PreMath, This is great! I liked it and subscribed!

@PreMath

2 ай бұрын

Welcome aboard! Glad to hear that! Thanks ❤️

Even though I studied Engineering, I never enjoyed maths, this is much more fun.

@PreMath

2 ай бұрын

🌹 Thanks for sharing ❤️

@lrdisco2005

2 ай бұрын

Interestingly, I am getting better at finding the solutions. Still some life in the old dog yet.

Tq sir ...

@PreMath

2 ай бұрын

You are very welcome! Thanks ❤️

Perfect! 🙂

@PreMath

2 ай бұрын

Glad to hear that! Thanks ❤️

😊❤❤❤😊

@PreMath

2 ай бұрын

Thanks dear ❤️

😊 thanks sir i solved it too

@PreMath

2 ай бұрын

Great 👍 Thanks for sharing ❤️

At 4:55, ΔEOC is similar to ΔABC by angle-angle (common angle

@PreMath

2 ай бұрын

Thanks for sharing ❤️

Thanks for watching this video

1) Let the Radius be "R" 2) (R + 7)^2 = R^2 + 14^2 ; R^2 + 14R + 49 = R^2 + 196 ; 14R = 196 - 49 ; 14R = 147 ; R = 147 / 14 ; R = 21 / 2 ; R = 10,5 lin un 3) Triangle [OEC] is similar to Triangle [ABC] 4) AB / BC = OE /EC 5) AB / 28 = (21/2) / 14 6) AB / 28 = 21 / 28 ; AB = (21 * 28) / 28 ; AB = 21 lin un 7) Triangle Area = (21 * 28) / 2 = 21 * 14 = 294 sq un 8) Area of Semicircle = (Pi * R^2) / 2 sq un 9) A.of S. = [(441/4)*Pi*] / 2 = (441 / 8) * Pi ~ 173,2 sq un 10 Green Region Area = 294 - 173,2 ~ 121 11) Answer: The Area of Green Region is approx. equal to 121 Square Units.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

An excellent Friday problem!

@PreMath

2 ай бұрын

Indeed! Glad to hear that! Thanks ❤️

Let's have a try: . .. ... .... ..... Since AC is a tangent to the semicircle, we known that ∠OEC=90°. So the triangle OCE is a right triangle and we can apply the Pythagorean theorem. With r being the radius of the semicircle we obtain: OC² = OE² + CE² (OD + CD)² = OE² + CE² (r + 7)² = r² + 14² r² + 14*r + 49 = r² + 196 14*r = 147 ⇒ r = 21/2 BD is the diameter of the semicircle, therefore AB is also a tangent to the circle. So we known that ∠ABC=90° and (according to the two tangent theorem) AB=AE. Now we apply the Pythagorean theorem to the right triangle ABC: AC² = AB² + BC² (AE + CE)² = AB² + (BD + CD)² (AE + CE)² = AB² + (2*r + CD)² (AB + 14)² = AB² + (2*21/2 + 7)² (AB + 14)² = AB² + 28² AB² + 28*AB + 196 = AB² + 784 28*AB = 588 ⇒ AB = 21 Now we are able to calculate the size of the green area: A(green) = A(ABC) − A(semicircle) = (1/2)*AB*BC − πr²/2 = (1/2)*AB*(2*r + CD) − πr²/2 = (1/2)*21*(2*21/2 + 7) − π*(21/2)²/2 = (1/2)*21*28 − 441*π/8 = 294 − 441*π/8 ≈ 120.82 Best regards from Germany

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Thank you❤, Please keep uploading videos like this, I find it interesting... Now I am going to binge watch them all and it will help me increase my math geometry knowledge. Know that every video helps me THANK YOU SO MUCH

❤❤❤ thanks dear teacher ❣️😊☺️

@PreMath

2 ай бұрын

You are very welcome! Thanks dear❤️

Draw radius OE. As CA is tangent to semicircle O at E, ∠CEO = 90°. Triangle ∆CEO: OE² + CE² = OC² r² + 14² = (r+7)² r² + 196 = r² + 14r + 49 14r = 196 - 49 = 147 r = 147/14 = 21/2 Note that OC = 21/2 + 7 = 35/2, OE = 21/2, and CE = 14 = 28/2. So by observation, ∆CEO is a 3.5:1 ratio 3-4-5 Pythagorean triple trisngle. As ∆CEO and ∆ABC are both right trisngles that share internal angle ∠C, they are similar, so ∆ABC will be a 3-4-5 Pythagorean triple triangle of some ratio as well. For any point outside of a circle, the two tangents drawn from that circle to the point will be equidistant. As AB and EC are both tangent to semicircle O and share point A in common, AB = EA. Let AB = EA = x. As ∆ABC is some version of a 3-4-5 Pythagorean triple triangle, the ratio of BC to AB is 4 to 3. Triangle ∆ABC: BC/AB = 14/(21/2) = 4/3 (2(21/2)+7)/x = 4/3 4x = 3(28) = 84 x = 21 A = bh/2 = 28(21)/2 = 294 Semicircle O: A = πr²/2 = π(21/2)²/2 = 441π/8 Green area: A = 294 - 441π/8 ≈ 120.82

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Draw radius EO. By the Circle Theorem, ∠ABC & ∠CEO are right angles. Label the radius of the semicircle as r. Apply the Pythagorean Theorem on △CEO. a² + b² = c² r² + 14² = (r + 7)² r² + 196 = r² + 14r + 49 14r + 49 = 196 14r = 147 r = 10.5 Then, BD = 21 & BC = 28. By the Two-Tangent Theorem, AB = AE. Apply the Pythagorean Theorem on △ABC. a² + b² = c² a² + 28² = (a + 14)² a² + 784 = a² + 28a + 196 28a + 196 = 784 28a = 588 a = 21 Green region area = △ABC Area - Semicircle Area A = 1/2 * b * h = 1/2 * 28 * 21 = 14 * 21 = 294 A = (πr²)/2 = [π * (10.5)²]/2 = [π * (21/2)²]/2 = (π * 441/4)/2 = (441π/4)/2 = 441π/8 Green region area = 294 - 441π/8 So, the area of the green shaded region is 294 - 441π/8 square units (exact), or about 120.82 square units (approximation).

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

Nice!

Even though I don't bother to do the calculations, I am happy if I can look at the diagram and see how it should be approached. I could see for this one that right-angle tangency and point-originated tangency were the key.

5:09 At this point, rather than use the two-tangent theorem (which I had forgotten!!), I realized that triangles OEC and ABC are similar - they share the angle at C and a right angle, so the third angle is also identical - and furthermore they are in the ratio 2:1, therefore AB is twice OE. At that point, I had the area of the green triangle from half base x height and the area of the semicircle from 0.5 x pi r^2, and though I worked it out with a calculator rather than generate a precise expression, they agree to two DP.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

A nice puzzle.🎉 r^2+14^2=(r+7)^2, 14r=14^2-7^2=3×49, r=21/2, so BC=21+7=28, AB=28×3/4=21, therefore the answer is 1/2×21×28-1/2×(21/2)^2 pi=294-(441/8) pi.😊

@PreMath

2 ай бұрын

Glad to hear that! Thanks for sharing ❤️

connect O to E OE^2+CE^2==OC^2 r^2+14^2=(7+r)^2 So r=21/2 units Let angle OCE=x Cos(x)=14/7+21/2 x=36.87° Tan(36.87°)=AB/BC=AB/28 (BE=2(21/2)+7=28units AB=(28)tan(36.87)=21units So Area of the green region=1/2(28)(21)-1/2(π)(21/2)^2=120.82 square units.❤❤❤ Thanks sir.

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Muitíssimo obrigado pelo problema!!!

Intanto r^2+14^2=(r+7)^2..14r=147..r=21/2..poi h^2+(2r+7)^2=(h+14)^2..784=28h+196..h=588/28=42/2=21..poi faccio differenza di aree

@PreMath

2 ай бұрын

Thanks for sharing ❤️

After finding the radius I was just sitting there hopelessly not knowing to to proceed. I knew I found the bottom side and the right angle triangle at the left but only if I had one more property could I use trigonometry. Thats when I saw "focus on triangle OEC" at the left. That came in clutch. I immdiately used the "law of sines" to find out the angle at the right side of the triangle and then proceeded to use trigonometry to finish rest of it.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

14*14=7(7+2r) ; r=21/2 ; b=7+2r=28; 28*28+h*h=(h+14)(h+14); h=21 Área verde =(28*21/2)-[π(21/2)^2/2]=294-π(441/8)~120,82. Saludos.

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

1.9k videos 1.9k+ knowledge to go🔥🔥🔥

BCxCD=CExCE => (2R+7)=14x14=196=>2R=21=>BC=28,OE=10.5,OC=17.5=>AB=OExBC/EC=21=> S_abc=21x28/2=294

@PreMath

2 ай бұрын

Thanks for sharing ❤️

r²+14²=(r+7)² r²+196=r²+14r+49 14r=147 r=21/2 ⊿ABC∞⊿OEC BC=21/2+21/2+7=28 AB /21/2 = 28 / 14 AB = 21 Area of Green region : 21*28/2 - 21/2*21/2*π*1/2 = 294 - 441π/8

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

294-pi*10.5^2/2

@PreMath

2 ай бұрын

Thanks for sharing ❤️

In my opinion it is not proven that the angle ABO is a right angle. You first have to prove that AB is a tangent to the semisercle. Furthermore: later on you apply the two tangent theorem.... However, I really enjoy your videos. Keep up the good work!

@PreMath

2 ай бұрын

Thanks ❤️

@ybodoN

2 ай бұрын

Since the semicircle is fully inscribed in the triangle, the angle ABO can't be less than 90°. But it is true that it could be more than 90° and the problem would then make no sense 🤔

S=294-441π/8≈120,82

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

I did et exactly the same way.

@PreMath

2 ай бұрын

Thanks ❤️

I did the whole thing without calculator and got 120.75 sq un. The small variation was probably because I used 22/7 for pi.

In summary: Draw the radius OE = r. Then r² + 14² = (r + 7)² ⇒ r = 21⁄2 ⇒ CEO ~ ABC is a 3:4:5 right triangle ⇒ BC = 28 ⇒ AB = 21. So the area of the triangle ABC = 294 and the area of the semicircle = ½ π (21⁄2)² ⇒ green area = 294 − 441⁄8 π sq. u. Thank you PreMath 🙏

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

120.82 answer

I used Pi = 22/7 and got 120.75 square units!

I found 120.2, not 120.82

1$t View 😂

@PreMath

2 ай бұрын

Excellent! Thanks dear❤️