt5, BC = 8/rt5 -rt5 = 3/rt5. Then area ABC =1/2of3rt5x4rt5 = 6/5.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Happy Holi sir

π = (πr²)/4 4π = πr² r² = 4 r = 2 So, the radius of the blue quadrant is 2 u. Then, because C is the midpoint of segment DO, CD = CO = 1. ∠AOC is a right angle because it is the central angle of the quadrant. a² + b² = c² 2² + 1² = (AC)² 4 + 1 = (AC)² (AC)² = 5 AC = √5 Reflect the quadrant across radius DO. A' = E. Thus we get a diameter AE. Draw segment CE. This backs up segment AE being a diameter by Thales' Theorem as ∠ABE is a right angle. AE = 4. But segment CE also forms isosceles △ACE. So, CE = √5. ∠COE is a right angle as reflections are rigid. Let α & β be the measures of complementary angles. Let m∠AEB = α. Then, m∠ECO = β. But then m∠BAO = β as well. So, △ABE ~ △COE by AA. There will be a proportion. AB/CO = AE/CE = BE/EO AB/1 = 4/(√5) AB = (4√5)/5 BE/2 = 4/(√5) BE/2 = (4√5)/5 BE = (8√5)/5 But BE = BC + CE, and CE = √5, so BC = BE - CE = (8√5)/5 - √5 = (3√5)/5. Segment BC is the base of the blue triangle and segment AB is the height. Find the area of △ABC. A = 1/2 * b * h = 1/2 * (3√5)/5 * (4√5)/5 = (2√5)/5 * (3√5)/5 = 30/25 = 6/5 So, the area of the blue triangle is 6/5 square units, or 1.2 square units.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

AC=√5=5/√5; AB=4/√5 That means we have a Pythagorean triple 3-4-5, BC=3/√5; Therefore the area of the blue triangle = 1/2*4/√5*3/√5 = 6/5 = 1,2

Complete the square AODF. Since OC = ½ OA, ACF is a 2 : √5 : √5 isosceles triangle. The radius of the quarter circle is ¼ π r² = π ⇒ r = 2. So OC = CD = 1 and AC = CF = √5. Extend AB to G on FD. Note that FBG ~ FCD ⇒ BG = 1/√5 and BF = 2/√5 ⇒ BC = 3/√5 and AB = 4/√5. Therefore, the area of the blue triangle is 3/√5 × 4/√5 = 1.2 square units (and ABC is a 3:4:5 triangle).

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

@quigonkenny

2 ай бұрын

I think you mean "Extend AB to G on FD"? Since B should already be on FC unless I'm misenvisioning your calculations...

@ybodoN

2 ай бұрын

@@quigonkennyExact! Thanks for pointing out! It's corrected now 😉

Thank you! Thinking out of the box with that 90 degree angle B.

@PreMath

2 ай бұрын

Perfect! You are very welcome! Thanks ❤️

A = πr²/4 π = πr²/4 r² = 4 r = √4 = 2 Draw OE colinear with and congruent to AO. Extend arc AD to AE, extending quarter circle to a semicircle. Finally, draw CE. This creates right triangle ∆ABE, as A and E are endpoints of a diameter and B is a point on the circumference. Draw BF, where F is the point on OA where BF is perpendicular to OA. By SAS, ∆OCE and ∆FBE are similar triangles. Let x = OF and y = FB. Triangle ∆FBE: FB/EF = OC/EO y/(r+x) = (r/2)/r = 1/2 y = (r+x)/2 = (2+x)/2 Draw OB. Triangle ∆OFB: OF² + FB² = OB² x² + ((2+x)/2)² = 2² x² + (4+4x+x²)/4 = 4 4x² + 4 + 4x + x² - 16 = 0 5x² + 4x - 12 = 0 5x² + 10x - 6x + 12 = 0 5x(x+2) - 6(x+2) = 0 (x+2)(5x-6) = 0 x + 2 = 0 | 5x - 6 = 0 x = -2 ❌ | 5x = 6 ==> x = 6/5 y = (2+(6/5))/2 = (16/5)/2 = 8/5 The area of the shaded triangle ∆ABC is the area of ∆ABE minus the area of ∆ACE. Triangle ∆ABC: A = (2r)(y)/2 - 2r(r/2)/2 A = (4)(8/5)/2 - 4(1)/2 A = 16/5 - 2 = 6/5 = 1.2

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

Radius we will obtain easily as 2 AC by Pythagorean theorem Sqroot of 5 because AO is 2 and OC is 1 Then because ABC is a RIGHT angle triangle and hypotenues is 5/sqroot 5 other sides will be 4/sqroot 5 and BC is 3/sqroot 5 Area of ABC Triangle is 1/2 x4/sqroot 5x3/sqroot 5 that is 6/5 equal to 1.2

@PreMath

2 ай бұрын

Thanks ❤️

Something different: Let's use an adapted orthonormal. O0;0) A(2;0) D(0;2) C(0;1) and B(2.cos(t); 2.sin(t)) with t =angleAOB (The radius of the circle beeing 2). Then VectorCB(2.cos(t); 2.sin(t) -1) and VectorAB(2.cos(t) -2; 2.sin(t)) These vectors are orthogona!, so: (2.cos(t)).(2.cos(t) -2) + (2.sin(t) -2). (2.sin(t)) =0 We develop: 4.(cos(t))^2 -4.cos(t) + 4.(sin(t))^2 -2.sin(t) = 0, we divide by 2 and use (cos(t))^2 + (sin(t))^2 = 1, then : 2 -2.cos(t) -sin(t) =0 or: sin(t) = 2 - 2.cos(t). Once more we use (cos(t))^2 + (sin(t))^2 = 1 and we have: (cos(t))^2 + 4 -8.cos(t) +4.(cos(t))^2 = 1 or 5.(cos(t))^2 -8.cos(t) +3 = 0. Deltaprime = (-4)^2 -(5).(3) =1, so cos(t) = (4 +1)/5 = 1 which is impossible as t 0° or cos(t) = (4 -1)/5 = 3/5. Having cos(t) = 3/5, we get (sin(t)^2 = 1 - (3/5)^2 = 16/25 and sin(t) = 4/5 (0

@PreMath

2 ай бұрын

Thanks ❤️

It is an interesting puzzle, but not difficult if you get the trick, extend the triangle to right-angled triangle, we see it a 1:2 right-angled triangle, so 4^2=16=5a^2, a^2=16/5, a=4/sqrt(5)=4/5 sqrt(5), and the area of the large triangle is 1/2 4/5 sqrt(5)×8/5 sqrt(5)=16/5, therefore the answer is 16/5-1×2=6/5.😊

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

using analytic geometry it is easy to find the equations of the lines then find the lengths of the sides then apply the area formula... same results

Hello! Because I couldn't see, that BE has to cross the mid point of OD, I used a different way to find the solution. Without Thales, but by constructing auxiliary lines in a different way: Dropped perpendicular lines from B on OD and from B on OA. Then just Pythagoras, system of equations with unknowns, solve algebraic. Same result, perhaps a little more work. Nice geometric challenge, fun to try - Thanks and wish You a happy Sunday 😊

شكرا لكم على المجهودات يمكن استعمال الدائرة المحيطة بOوAوBوC AOB= ACB=a OA=OB=2 AC=جذر5 ......... BC=3/(جذر5) AB=4/(جذر5) S=6/5

The part difficult for me to see is that the extension of BC to E is the same line (i.e., does not catch an angle). Once this is done, we can see that OEC=arctan (1/2) = 26.565=OAC --> OCE=63.435=OCA --> ACE=2*63.435=126.87 --> ACB=180-ACE=53.13 --> ABC=90-53.13=36.87 (so ABC is a 3-4-5 triangle). AC=sqrt(1^2+2^2) = sqrt(5) --> BC=3/5*AC=3/sqrt(5) --> AB=4/5*AC=4/sqrt(5) --> [ABC]=AB*BC/2 = 4*3/5/2 = 6/5 = 1.2

@PreMath

2 ай бұрын

Thanks ❤️

@juanalfaro7522

2 ай бұрын

@@PreMath Now I see why B, C, and E all need to be on the same line. The extension of BC into a point in the diameter line cannot be a point inside nor outside the diameter because it would make an angle with the line from the circumference end to B which would negate that that line makes a right angle with BA, so it's a contradiction.

Alternate Solution: let a=AB, b=BC, r=radius of circle and T = angle OAC = angle DCB (pi * r ^ 2) / 4 = pi or r=2 using law of cosines on OAB: 2^2 +a^2 - 2(a)(2) * cos(T) = 2^2 cos(T) = a/4 (1) a * sin(T) = b * cos(T) + 1 (2) a * cos(T) + b * sin(T) = 2 a * cos(T) = 2-b * sin(T) (3) cos(T) * (2) + sin(T) * (3) : a = sin(T) + 2 * cos(T) a = sin(T) + 2 * a/4 sin(T) = a/2 (4) (1) & (4): sin(T)^2 + cos(T)^2 = 1 (a/2)^2 + (a/4)^2 = 1 a = 4/(5 ^ .5) AC^2 = 1^2 + 2^2 =5 a^2 + b^2 = AC^2 = 5 b = 3/(5 ^ .5) A=1/2 * a *b = 6/5 = 1.2

All Hail! Thales! 🙂

@PreMath

2 ай бұрын

Well said! Thanks ❤️

شكرا لكم على المجهودات يمكن استعمال المعلم المتعامد الممنظم (O,A,D)

Did it with trig.

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

1) Triangle [OAC] and Triangle [ABC], as they share the same Hypotenuse are Geometrically identical. But not Isometric (same measures of the cathetus). Proof: 2) OA^2 + OC^2 = AC^2 ; 2^2 + 1^2 = AC^2 ; 4 + 1 = AC^2 ; AC^2 = 5 ; AC = SQRT(5). 3) AB^2 + BC^2 = AC^2 4) OA^2 + OC^2 = AB^2 + BC^2 5) As : OA^2 + OC^2 = 5 6) Then : AB^2 + BC^2 = 5 7) Now draw a Line between Point O and Point B. This Line is perpendicular to Line AC. Let's call the Point of Interception E. 8) The Length of that Line OB, equal the Radius = 2. But Length OE different from Length EB. Now it's quite easy!! 9) Area of Triangle [OAC] = 1 10) [OE * SQRT(5)] / 2 = 1 ; OE = 2/SQRT(5) ; OE = 2*SQRT(5)/5 OE ~ 0, 8944 lin un 11) EB = [2 - (2* SQRT(5))/5] lin un ~ 2 - 0,8944 12) Blue Triangle Area = A = (EB * AC) / 2 ; A = 1,1056 * SQRT(5) / 2 ; A ~ 2,472 / 2 ; A ~ 1,24 sq un. 13) Answer: The Blue Triangle Area is equal to approx. 1,24 Square Units. Checking the Answer: 1) AB^2 + BC^2 = 5 2) (AB * BC)/2 = 1,236 3) AB ~ 1.69503 and BC ~ 1.45838 4) 1,69503^2 + 1,45838^2 = 5 ; 2,873 + 2,127 ; 5 = 5

@PreMath

2 ай бұрын

Thanks ❤️

πr²/4=π→ r=2→ AO=2 y OC=CD=1→ AC=√(2²+1²)=√5=5√5/5 →→ Tomando como eje OD, hacemos la simetría de la figura→ CB=CB´ y AC y CB´son colineales→ Potencia del punto C respecto a la circunferencia =1*(1+2)=√5*CB´→ CB´=CB=3/√5=3√5/5→ El triángulo CBA es dr tipo 3/4/5→ Área CBA=(3*4/2)*s²=6*(√5/5)²=6/5 ud². Gracias y un saludo cordial.

I missed this - area of a 1/4 circle is pi??? Help!

R=2...ipotenusa=√5...C1=3/√5..C2=4/√5...Ab=(12/5)/2=6/5

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

@giuseppemalaguti435

2 ай бұрын

​@@PreMathciao

Let's find the area: . .. ... .... ..... First of all we calculate the radius r of the quarter circle: A = πr²/4 π = πr²/4 4 = r² ⇒ r = 2 Now let's assume that O is the center of the coordinate system and that OA is located on the x-axis. With OC=CD=r/2=1 we obtain: O: ( 0 ; 0 ) A: ( −2 ; 0 ) B: ( xB ; yB ) C: ( 0 ; −1 ) D: ( 0 ; −2 ) Now let's think outside the quarter circle. We mirror the quarter circle along CD resulting in a semicircle. May E with xE=+2 and yE=0 be the mirror point of A. According to Thales theorem the triangle ABE must be a right triangle. This triangle consists of the two congruent right triangles OAC and OCE on one hand and of the blue ABC on the other hand, so B, C and E are located on the same line. This line can be represented by the function: y = x/2 − 1 B is located on the quarter circle, therefore we can conclude: x² + y² = r² = 4 x² + (x/2 − 1)² = 4 x² + x²/4 − x + 1 = 4 5*x²/4 − x − 3 = 0 5*x² − 4*x − 12 = 0 x = {4 ± √[4² − 4*5*(−12)]}/(2*5) = [4 ± √(16 + 240)]/10 = (4 ± √256)/10 = (4 ± 16)/10 x = (4 + 16)/10 = 2 ⇒ Point E x = (4 − 16)/10 = −6/5 ⇒ Point B xB = −6/5 yB = xB/2 − 1 = −3/5 − 1 = −8/5 Now we are able to calculate the area of the blue right triangle: A(ABC) = A(ABE) − A(OAC) − A(OCE) = (1/2)*AE*h(AE) − (1/2)*OA*OC − (1/2)*OE*OC = (1/2)*(2*r)*|yB| − (1/2)*r*(r/2) − (1/2)*r*(r/2) = r*|yB| − r²/2 = 2*|−8/5| − 2²/2 = 16/5 − 2 = 16/5 − 10/5 = 6/5 Best regards from Germany

@PreMath

2 ай бұрын

Excellent! Thanks ❤️

3(15°)=45° (15°)=45° {45°+45°+90°}=180° 3(15°)=45° 3(15°)=45° {45°+45°}=90° {180°/90°}=2 (x+2x-2)!

@PreMath

2 ай бұрын

Thanks ❤️

Alert! Alert! ⚠️⚠️ PREMATH is near to 400k family Please sub him cuz you will learn new everyday 😊❤😊

@PreMath

2 ай бұрын

Thanks dear for your continued love and support!❤️🌹

May you make harder than this?

@PreMath

2 ай бұрын

Keep watching... Thanks ❤️