Can you calculate area of the Yellow Semicircle? | (Triangle) |

[@MrPaulc222]

By Heron's Formula, the triangle area is sqrt((21)(6)(7)(8)) = 84. The triangle can also be split into two triangles of (13/2)r and 7r, so 84 = (27/2)r. 168 = 27r, so 56 = 9r r = 56/9 r^2 = 3136/81 3136pi/81 for a full circle. Semicircle is 1568pi/81 approximates to 60.815 un^2. Yes, we did it the same way, but only because you taught me that way. Thanks.

[@sorourhashemi3249]

Thanks easy❤

[@PrithwirajSen-nj6qq]

We may find the area of 🔺 ABC =1/2*12*14=84 sq -- units (1) [as it is a special type of triangle.] 🔺 ACO =1/2 *r *13 sq units -(2) 🔺 BOC =1/2*r*14 sq unitsy--(3) From (1) (2).(3) r(13+14)/2=84 >r=84*2/27 =6.22(approx) Area of semicircle = π*(6.22)^2/2 =(22*6.22*6.22)/2*7 =61.089 sq units (approx)

[@Thampuran-o9o]

👍👍👍

[@-wx-78-]

Mirror △ABC (and semicircle) down and you get deltoid/kite ACBC₁ with inscribed circle. S[ACBC₁] = 2·84 = pr = (13+14)·r, therefore r = 168/27 = 56/9.

[@jamestalbott4499]

Thank you!

[@alanthayer8797]

I Like these Semi Circle ,! Just checking n Sir thanks as usual!

[@alanthayer8797]

I Like these Semi Circle ,! Just checking n thanks as usua Sir!

[@zawatsky]

Если не ошибаюсь, СО лежит на биссектрисе. Тогда он делит АВ в пропорции АО/ОВ=13/14. Т. е. АО составляет 13/27 от АВ, а ОВ - 14/27. 15*13/27=195/27=65/9=7²/₉. 15*14/27=(7²/₉)+2/27=7+6/27+2/27=7+8/27=7⁸/₂₇.

[@alexniklas8777]

Good afternoon sir. Your solution is simpler. My solution is more complicated, defined 'AO' and 'AP', then 'r'. Thanks sir!

[@Abdelfattah-hr8tt]

شكرا لك من القلب

[@gnanadesikansenthilnathan6750]

Why don't we draw a line in the middle? Use triplets concept ?

[@uwelinzbauer3973]

I used trigonometry, the law of cosine, to find out the angles CAB and ABC. Then two equations with two unknowns concerning the two triangles AOP and OBQ.

[@uwelinzbauer3973]

@lmkkk9398 I have to apologize, my comment was not finished, because I had not enough time to complete it. I had the intention to report, that my calculations lead me in the end to the same result, Radius r≈6.22. I have the impression, that the calculations shown in this video are correct. I'm not able to review your calculations, but perhaps you find a wrong assumption. Did you observe that the center point of the circle is not the mid point of AB?

[@cyruschang1904]

Triangle base and height = 15 & h √(13^2 - h^2) + √(14^2 - h^2) = 15 (14^2 - h^2) = 15^2 + (13^2 - h^2) - 30√(13^2 - h^2) 30√(13^2 - h^2) = 15^2 + 13^2 - 14^2 = 29 + 13^2 = 198 900(13^2 - h^2) = 198^2 30h =√(900(13^2) - 198^2) = √(390^2 - 198^2) = √[(588)(192)] = 336 h = 336/30 = 112/10 = 56/5 Triangle area = 13r + 14r = 15h = 15(56/5) = 3(56) r = 3(56)/27 = 56/9 Semicircle area = (56/9)(56/9)π ÷ 2 = (28/9)(56/9)π

[@wasimahmad-t6c]

60.43

[@pwmiles56]

Without using areas... By equal tangents theorem CP=CQ=x say. And angle OCP = angle OCQ = arctan(r/x) = theta say By cosine rule 15^2 = 13^2 + 14^2 - 2.13.14 cos(2 theta) cos(2 theta) = (13^2 + 14^2 - 15^2) / 2.13.14 = 140 /2.13.14 cos(2 theta) = 5/13 But cos 2 theta = cos^2 theta - sin^2 theta cos^2 theta - sin^2 theta = -------------------------------------- cos^2 theta + sin^2 theta 1 - tan^2 theta = ----------------------- 1 + tan^2 theta Solve for tan^2 theta tan^2 theta = (1 - cos 2 theta) / (1 - cos 2 theta) = =(1 - 5/13) / (1 + 5/13) = 8 / 18 = 4 / 9 tan theta = 2/3 = r/x x = (3/2) r By angle bisector theorem AO/13 = BO/14 14 AO = 13 BO 196 AO^2 = 169 BO^2 By Pythagoras for AO, BO and using x=(3/2) r 196((13 - 3r/2)^2 + r^2) = 169((14-3r/2)^2 + r^2) 196(169 - 39r + (13/4)r^2) = 169(196 - 42r + (13/4)r^2) (13/4)(196 - 169) r^2 = (196.39 - 169.42) r 27 (13/4) r = 196.39 - 169.42 (27/4) r = 196.3 - 13.42 = 42 r = 4.42 / 27 = 56 / 9

[@marcgriselhubert3915]

I propose something else (but I know it is less good). We use an orthonormal center A and first axis (AB). We have A(0; 0) and B(0; 15) The equation of the circle center A and radius 13 is x^2 + y^2 = 169 and the equation of the circle center B and radius 14 is (x - 15)^2 + y^2 = 196 or x^2 + y^2 -30.x +29 = 0. At the intersection of these circles we have 169 -30.x + 29 = 0 and so x = 198/30 = 33/5. So at the intersection we also have (33/25)^2 + y^2 = 169 which gives y^2 = (169.25 - 1089)/25 = 3136/25 and y = 56/5. Finally we have C(33/5; 56/5) VectorAC is colinear to VectorU(33; 56), the equation of (AC) is: (x).(56) - (y).(33) = 0 or 56.x - 33.y = 0 VectorBC(-42/5; 56/5) is colinear to VectorV(-3; 4), the equation of (BC) is (x - 15).(4) - (y).(-3) = 0 or 4.x + 3.y -60 = 0. Be a the abscissa of O, then O(a; 0). The distance from O to (AC) is abs(56.a -33.0)/sqrt(56^2 + 33^2) = (56.a)/sqrt(4225) = (56/65).a The distance from O to (BC) is abs(4.a - 60)/sqrt(4^2 + 5^2) = (60 - 4.a)/5 (as a

[@alexundre8745]

Bom dia Mestre Darei uma pausa p resolver, já passo p o Sr o feedback Grato Acertei!!!

[@Algcleveralways]

I am the number one ❤ Thanks my teacher Do you mind helping me ?

[@giuseppemalaguti435]

r/sinα+r/sin β=15...cosα/2=7/√65,cosβ/2=√(4/5)..(Briggs formule)..sinα=56/65..sinβ=4/5...(65/56)r+(5/4)r=15..135r=56*15..9r=56...r=56/9

[@countysecession]

What is Briggs formula. I can't find anything about it on Google.

[@আব্দুলকুদ্দুস-ল৮ন]

why OP & OQ are perpenducular on AC and BC? which theorem says it?

[@ayushchandrachaudhary5237]

Tangents and radius are perpendicular to each other AC and BC are tangets

[@countysecession]

@@lmkkk9398 The video is correct. The side lengths are different. Angle PAO is 59.49 degrees. Angle QBO is 53.13 degrees.

[@santiagoarosam430]

Perímetro ABC=42→ 42/2=21→ Área ABC=√(21*8*7*6)=84 =13r/2 +14r/2 =27r/2→ r=168/27=56/9→ Área semicírculo =(π/2)(56/9)² =1568π/81 ≈60,81502... u². Gracias y un saludo cordial.

[@PrithwirajSen-nj6qq]

Triangles COP and COQ are congruent. Hence CO is the bisector of angle ACB According to angle bisector Theorem AO/OB =13/14 Then AO =15*13/(13+14) = 15*13/27=65/9 Angle CAB =β Cos β=(13^2 +15^2 - 14^2)/2*13*15 =99/13*15 Sin^2β=1- 99^2/(13*15)^2 sinβ=√[(13*15)^2-99^2]/13*15 In 🔺 AOP sinβ=r/AO=9r/65 =[√(13*15)^2 -(99)^2]/13*15 = 168/13*15 > 13*15*9r=168*65 > r = 168*65/13*15*9=168/27 Area of semicircle =π(168/27)^2/2 =60.8394 sq units (approx)

[@sergioaiex3966]

Solution: Let's calculate ∆ ABC Area by Heron's Formula A = √ [s (s - a) (s - b) (s - c)] s = (13 + 14 + 15)/2 s = 21 ∆ ABC Area = √ [s (s - a) (s - b) (s - c)] A = √ 21 (21 - 13) (21 - 14) (21 - 15) A = √ 21 . 8 . 7 . 6 A = √ 3 . 7 . 2² . 2 . 7 . 2 . 3 A = 3 . 7 . 2 . 2 A = 84 Connecting "O" to "P" and "O" to "Q", where "P" and "Q" are the points of tangency, being OP and OQ the radii "r", and also connecting "O" to "C", we form the ∆ CAO and the ∆ CBO Therefore: ∆ ABC Area = ∆ CAO Area + ∆ CBO Area ... ¹ ∆ CAO Area = ½ b h ∆ CAO Area = ½ 13 . r ∆ CAO Area = 13r/2 ∆ CBO Area = ½ b h ∆ CBO Area = ½ 14 . r ∆ CBO Area = 14r/2 Substituting in ¹ 84 = 13r/2 + 14r/2 84 = 27r/2 r = 168/27 (÷3) r = 56/9 Semicircle Area = π r²/2 Semicircle Area = π (56/9)²/2 Semicircle Area = π (56/9)²/2 Semicircle Area = π (3,136/81)/2 Semicircle Area = π 3,136/162 Semicircle Area = 1,568/81 π Square Units ✅ Semicircle Area = 60.8150 Square Units ✅

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... First of all we calculate the area of the triangle ABC with Heron's formula: s = (AB + AC + BC)/2 = (15 + 13 + 14)/2 = 42/2 = 21 = 3*7 s − AB = 21 − 15 = 6 = 2*3 s − AC = 21 − 13 = 8 = 2³ s − BC = 21 − 14 = 7 A(ABC) = √[s*(s − AB)*(s − AC)*(s − BC)] = √(3*7 * 2*3 * 2³ * 7) = √(2⁴ * 3² * 7²) = 2² * 3 * 7 = 84 The triangle ABC can be divided into the triangles ACO and BCO. Since AC and BC are tangents to the semicircle, we know that OP is perpendicular to AC and that OQ is perpendicular to BC. So with r being the radius of the semicircle we can conclude: A(ABC) = A(ACO) + A(BCO) = (1/2)*AC*OP + (1/2)*BC*OQ = (1/2)*AC*r + (1/2)*BC*r = (1/2)*(AC + BC)*r 84 = (1/2)*(13 + 14)*r = (27/2)*r ⇒ r = 2*84/27 = 56/9 Now we are able to calculate the area of the yellow semicircle: A(yellow semicircle) = πr²/2 = π*(56/9)²/2 = 1568π/81 ≈ 60.82 Best regards from Germany

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Area of the Triangle [ABC] (A) = 84 sq un ; using Heron's Formula with Sides = (13 ; 14 ; 15) 02) OP = OQ = R lin un 03) 13*R + 14*R = 2A ; 27*R = 168 04) R = 168 / 27 ; R = 56 / 9 ; R ~ 6,2222(2) lin un 05) Semicircle Area (SA) = Pi*R^2 / 2 06) SA = (56/9)^2 * Pi / 2 ; SA = (3.136 * Pi / 81) / 2 ; SA = 3.136 * Pi / 162 ; SA = 1.568Pi / 81 sq un ; SA ~ 60,6 sq un Therfore, OUR BEST ANSWER IS : The Semicircle Area is equla to 1.568Pi/81 Square Units ot Approx. equal to 60,6 Square Units.