We complete the circle and call E and F as the intersection points of CD and BD, respectively, with the circle. We have DC*DE=DB*DF, from which 5*5=3*(2R-3), so R=17/3, and from which the required area is equal to π*((17/3)²)/4-(17/3)²/2=(289π)/36-(289/18).
We extend BO by another radius distance downwards, to intersect the semi-circle at a point E. Thus angle BCE is 90 degrees. So BD x DE = square of CD. Thus 3 x (2r -3) = 25. So 6r - 9 = 25. Thus r = 17/3.
Intersecting chords theorem: a.b=2.(3+5) --> a = 2*8/b a = 16/(3√2)= 8/3 √2 cm Radius of quarter circle: c= a+b = 8/3 √2 + 3√2 = 17/3 √2 cm R = c/√2 = 17/3 cm Area of circular segment: A = ½R²(α-sinα) = ½R²(90°-sin90°) A = 9,16445 cm² ( Solved √ )
@michaelkouzmin281 1) You know that 90°=π/2 radians. Here is "π" 2) Formula for circular segment area, when α=90° : A= ½R²(α-sinα)= ½R²(π/2-1) A= ¼πR²-½R² = A₁-A₂ (exactly as video) Remember this formula of circular segment area, is very useful when α is not 90° Of course I didn't deduct sin90° from 90°, you can't mix angle units !!! NEVER !!! Everytime you see mixed units (For example "90°-sin90°"), it is only for viewing, but you have to make calculations in radians, Never mix sexagesimal degrees with sine or cosine or tangent !!!
Let's find the area: . .. ... .... ..... The triangles ABO and BDP are similar right triangles. With r being the radius of the quarter circle we can conclude: BD/DP = OB/OA = r/r = 1 ⇒ BD = DP = 3 ⇒ OD = OB − BD = r − 3 The triangle CDO is a right triangle, so we can apply the Pythagorean theorem: OD² + CD² = OC² OD² + (CP + DP)² = OC² (r − 3)² + (2 + 3)² = r² r² − 6r + 9 + 25 = r² 34 = 6r ⇒ r = 34/6 = 17/3 Now we are able to calculate the area of the green region: A(green) = A(quarter circle) − A(triangle ABO) = πr²/4 − (1/2)*OA*OB = πr²/4 − r²/2 = π*(17/3)²/4 − (17/3)²/2 = 289π/36 − 289/18 ≈ 9.164 Best regards from Germany
@@unknownidentity2846 chat bots? Bro you are not getting to me . I saw your comments on every vids with full explanation and that one line in the end and all ur comments are in patterns.????👍👍
BO=3+(r-3) ; CD=2+3=5 ; CO=r---> 5²+(r-3)²=r²---> r=17/3 ---> Área del segmento circular AB =[π(17/3)²/4]-[(17/3)²/2] =289(π-2)/36 =9,16445...u². Gracias y un saludo cordial.
DP = DB due to 45deg angle. ODC has sides r-3, 5, and r. (r-3)^2 + 5^2 = r^2 r^2 - 6r + 9 + 25 = r^2 Simplify: 34 - 6r = 0, so r = 34/6 = 17/3. White triangle area = ((17/3)^2)/2 so 289/18. Quadrant area = (289/36)pi. Green area = (289/36)pi - 289/18. 9.164 is a rough rounded solution. Yes, I see we went the same way with this one.
You know how you can mix flour eggs sugar and a few other things like chocolate and leave it sit in pan in a fridge for days, ...nothing! But put it in the oven and whalla, ... cookies! Today's problem kinda like that...😊
NICE simplification! Thanku for daily content sir ! U consistently doin different problems is cool
Excellent! Glad you are enjoying the channel. ❤️
You are very welcome!
Thanks for the feedback ❤️🙏
We complete the circle and call E and F as the intersection points of CD and BD, respectively, with the circle. We have DC*DE=DB*DF, from which 5*5=3*(2R-3), so R=17/3, and from which the required area is equal to π*((17/3)²)/4-(17/3)²/2=(289π)/36-(289/18).
Thanks for your detailed solution! 👍🙏
Daily challenges are unique. We got epic teacher❤
Glad to hear that!
Thanks for the feedback ❤️🙏
We extend BO by another radius distance downwards, to intersect the semi-circle at a point E. Thus angle BCE is 90 degrees. So BD x DE = square of CD. Thus 3 x (2r -3) = 25. So 6r - 9 = 25. Thus r = 17/3.
r^2 = 5^2 + (r - 3)^2
6r = 25 + 9 = 34
r = 17/3
Shaded area = (π/4)(r^2) - (1/2)(r^2) = (17/3)(17/3)(π - 2)/4 = (289)(π - 2)/(36)
Nice... I took the crazy train using the intersecting chord theorem, missing the right triangle... but all worked out in the end...
I think that's a little easier; 3*(2r-3)=25.
Intersecting chords theorem:
a.b=2.(3+5) --> a = 2*8/b
a = 16/(3√2)= 8/3 √2 cm
Radius of quarter circle:
c= a+b = 8/3 √2 + 3√2 = 17/3 √2 cm
R = c/√2 = 17/3 cm
Area of circular segment:
A = ½R²(α-sinα) = ½R²(90°-sin90°)
A = 9,16445 cm² ( Solved √ )
2 questions:
1) where pi has gone?
2) Is it correct to deduct sin(D90) from D90?
@michaelkouzmin281
1) You know that 90°=π/2 radians. Here is "π"
2) Formula for circular segment area, when α=90° :
A= ½R²(α-sinα)= ½R²(π/2-1)
A= ¼πR²-½R² = A₁-A₂ (exactly as video)
Remember this formula of circular segment area, is very useful when α is not 90°
Of course I didn't deduct sin90° from 90°, you can't mix angle units !!! NEVER !!!
Everytime you see mixed units (For example "90°-sin90°"), it is only for viewing, but you have to make calculations in radians, Never mix sexagesimal degrees with sine or cosine or tangent !!!
Excellent!
Thanks for sharing ❤️
Let's find the area:
.
..
...
....
.....
The triangles ABO and BDP are similar right triangles. With r being the radius of the quarter circle we can conclude:
BD/DP = OB/OA = r/r = 1 ⇒ BD = DP = 3 ⇒ OD = OB − BD = r − 3
The triangle CDO is a right triangle, so we can apply the Pythagorean theorem:
OD² + CD² = OC²
OD² + (CP + DP)² = OC²
(r − 3)² + (2 + 3)² = r²
r² − 6r + 9 + 25 = r²
34 = 6r
⇒ r = 34/6 = 17/3
Now we are able to calculate the area of the green region:
A(green) = A(quarter circle) − A(triangle ABO) = πr²/4 − (1/2)*OA*OB = πr²/4 − r²/2 = π*(17/3)²/4 − (17/3)²/2 = 289π/36 − 289/18 ≈ 9.164
Best regards from Germany
You r bot
@@zupitoxyt A bot does not answer.🙂
@@unknownidentity2846 chat bots?
Bro you are not getting to me . I saw your comments on every vids with full explanation and that one line in the end and all ur comments are in patterns.????👍👍
Excellent!
Thanks for sharing ❤️
BO=3+(r-3) ; CD=2+3=5 ; CO=r---> 5²+(r-3)²=r²---> r=17/3 ---> Área del segmento circular AB =[π(17/3)²/4]-[(17/3)²/2] =289(π-2)/36 =9,16445...u².
Gracias y un saludo cordial.
Excellent!
Thanks for sharing ❤️
Great. Thank you, Sir.
Excellent!
You are very welcome!
Thanks for the feedback ❤️🙏
Bom dia Mestre
Deus abençoe-o por essa obra de ensino
Grato
You are very welcome!
Thanks for the feedback ❤️🙏
S=289(π-2)/36≈9,17
Excellent!
Thanks for sharing ❤️
DP = DB due to 45deg angle.
ODC has sides r-3, 5, and r.
(r-3)^2 + 5^2 = r^2
r^2 - 6r + 9 + 25 = r^2
Simplify: 34 - 6r = 0, so r = 34/6 = 17/3.
White triangle area = ((17/3)^2)/2 so 289/18.
Quadrant area = (289/36)pi.
Green area = (289/36)pi - 289/18.
9.164 is a rough rounded solution.
Yes, I see we went the same way with this one.
Glad to hear that!
Thanks for sharing ❤️
BD= 3.
(2r-3)3= 5^2; 6r= 34; r= 17/3.
S= π×r^2/4- r^2/2=r^2/2(π/2- 1)= 9,16
Excellent!
Thanks for sharing ❤️
Just another solution:
1. BD =3;
2. PB = 3*sqrt(2);
3. According to intersecting chords theorem: AP*PB = 2* (PD+CD) => AP = 2*(3+5)/(3*sqrt(2)) = 17*sqrt(2)/3;
4. r = AB/sqrt(2) = (17*sqrt(2)/3)/sqrt(2) = 17/3;
5. A(sectOAB) = pi*r^2/4 = 289*pi/36;
6. A(trOAB) r^2 = 289/18;
7. A(green) = (289/18)*(pi/2-1) = approx 9.164 sq units.
Excellent!
Thanks for sharing ❤️
Property of intersecting chords.
5•5=3(2r-3).r=17/3.
Excellent!
Thanks for sharing ❤️
α=ACO angle
(1) DO=r*sin α=r-3; (2) CD=r*cos α=5
square both
(1) => r^2*sin^2 α=r^2+9-6r
(2)=> r^2*cos^2 α=25 => r^2*(1-sin^2 α)=25 => r^2 - r^2*sin^2 α=25
(1)+(2) => r^2=r^2+9-6r+25 => 6r=34 => r=17/3
Excellent!
Thanks for sharing ❤️
Fine.
Excellent!
Thanks for the feedback ❤️
CD = CP + DP = 2 + 3 = 5 = DE → CE = 10; BP = 3√2; AP = k → 3 k√2 = 16 → k = 8√2/3 →
AB = r√2 = 3√2 + k = 17√2/3 → r = 17/3 → green shaded area = (17/6)^2 (π - 2)
𝑏𝑡𝑤: ∆ 𝑂𝐷𝐶 = 𝑝𝑦𝑡ℎ. 𝑡𝑟𝑖𝑝𝑙𝑒 = (1/3)(8 − 15 − 17)
Excellent!
Thanks for sharing ❤️
Достраиваем отрезок CD до хорды, а BO до диаметра. DB=3, CD=5. 3*(2r-3)=25. r=17/3.
Excellent!
Thanks for sharing ❤️🙏
Solution:
AOB is an isosceles, right-angled triangle, like PDB, so
PD = 3 = BD. Then the Pythagorean theorem applies to the triangle ODC:
OD²+CD² = OC² ⟹
(r-3)²+5² = r² ⟹
r²-6r+9+25 = r² |-r²+6r ⟹
6r = 34 |/6 ⟹
r = 34/6 = 17/3 ⟹
Green area = quarter circle - isosceles, right-angled triangle
= π*r²/4-r²/2 = (π/2-1)*r²/2= (π/2-1)*(17/3)²/2 ≈ 9.1645
Integrate and subtract the area of triangle?
Is BD / DP = BO / OA correct ?
Yes, it it.
@@unknownidentity2846 thanks !
[289/18][π/2-1]
What's your education qualification?
Please see the bio in "about"...
(PiRsq/4)-(1/2Rsq)
You know how you can mix flour eggs sugar and a few other things like chocolate and leave it sit in pan in a fridge for days, ...nothing! But put it in the oven and whalla, ... cookies! Today's problem kinda like that...😊
😀
Thanks for sharing ❤️