วีดีโอ

A line segment could be drawn from point B to the midpoint of AC. This line segment is also a median and. according to the centroid theorem, will pass through the intersection of the other 2 medians. This median will divide ΔACR into 2 triangles of equal area (if the sides along AC are the bases, they have equal length and the height is common.) Let the areas each be designated z. Using the method given in the video, we can prove x = y = z. Therefore, the medians divide ΔABC into 6 triangles of equal area. (There must be a theorem for this.) The blue area equals the sum of areas of 2 of the triangles, so 1/3 of the area of ΔABC. Area ΔABC was found by Heron's formula to be 756 cm², so the blue area = (1/3)(756 cm²) = 252 cm².

Thanks for the video as always 👍 btw I was wondering, wouldn’t it be easier to write the problem without units? Maybe it’s just me being lazy 😅

Here is perhaps a less smart method inspired by yours. Same Method as you with Heron Formula to find : Area (ABC)=756 Let's call point S intersection point of (BR) and (AC) R is the isobarycenter of ABC (intersection of the 3 medians of ABC) Then : BR=2/3*BS Area(triangle BPR)=1/2*BP*BR*sin(angle PBR) Area(triangle BPR)=1/2*(1/2*AB)*(2/3*BS)*sin(angle PBR) Area(triangle BPR)=1/3*(1/2*AB*BS*sin(angle PBR)) Area(triangle BPR)=1/3*(1/2*AB*BS*sin(angle ABS)) Area(triangle BPR)=1/3*Area(triangle ABS) Area(triangle BQR)=1/2*BQ*BR*sin(angle QBR) Area(triangle BQR)=1/2*(1/2*BC)*(2/3*BS)*sin(angle QBR) Area(triangle BQR)=1/3*(1/2*BC*BS*sin(angle QBR)) Area(triangle BQR)=1/3*(1/2*BC*BS*sin(angle CBS)) Area(triangle BQR)=1/3*Area(triangle CBS) Blue area=Area(quadrilateral BPRQ) Blue area=Area(triangle BPR)+Area(triangle BQR) Blue area=1/3*Area(triangle ABS)+1/3*Area(triangle CBS) Blue area=1/3*Area(triangle ABC) Blue area=1/3*756 = 252 cm²

Las líneas interiores de la figura son medianas; las tres que se pueden trazar se cortan en el baricentro→ Las medianas dividen ABC en seis celdas triángulares de áreas: a ; a ; b ; b ; c ; c→ c+2a=c+2b→a=b ; 2a+b=b+2c→a=c → a=b=c → Área ABC =6a→ Área azul =2a =ABC/3 → Según fórmula de Herón, ABC=√(63*24*21*18)=756→ Área azul =756/3=252 ud². Gracias y saludos.

AR=BP=½AB=5; BR=vʼ(AB²+AR²)=5vʼ5 BQ=BP²/BR=vʼ5; QR=BR-BQ=4vʼ5 [BPR]=½BP•AR=25/2 sq.un (km² 😳) [PQR]=[BPR](QR/BR)=10 sq.un 😁

AC*CD = CQ*CP = BC*CE --> (18+9) * x = (12+x)9 --> 27x = 108+9x --> 18x=108 --> CD=x=6m

1/ R= 7 2/ Label ED= x and CD= a 3/ Draw the bisector (the diameter) of CD and intersecting the chord and arc AF at HI. Note that HI=DE= x -->a+x= 2R = 14 (1) And sq (a/2) = DE.DF=14x (2) From (1) and (2) we have: X= 14/5 and a= 4x= 4.14/5 =11.2 m 3/ Area of triangle OBE Height from O to BE= 7-14/5= 4.2m Area=23.52 4 /Area of sector OBE: By using cosine law: Cos angle BOE= -0.28-> angle BOE=106.26 degress --> Area of sector= pi. 49. 106.26/360 =45.412 Area of segment= 45.412-23.52=21.89 sq m Area of yellow region = area of rectangle BCDE-area of segment= 9.47 sq m😅😅😅

(18+9).x = (x+12).9 27 x / 9 = x + 12 3x - x = 2x = 12 x = 6 m ( Solved √ )

La potencia de C respecto a ambas circunferencias es CP*CQ---> CD*(CB+BA) =CP*CQ= CB*(CD+DE)---> CD*(9+18)=9*(CD+12)---> CD=6. Gracias y saludos.-

M es punto medio de AB ; N lo es de BE y P lo es de CD. ACDF=2a*2a. ---> AM²+MO²=OA²---> (2a-r)²+a²=r²---> a=4r/5=MO=BN=NE---> BE=8r/5 ; AM=MB=3r/5=ON ; NP=flecha del arco BPE=2r/5. ---> tg(BON)=BN/NO=4/3---> Ángulo BON=53,13º--->Ángulo BOE=2*BON=106,26º --->. Área amarilla =(Rectángulo BCDE) - (Segmento circular BPE) =[(2r/5)(8r/5)]-[(πr²*106,26º/360º)-(8r*3r/5*5*2)]= r²[(28/25)-(π*106,26º/360º)]=(154/π)*[(28/25)-(π106,26°/360°)]= 9,4464..m². Gracias y un saludo cordial. -

Radius of circle : A = 154m² = πR² --> R=7,00141m Side of square : (Inters.chords theorem) s. (2R-s) = (½s)² = ¼s² 2R - s = ¼ s --> s=8/5 .R s = 11,20225m Angle of circular sector : sin ½α = ½s / R = 0,8 --> α=106,26° Yellow shaded area : A = A₁ - A₂ - A₃ A = s² - s.2(s-R) - ½R²(α-sinα) A = 125,4904 - 94,1177 - 21,9262 A = 9,4465 m² ( Solved √) ±0.0001m

Yellow=(56/5)^2-(6/5*7*8/5*7+7^2*arctan(4/3)-7*3/5*7*4/5) Yellow = 9,44 m² (+/- 0,01 m²) (with approximation : Pi=22/7)

5:50 16÷(1/vʼ10)=16÷(vʼ10/1) ??? 😯

Great job. But : From 0:00 to 1:52, you write : Blue area=Area of overlapping circles = 1054,52 m² From 1:54 to the end of the video, you write : Blue area=Area of overlapping circles = 1056,52 m² I continue with the value : Blue area = 1056,52 m² 13:11 sin(1/2*theta)=4/5 13:19 1/2*theta=arcsin(4/5) 13:22 1/2*theta=53,1301° (approximation) 13:28 1/2*theta=53° (approximation) 13:53 sin(theta)=sin(106°) 14:15 sin(theta)=0,9613 (approximation) In reality, sin(theta)=24/25=0,96 without approximation Because : cos(theta/2)=sqrt(1-(sin(theta/2))^2)=sqrt(1-(4/5)^2)=3/5 sin(theta)=2*sin(theta/2)*cos(theta/2) sin(theta)=2*(4/5)*(3/5)=24/25=0,96 15:28 r=13,9974=14 (approximation) ************** Without approximation : A(rectangle)=40*(Blue area)/(25*Pi/2+6-25/2*arctan(4/3)) If Blue area = 1056,52 m², A(rectangle) = 1254,82 m² (+/- 0,01 m²)

Angle of circular segment: tan ½α = 4/3 --> α = 106,26° Blue shaded area: A = A₁ -A₂ = 2.πR² - 2[½R²(α-sinα)] A = R² (2π - 0,8946) = 1054,52m² Radius of circle: R = 13,989 m = h/2 R² = (2x)²+(1,5x)² = 6,25x² x = 5,5956 m Rectangle area: b = 2R+3x = 44,765 m A = b.h = 44,765 x 27,978 A = 1252,43 m² ( Solved √ )

6:50 ½x(x-9)=99+½x(x-15) !!! 😬 8:18-8:55 33²=(30+3)²=30²+2•30•3+3²= =900+180+9=1089 !!! 😁

1/ AQ= 24 -> AP=AR= 24 Focus on the triangle BAR: AR= 4x6= 24, BR=5x6=30 so, AB = 3x6= 18 ( triangle BAR is a 3-4-5 triple) --> Area of BAR= A1=1/2 x18x24=9x24 2/ The triangle PQR is similar to the triangle BAR . Label Area of PQR as A2 We have: A2/A1= sq (PR/BR)=sq(48/30)=sq(8/5) -> A2= 9x24x64/25 A2=552.96 sq units😅😅😅

Another tricky problem!😅 1/Label O as the center of the circle, OA and OB as 2 radiuses, vertical and horizontal respectively, and point C on the quadrant circle ( forming the chord AC= 48) 1/ Draw the height OH to AC, we have: AH=HC=24 2/ From B extend this base line perperdicular to OH intersecting OH at point I. Focus on the 2 triangles AHO and OIB, they are congruent( a/s/a) so, AH=OI=24-> OH=46+24= 70 --> by Pythagorean theorem R=74 Area of the yellow area = Area of the quadrant- ( 2 Area of the triangle OHA+ Area of the rectangle of which the 2 sides=24and 46) = 1520.85 sq units 😅😅😅

very nice!

Area of right triangle BPR: A = ½ y.z = ½ .x/2 .108/x = 27 m² Now, let's find the square adjacent to previous right triangle, with same area. A = 27 m² = s² --> s =√27m This square has opposite vertices B and Q, and its side is equal to height of blue shaded triangle: h = s = √27 = 3√3 m Area of blue shaded triangle: A = ½b.h = ½ 26 3√3 A = 67,54 m² ( Solved √ ) [ There's a little mistake in video, y=x/2 m and not y=x/(2m) ]

1/ Label angle PDA= alpha and PQ= a, and QC = d ( diameter of the circle) We have: Angle PCQ = alpha ( having the same arc PQ) So PC= 3a) --> sqa + sq(3a)= sq d 1O sqa= sqd Area of the circle = pi sqd/4 = pi x 10 (sq 32)/4=pi x sq 32x 2.5=2560 pi sq m 2/ Angle PQR= Angle PQC- angle RQC Because angle PQC= 90-alpha so, tan (PQC)= 3 and tan (RQC)=1/2 -> tan (PQR)= (3-1/2)/(1+3/2)= 1 Angle PQR= 45 degrees

Good demonstration. Thanks too to @maisonville7656 for these interesting remarks. Here another demonstration, a bit different from 48 hours ago before you deleted a similar video. Like you at 8:22, Q is the incenter of the right-angled triangle ABC (Q center of the inscribed circle with radius=h) Then : h=(AB+BC-AC)/2 AB=AC*cos(2*a) and BC=AC*sin(2*a) h=AC/2*(cos(2*a)+sin(2*a)-1) PB=x/2 and BR=108/x then we must understand that : PB*BR=x/2*108/x=54 PB=BC*tan(b)=AC*sin(2*a)*tan(45°-a) BR=AB*tan(a)=AC*cos(2*a)*tan(a) PB*BR=AC^2*sin(2*a)*tan(45°-a)*cos(2*a)*tan(a) We can notice that : sin(2*a)*tan(a)=1-cos(2*a) cos(2*a)*tan(45°-a)=1-sin(2*a) PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a)) PB*BR=1/2*AC^2*(cos(2*a)+sin(2*a)-1)^2 PB*BR=1/2*(2*h)^2 PB*BR=2*h^2 h=sqrt(PB*BR/2) h=sqrt(54/2) h=3*sqrt(3) A|ACQ|=1/2*AC*h A|ACQ|=1/2*26*3*sqrt(3) A|ACQ|=39*sqrt(3) ************************ Nota Bene : We can demonstrate in other way, geometrically that : PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a)) Let's introduce 2 points : S orthogonal projection of P on (AC) T orthogonal projection of R on (AC) Then, PBC and PSC are 2 identical right-angled triangle (same angles b, 90°-b, 90° and same hypothenuse PC) Then, ABR and ATR are 2 identical right-angled triangle (same angles a, 90°-a, 90° and same hypothenuse AR) Then : PB=PS ; BR=RT ; SC=BC ; AT=AB TC=AC-AT=AC-AB=AC-AC*cos(2*a)=AC*(1-cos(2*a)) AS=AC-SC=AC-BC=AC-AC*sin(2*a)=AC*(1-sin(2*a)) AS*TC=AC^2*(1-cos(2*a))*(1-sin(2*a)) APS and RTC are 2 right-angled triangle. PS=AS*tan(2*a) and RT=TC*tan(2*b)=TC/tan(2*a) PS*RT=AS*tan(2*a)*TC/tan(2*a) PS*RT=AS*TC PB*BR=AS*TC PB*BR=AC^2*(1-cos(2*a))*(1-sin(2*a))

[PRO]=½PR•OH1 [PRS]=½PR•SH2=½RS•PS=½PQ² OS || PR => OH1=SH2 => [PRO]=[PRS] [Blue]=½(100x)²=5000x²……… 😁

Q is the center of the in-circle. angle B being 90 degrees, BQ must be equal to sqr(2)*radius of in-circle. Figuring out that triangles BQR and BQP are similar was more work, but can easily be done. BR/BQ=BQ/BP gives 2*radius^2=54. The radium of the in-circle is the height of triangle AQC. Nice problem. Most of the work was to demonstrate BQR and BQP triangles are similar.

Thank you sir for this amazing presentation

Aceptando los redondeos de la propuesta, r=7; PQRS es un rectángulo compuesto por dos cuadrados adosados de dimensiones 2r*2r y RT es la hipotenusa de un triángulo rectángulo semejante a otro de lados 1/2/√5 ---> Si r=7 cm---> RT=r√5=7√5 cm y área PQRS =8*r²=8*7² =392 cm². Gracias y un saludo cordial.

Si copiamos y giramos 90º sucesivamente la figura propuesta, obtenemos una circunferencia con una cruz griega inscrita, formada por un cuadrado central de dimensiones 48x48 y cuatro brazos iguales de 48x46. La diagonal de la cruz será el diámetro de la circunferencia → ɸ=√[(46+48+46)²+48²]=148→ Radio =r=148/2=74 → Área amarilla de la figura =[(π*74²)-(48²+4*48*46)]/4 =(5476π-11136)/4 =1516,84034...ud². Interesante rompecabezas. Gracias y un saludo cordial.

La pendiente de la línea inclinada es 1/2→ Área azul =(8²/4)+[(12²/4)+2*12]+(20²/4)=176 cm². Rompecabezas divertido. Gracias y un saludo cordial.

Tg X=[2a+(2a/√3)]/a =4/(3-√3)=3,1547..---> X=72,4120°.. Gracias y saludos

Similarity of triangles: (½S-½s)/s = ½s/(Scos30°-s) (S-s)(S√3/2 -s)=s² S²√3/2 - Ss - Ss√3/2 +s²=s² S²√3/2 = Ss(1+√3/2) S/s= (1+√3/2)/(√3/2) S/s =2/√3+1 = 2,1547 tan x= (½S+½s) / ½s = (S+s)/s = S/s + 1 x = 72,412° ( Solved √ ) Given data S=100mm was just for distraction, is a redundancy, is not necessary !!!

❤❤ nice one my child 😅

118

Very tricky but fun. 1/ Consider the two triangles PRT and QRV, they are congruent (s-a-s) -> angle RPS=SPR= alpha -> the quadrilteral SQPR is cyclic Similarly the angle STR =SVR = beta --> the quadrilateral STVR is cyclic too -> angle ASP=PSR=RSV=VST= 60 degrees 2/ Consider the two triangles QSR and RST, they are similar -> QR/ SR = SR/ST or 6/x = x/16 -> sq x= 6. 16 X = 4 sqrt 6😅😅😅

Thank you so much sir for this useful exercise. I've given it a try and I've found that: Tan(X)=(6+2√3)/3 so X=72°.412 keep going sir because you're doing great.

1/ We have: the triangle PAQ is a right isosceles and PTQ is an equilateral one. -> QA= 8 and PA=8sqrt2 2/ AT intersects PQ at point I. Note that IT is the perpendicular bisector of PQ so, point I is the midpoint of PQ and IT//QS 3/ Demonstrating that a=b Label the area of triangle ABT=c We have: Area of QAT=Area of BAT ( same base, same height) -> a+c= b+c --> a=b --> Area of the blue are= 2 times area of the triangle QAB= 8 x AB😅 4/ Calculating the base AB Focus on the triangle QBS The angle BQS= 30 degrees, QSB=45 degrees --> angle QBS= 105 degrees. Drop the height BH = h to QS. We have two special triangles: BHS is a 45-90-45 triangle and BHA is a 60-90-30 triangle So, SH= h and HA=h. sqrt3 -> h + hsqrt3 = QS =8 sqrt2 --> h = 8sqrt2/(sqrt3+1) --> BS= hsqrt2=16/(sqrt3+1) AB= AS-BS= 8- 16/(sqrt3+1) AB= 8(sqrt3-1)/(sqrt3+1) --> Area of the blue= 8 AB =64(sqrt3-1)/(sqrt3+1) = 64(2-sqrt3) 😅😅😅

Awesome! Thank you professor.

Very nice solution, I always forgot about cyclic quadrilaterals hope you post more about them. Thanks

I think your method is the best. Here is my method (with too many calculus) : Same method as you from 0:00 to 3:08 I use your notations : vertex P,Q,R,S,T,V and L(red line)=x=RS and PR=a and RV=b I decide to use an orthonormal repair where P is the center. Equation of line (PT) : y=(b*sqrt(3)/2)/(a+b/2)*x Equation of line (QV) : y=(-a*sqrt(3)/2)/(a/2+b)*(x-(a+b)) S (xS ; yS) intersection point of (PT) and (QV) Then : yS=(b*sqrt(3)/2)/(a+b/2)*xS=(-a*sqrt(3)/2)/(a/2+b)*(xS-(a+b)) After many calculus : xS=a/2*(2*a+b)*(a+b)/(a^2+b^2+a*b) yS=b*sqrt(3)/(2*a+b)*xS Equation 1 : QS^2=(xS-xQ)^2+(yS-yQ)^2 with xQ=a/2 and yQ=sqrt(3)/2*a 6^2=(xS-a/2)^2+(yS-sqrt(3)/2*a)^2 36=(xS-a/2)^2+(b*sqrt(3)/(2*a+b)*xS-sqrt(3)/2*a)^2 Equation 2 : ST^2=(xS-xT)^2+(yS-yT)^2 with xT=a+b/2 and yT=sqrt(3)/2*b 16^2=(xS-(a+b/2))^2+(yS-sqrt(3)/2*b)^2 256=(xS-(a+b/2))^2+(b*sqrt(3)/(2*a+b)*xS-sqrt(3)/2*b)^2 In fact, Equation 1 and Equation 2 are 2 equations in function of 2 variables "a" and "b". Because xS is expressed in function of "a" and "b". After many calculus, Equation 1 become : 36=a^4/(a^2+b^2+a*b) 6=a^2/sqrt(a^2+b^2+a*b) Equation 2 become : 256=b^4/(a^2+b^2+a*b) 16=b^2/sqrt(a^2+b^2+a*b) What is the value of x ? x^2=RS^2=(xS-xR)^2+(yS-yR)^2 with xR=a and yR=0 x^2=(a-xS)^2+yS^2 After many calculus : a-xS=a*b/2*(b-a)/(a^2+b^2+a*b) After many other calculus, we obtain : x=a*b/sqrt(a^2+b^2+a*b) x^2=a^2*b^2/(a^2+b^2+a*b) x^2=a^2/sqrt(a^2+b^2+a*b)*b^2/sqrt(a^2+b^2+a*b) x^2=6*16 x=sqrt(6*16) x=4*sqrt(6) Nota bene : sqrt(a^2+b^2+a*b)=a^2/6=b^2/16=a*b/x a^2/b^2=6/16 a/b=sqrt(6)/4 b/16=1/b*sqrt(a^2+b^2+a*b) b=16*sqrt(1+(a/b)+(a/b)^2) b=16*sqrt(1+sqrt(6)/4+6/16) b=4*sqrt(22+4*sqrt(6))=22,5 m on average a=sqrt(6)/4*b a=sqrt(6)*sqrt(22+4*sqrt(6))=13,8 m on average

Lado del cuadrado AQRS: 32/4=8 → QPA=45º→ AP=AQ→ PQ=8√2=QS. QPT=45º+15º=60º y PQ=PT=QT=QS→ PQT es triángulo equilátero con ángulos 60º/60º/60º→ AQB=60º-45º=15º→ ABQ=90º-15º=75º=TBS. QP y QS son diagonales de cuadrados de 8*8 adosados y por tanto son perpendiculares→ TQS=90º-60º=30º→ Cómo QT=QS→QTS=QST=75º→ Los triángulos SBT y SQT son isósceles y semejantes → Si M es punto medio de BT y QS=8√2→ MS=4√2→ QM=4√2√3→ MT=MB=QT-QM=8√2-4√2√3=4√2(2-√3) → Razón de semejanza entre MBS y BAQ =s=MS/AQ=4√2/8=√2/2→ Razón entre áreas =s² =2/4=1/2→ Áreas SBT=QTS → Área azul =BAQ+SBT =2*MT*MS*2/2=2MT*MS =2[4√2(2-√3)]*4√2 =128-64√3 m². Sencillo pero farragoso. Gracias y un saludo cordial.

1/ Consider the triangle of which the area= 99 sq m, label the left lower vertex as E and the upper vertex as F. Focus on the two triangles AEB and CED, the sum of the their area= 1/2 that of the square So, area of ABE+ 99+ area of CFD = sqx/2 Or 9x/2 +99+ area of CFD= sqx/2 (1) Moreover, we have: area of BFC + Area of CFD= 1/2 sqx/2 Or 15x/2 + area of CFD= sqx/2. (2) (1)=(2) so 9x/2 + 99= 15x/2 -> x = 33 Area of the square =sq33=1089 sq m😅😅😅

I think I could solve it . a and b are equal as a=64-32√3 and b=64-32√3 so a+b=128-64√3 Thank you sir for your great efforts.

Second method : Same method like you from 0:00 to 6:40 Blue area = a+b a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3)) a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3)) b=area(BST)=BS*H/2 and BS=AS-AB=8-8*(2-sqrt(3))=8*(sqrt(3)-1) H=PT*sin(15°)=8*sqrt(2)*(sqrt(6)-sqrt(2))/4=4*(sqrt(3)-1) Then, b=BS*H/2=8*(sqrt(3)-1)*4*(sqrt(3)-1)/2=16*(sqrt(3)-1)^2=16*(4-2*sqrt(3))=32*(2-sqrt(3)) Blue area=a+b=32*(2-sqrt(3))+32*(2-sqrt(3))=64*(2-sqrt(3))

Cool ! My first method : Same method like you from 0:00 to 12:35 But, with : a+c=b+c=32, i find a=b (i won't calculate c, i choose to calculate a) Blue area = a+b=2*a a=area(QAB)=QA*AB/2 and QA=8 and AB=QA*tan(15°)=8*(2-sqrt(3)) a=8*8*(2-sqrt(3))/2=32*(2-sqrt(3)) Blue area=2*a=64*(2-sqrt(3))

1/ Drop the height PH to BR. Label PH= h Because the 2 trianhles BPH and BRA are similar, so: PH/BH= RA/BA= 1/2 --> BH= 2PH -> sq h + sq (2h) =25 -> h= sqrt5 2/ we also have: sq BP =BQxBR BQ= 25/ BR = 25/(5sqrt5) = sqrt5 3/ Area of triangle PQR = Area of BRP- Area of PQB =1/2( 5x5 - sqrt5 x sqrt5) =20/2= 10 sq km😅😅😅

A = ½ b.h b = 5/cos45° = 5√2 km h = 2/5 b = 2√2 km A = 10 km² ( Solved √ ) !! without calculador !!

A|PQR| = A|BPR| - A|BPQ| A|BPR|=BP*AR/2=5*5/2=25/2 km² A|BPQ|=BP*xQ/2=5*xQ/2 We must find coordinates of Q (xQ ; yQ) A center of the orthonormal frame. Equation of line (BR) : y=10-2*x ; Equation of the circle : (x-5)^2+(y-5)^2=5^2 We combine these 2 equations and we obtain the coordinates of the 2 intersection points Q and R. (x-5)^2+(5-2*x)^2=25 5*x^2-30*x+50=25 5*x^2-30*x+25=0 x^2-6*x+5=0 (x-1)*(x-5)=0 x=1=xQ then y=yQ=8 OR x=5=xR then y=yR=0 A|PQR| = A|BPR| - A|BPQ| = 25/2 - 5*1/2 = 10 km²

BR=√(5²+10²)=5√5---> BQ*BR=BP²--->BQ=√5---> QR=BR-BQ=4√5 --->. PC Y BR son perpendiculares y se cortan en S---> Razón de semejanza entre PSB y RAB = s=5/5√5=√5/5--> PS=√5---> Área sombreada PQR=QR*PS/2=10 km². Gracias y saludos

Use the designations given at 6:20. In addition, let the center of the circle be designated as point O. Construct OP and let the intersection with BR be designated as point S. Note that ΔBPS and ΔBAR are similar, corresponding sides of ΔBPS half as long as ΔBAR, so length PS = (1/2)AR = 2.5. ΔPSR has base PS = 2.5, height OR = 5, therefore area (1/2)(2.5)(5) = 6.25. Angle a is an inscribed angle intersecting arc PQ. <POQ is a central angle intersecting the same arc, so is twice as large. tan(a) = 1/3 as computed in the video. tan(2a) = 3/4 from the tangent double angle formula. Drop a perpendicular from Q to OP and label the intersection as point T. Note that tan(2a) = QT/OT = 3/4. We recognize ΔQTO as a 3-4-5 right triangle, so ratio, QT/OQ = 3/5. However, OQ is a radius, length 5, so QT has length 3. ΔQPS has base PS and height QT, so area (1/2)(2.5)(3) = 3.75. Area ΔPQR = area ΔPSR + area ΔQPS = 6.25 + 3.75 = 10 km², as Math and Engineering also found.

Here is another method : x^2*y^2=7203 (2*r)^2=4*r^2=x^2+y^2 P (0;0) center of an orthonormal repair C center of the semi-circle R (xR;0) A (xA;0) B (xB;yB) Q (xQ;yQ) C (r;0) (same points as 1:30) cos(k)=cos(angle BRP)=x/(2*r) then : xR-xB=BR*cos(k)=x*cos(k)=x^2/(2*r) 2*r-xB=x^2/(2*r) then xB=2*r-x^2/(2*r) yB=BR*sin(k)=x*sin(k)=x*y/(2*r) Equilateral triangle then : xQ=(2*r)*cos(60deg)=2*r*1/2=r and yQ=(2*r)*sin(60deg)=2*r*sqrt(3)/2=sqrt(3)*r 7^2=(xB-xQ)^2+(yQ-yB)^2 49=(2*r-x^2/(2*r)-r)^2+(sqrt(3)*r-x*y/(2*r))^2 49=(r-x^2/(2*r))^2+(sqrt(3)*r-x*y/(2*r))^2 49=r^2-x^2+x^4/(2*r)^2+3*r^2-sqrt(3)*x*y+(x*y)^2/(2*r)^2 49=(x^4+x^2*y^2)/(2*r)^2+4*r^2-x^2-sqrt(3)*x*y 49=x^2*(x^2+y^2)/(2*r)^2+y^2-sqrt(3)*x*y 49=x^2+y^2-sqrt(3)*x*y 49=x^2+y^2-sqrt(3)*sqrt(7203) 49=x^2+y^2-147 4*r^2=x^2+y^2=147+49=196=4*49 r^2=49 (then r=7) A(semi-circle)=Pi/2*r^2=49/2*Pi Nota Bene : x^2+y^2=x^2+7203/x^2=y^2+7203/y^2=196 x^2+7203/x^2-196=0 (x^2-49)*(1-147/x^2)=0 x=sqrt(49)=7 OR x=sqrt(147)=7*sqrt(3) x<y then x^4<x^2*y^2=7203 then x=7*sqrt(3) impossible Then x=7 and y=sqrt(7203)/x=7*sqrt(3)

Here is my version with trigonometry to share with. <QBP+<QBR=270 hence cos(<QBR))=cos(270-<QBP)=-sin(QBP) Let the side of the equilateral triangle be a=2*r, then x^2+y^2=a^2 ->y^2-a^2=-x^2 and x^2-a^2=-y^2 cos(<QBP)=y^2+7^2-a^2)/(2*7*y)=(y^2+49-a^2)/(14*y)=(49-x^2)/(14*y)….(1) cos(<QBR)=x^2+7^2-a^2)/(2*7* )=(x^2+49-a^2)/(14*x)=(49-y^2)/(14*x)=-sin(QBP) …(2) Squaring (1) and (2) and adding and simplifying x^6+y^6-98*(x^4+y^4)+49^2*(x^2+y^2)=196*x^2*y^2 (x^2+y^2)^3-3*x^2*y^2*(x^2+y^2)-98*((x^2+y^2)^2-2*x^2*y^2)+49^2*(x^2+y^2)= 196*x^2*y^2 canceling out 196*x^2*y^2 on both sides and let x^2+y^2=49*t to scale down t^3-2*t^2-8*t=0 we get t=4 discarding t=0 and t=-2 Hence x^2+y^2=t*49=4*49=4*r*r then r=7 Semicircle area=1/2*pi*7*7=49*pi/2 that’s our answer.