Just another way using secant theorem: 1. Let b= BP; a = CP, t= angle BPC; 2. Secant theorem: PB*PA = PC*PD; => b*(6+b) = a(9+a); (1) 3. Let us apply Cosine theorem to triangle ADP: (6+b)^2=6^2 + (9+a)^2 - 2*6*(9+a)*cos(60); (6+b)^2 = 36+(9+a)^2 - 6*(9+a); (2) 4. System of equations (1) and (2) => {a=7;b=8} and {a=-9;b=0} - rejected 5. Let us apply Sine theorem to triangle ADP: 6/sin(t)=(6+8)/sin(60) => sin(t) = 3*sqrt(3)/14; 6. Ayellow = a*b*sin(t)/2 = 7*8*(3*sqrt(3)/14)/2 = 6*srt(3) sq.units.
ADC=PBC=60° Connect A to C In∆ ADC Law of cosines AC^2=AD^2+DC^2-2(AD)(DC)cos(60°} AC^2=6^2+9^2-2(6)(9)(1/2) so AC=3√7 ABC=180°-60°=120° In∆ ABC Law of cosines AC^2=AB^2+BC^2-2(AB)(BC)cos(120°) (3√7)^2=6^2+BC^2-2(6)(BC)(-1/2) So BC=3 ∆ADP ~ ∆CBP AD/CB=AP/CP=DP/BP 6/3=AP/CP AP/CP=2/1 So AP=2a ; CP=a In ∆ABP AP^2=AB^2+DP^2-2(AB)(DP)cos(60°) (2a)^2=6^2+(9+a)^2-2(6)(9+a)(1/2) So a=7 So AP=14 and BP=14-6=8 So yellow area=1/2(BC)(BP)sin(60)=1/2(3)(8)sin(60°)=6√3 square units.❤❤❤ Thanks sir.
At 2:25, as an alternative to law of cosines, drop a perpendicular from A to CD and label the intersection as point E. Note that ΔADE is a special 30°-60°-90° right triangle. Using its ratio of sides, DE = 3 and AE = 3√3. ΔACE is a right triangle. CE = CD - DE = 9 - 3 = 6 and AE = 3√3, so apply the Pythagorean theorem to find t = AC = √(63). However, I find no similar strategy for finding x = BC. Law of cosines seems to be the most straightforward way. At 7:18, once we find that ΔADP and ΔBCP are similar and side AD is twice as long as corresponding side BC, then AP = 2(PC) and DP = 2(BP), from which the equations m + 6 = 2n and n + 9 = 2m can be written and rearranged as PreMath's m = 2n - 6 and 2m = n + 9. Proceed to find m, n and area ΔBCP as in the video.
Very nice solution! I used Ptolemy theorem, Heron’s formula along with several applications of the laws of sines and cosines and lots of algebraic expressions - rather involved but finally ended with the same solution.
Another method a little bit lengthy and difficult 1. Find diagonal AC and side BC using cosine 2. Find diagonal BD using Ptolemy theorem 3.find angle BCD 4.convert angle BCD TO ANGLE BCP USING COS(π-θ) 5. Find angle BPC using cos(120-θ) since we know value of θ 6. Appy sine to find side BP 7. Find required area using 1/2* BP*BC*sin60° 8. Which gives 6√3 as answer 👀
I used simply means of Geometry. 01) Let Point D has the Coordinates (0 ; 0) 02) Point P has Coordinates (16 ; 0) 03) Pont D has Coordinates (9 ; 0) 04) Area of Triangle [ADP] = 16 * 5,196 / 2 ; A = 41,568 Sq un. Height (h) of Triangle ADP = AA' = h = 6 * sin (60º) = 6 * (sqrt(3) / 2) ; h = 3*sqrt(3) lin un ; h ~ 5,1962 lin un. 05) So : Segment CP = (16 - 9) = 7 lin un 06) Vertical Distance between B and Line DP is equal to 2,969 lin un 07) Triangle [BCP] Area = (2,969 * 7) / 2 ; A = 20,783 / 2 ; A = 10,3915 sq un Thus, OUR ANSWER IS : Yellow Area is equal to approx. 10,4 Square Units. NOTE: I think that should be more Geometry and less Algebra in the Process of Solving these kind of Problems. To find the Center of the Circle I used a simple Method, drawing two Straight Lines Perpendicular and at the Middle Point of Line AD and Line DC, and I found its Intersect Point with Coordinates of C (4,5 ; sqrt(3)/2) and Radius equal to sqrt(21). 4,5^2 + (sqrt(3)/2)^2 = 20,25 + 0,75 = 21. R = sqrt(21). Then I got all the Coordinates and Distances I need to solve the given Problem. Greetings from The Islamic State Caliphate in Cordoba.
1. DE = 6 이 되도록 CD 위에 점 E를 설정합니다 2. △ADE 는 정삼각형입니다 3. △ABE 는 이등변삼각형입니다 4. △BCE 도 이등변삼각형이 됨을 파악합니다. (∠BEC 와 ∠EBC 크기가 왜 같을까요?) 5. CE =3 이므로 BC = 3 임을 얻습니다. 6. 이후의 풀이는 영상의 진행과 동일합니다. Translation by Translator 1. Set point E above the CD to DE = 6 2. △ADE is an equilateral triangle 3. △ABE is an isosceles triangle 4. Understand that the △BCE is also an isosceles triangle. (Why is the ∠ BEC and ∠ EBC the same size?) 5. Get BC = 3 because CE = 3. 6. The subsequent solution is the same as the progress of the video.
Cosine rule, twice: x²+6²-2.6.x.cos120°=9²+6²-2.6.9.cos60° x²+6x=27 x = 3cm Similarity of triangles: n/3 = (m+6)/6 n = (m+6)/2 Secant-secant theorem: m(m+6)=n(n+9) Replacing: m(m+6)=(m+6)/2 . [(m+6)/2 + 9] m = (m+6)/4 + 9/2 = m/4 + 6 m = 8 cm Area of triangle: A = ½.b.h = ½ m. x. sin60° A = ½ 8.(3.sin60°) A = 10,39 cm² ( Solved √ )
there is only one solution. the area has been calculated using the gauss area formula: 10 print "premath-can you find area of the yellow shaded triangle":dim x(1,3),y(1,3) 20 l1=9:l2=6:sw=l2/30:x1=0:y1=0:x2=l1*cos(rad(20)):y2=l1*sin(rad(20)) 30 x3=l2*cos(rad(80)):y3=l2*sin(rad(80)):a11=2*(x2-x1):a12=2*(y2-y1):a13=x2^2+y2^2-x1^2-y1^2 40 a21=2*(x3-x2):a22=2*(y3-y2):a23=x3^2+y3^2-x2^2-y2^2:goto 100 50 ngl1=a12*a21:ngl2=a22*a11 60 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 70 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 80 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 90 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl:r=sqr((x1-xl)^2+(y1-yl)^2):return 100 gosub 50:xm1=xl:ym1=yl:r1=r:xm2=x3:ym2=y3:r2=l2:xs=-l1 110 if xm2=xm1 or ym2=ym1 then 210 120 gosub 160:goto 200 130 ys=(r1^2-xm1^2-ym1^2-r2^2+xm2^2+ym2^2-2*xs*(xm2-xm1))/2/(ym2-ym1) 140 dg=((xs-xm2)^2+(ys-ym2)^2)/r2^2:dg=dg-1:return 150 if xm1=xm2 then else 150 160 gosub 130 170 dg1=dg:xs1=xs:xs=xs+sw:gosub 130:xs2=xs:if dg1*dg>0 then 170 180 xs=(xs1+xs2)/2:gosub 130:if dg1*dg>0 then xs1=xs else xs2=xs 190 if abs(dg)>1E-10 then 180 else return 200 print xs,ys:xs=xs+sw:gosub 160:print xs,ys:goto 240 210 if xm2=xm1 then 230 220 xs=xm1:ys=(r1^2-ym1^2-r2^2+ym2)^2/2/(ym2-ym1):goto 240 230 ys=ym1:xs=(r1^2-xm1^2-r2^2+xm2)^2/2/(xm2-xm1):rem den geradenschnittpunkt berechnen *** 240 xg11=0:yg11=0:xg12=x2:yg12=y2:xg21=x3:yg21=y3:xg22=xs:yg22=ys 250 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12) 260 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22) 270 a13=a131+a132:a23=a231+a232 280 gosub 50:x(0,0)=x2:y(0,0)=y2:x(0,1)=xl:y(0,1)=yl:x(0,3)=xs:y(0,3)=ys:x(0,2)=(x(0,1)+x(0,3))/2 290 y(0,2)=(y(0,1)+y(0,3))/2:ar=0 300 for a=0 to 3:iam=a-1:if a=0 then iam=3 310 iap=a+1:if a=3 then iap=0 320 da=x(0,a)*(y(0,iam)-y(0,iap))/2:ar=ar+da:next a 330 print "die gesuchte flaeche ist=";ar 340 x(1,0)=x1:y(1,0)=y1:x(1,1)=x2:y(1,1)=y2:x(1,2)=xs:y(1,2)=ys:x(1,3)=x3:y(1,3)=y3 350 masx=1200/(xl-x3):masy=850/(yl-y1):if masx
Let's find the area: . .. ... .... ..... The sum of opposite angles in a cyclic quadrilateral is 180°. So we obtain: ∠ADC + ∠ABC = 180° ⇒ ∠ABC = 180° − ∠ADC = 180° − 60° = 120° Now we apply the law of cosines to the triangles ACD and ABC: AC² = AD² + CD² − 2*AD*CD*cos(∠ADC) AC² = 6² + 9² − 2*6*9*cos(60°) AC² = 6² + 9² − 2*6*9*(1/2) AC² = 36 + 81 − 54 = 63 ⇒ AC = 3√7 AC² = AB² + BC² − 2*AB*BC*cos(∠ABC) 63 = 6² + BC² − 2*6*BC*cos(120°) 63 = 6² + BC² − 2*6*BC*(−1/2) 63 = 36 + BC² + 6*BC 0 = BC² + 6*BC − 27 BC = −3 ± √(3² + 27) = −3 ± √(9 + 27) = −3 ± √36 = −3 ± 6 Since BC>0, the only useful solution is BC=−3+6=3. Now we are able to apply the theorem of Ptolemy: AC*BD = AB*CD + AD*BC (3√7)*BD = 6*9 + 6*3 = 54 + 18 = 72 ⇒ BD = 72/(3√7) = 24/√7 Let's have a look at the triangles ACP and BDP. Both triangles share the angle BPC. Since A and D are both located on the circle and on the same side of the chord BC, we know that the angles ∠BAC(=∠PAC) and ∠BDC(=∠BDP) are identical. Therefore these two triangles are similar and we can conclude: AC/BD = AP/DP = CP/BP AC/BD = (AB + BP)/(CD + CP) = CP/BP (3√7)/(24/√7) = (6 + PB)/(9 + CP) = CP/BP 7/8 = (6 + PB)/(9 + CP) = CP/BP 7/8 = CP/BP ⇒ CP = (7/8)*BP 7/8 = (6 + BP)/(9 + CP) 7*(9 + CP) = 8*(6 + BP) 63 + 7*CP = 48 + 8*BP 63 + (49/8)*BP = 48 + 8*BP 15 = (64/8)*BP − (49/8)*BP 15 = (15/8)*BP ⇒ BP = 8 ⇒ CP = 7 Now we are able to calculate the area of the yellow triangle by applying the formula of Heron: BC = 3 BP = 8 CP = 7 s = (3 + 8 + 7)/2 = 18/2 = 9 A = √[s*(s − BC)*(s − BP)*(s − CP)] = √(9*6*1*2) = = √(3²*2²*3) = 6√3 Best regards from Germany
6√3....non lo controllo più..con il teorema del coseno risulta BC=3..DAB=arccos(-1/7)...quindi del triangolo giallo conosco tutti gli angoli e un lato...percio calcolo l'area che risulta 6√3
This one was a lot of fun to solve. Thanks for sharing.
Glad you enjoyed it!
You are very welcome!
Thanks for the feedback ❤️
Just another way using secant theorem:
1. Let b= BP; a = CP, t= angle BPC;
2. Secant theorem: PB*PA = PC*PD; => b*(6+b) = a(9+a); (1)
3. Let us apply Cosine theorem to triangle ADP:
(6+b)^2=6^2 + (9+a)^2 - 2*6*(9+a)*cos(60);
(6+b)^2 = 36+(9+a)^2 - 6*(9+a); (2)
4. System of equations (1) and (2) => {a=7;b=8} and {a=-9;b=0} - rejected
5. Let us apply Sine theorem to triangle ADP:
6/sin(t)=(6+8)/sin(60) => sin(t) = 3*sqrt(3)/14;
6. Ayellow = a*b*sin(t)/2 = 7*8*(3*sqrt(3)/14)/2 = 6*srt(3) sq.units.
This one is worth searching (a lot) ! Highly intricate - Thanks
Glad it was helpful!
Thanks for the feedback ❤️
Well done, sir! For one, how can a person be sure such a problem is solvable? Also, I had no clue how to do it without your explanation.
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
ADC=PBC=60°
Connect A to C
In∆ ADC
Law of cosines
AC^2=AD^2+DC^2-2(AD)(DC)cos(60°}
AC^2=6^2+9^2-2(6)(9)(1/2)
so AC=3√7
ABC=180°-60°=120°
In∆ ABC
Law of cosines
AC^2=AB^2+BC^2-2(AB)(BC)cos(120°)
(3√7)^2=6^2+BC^2-2(6)(BC)(-1/2)
So BC=3
∆ADP ~ ∆CBP
AD/CB=AP/CP=DP/BP
6/3=AP/CP
AP/CP=2/1
So AP=2a ; CP=a
In ∆ABP
AP^2=AB^2+DP^2-2(AB)(DP)cos(60°)
(2a)^2=6^2+(9+a)^2-2(6)(9+a)(1/2)
So a=7
So AP=14 and BP=14-6=8
So yellow area=1/2(BC)(BP)sin(60)=1/2(3)(8)sin(60°)=6√3 square units.❤❤❤ Thanks sir.
Thank you!
Fine!
Thanks for the feedback ❤️
A very tricky one ! But very fun !
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
At 2:25, as an alternative to law of cosines, drop a perpendicular from A to CD and label the intersection as point E. Note that ΔADE is a special 30°-60°-90° right triangle. Using its ratio of sides, DE = 3 and AE = 3√3. ΔACE is a right triangle. CE = CD - DE = 9 - 3 = 6 and AE = 3√3, so apply the Pythagorean theorem to find t = AC = √(63). However, I find no similar strategy for finding x = BC. Law of cosines seems to be the most straightforward way.
At 7:18, once we find that ΔADP and ΔBCP are similar and side AD is twice as long as corresponding side BC, then AP = 2(PC) and DP = 2(BP), from which the equations m + 6 = 2n and n + 9 = 2m can be written and rearranged as PreMath's m = 2n - 6 and 2m = n + 9. Proceed to find m, n and area ΔBCP as in the video.
Very nice.Thanks.
Very nice solution! I used Ptolemy theorem, Heron’s formula along with several applications of the laws of sines and cosines and lots of algebraic expressions - rather involved but finally ended with the same solution.
Another method a little bit lengthy and difficult
1. Find diagonal AC and side BC using cosine
2. Find diagonal BD using Ptolemy theorem
3.find angle BCD
4.convert angle BCD TO ANGLE BCP USING COS(π-θ)
5. Find angle BPC using cos(120-θ) since we know value of θ
6. Appy sine to find side BP
7. Find required area using 1/2* BP*BC*sin60°
8. Which gives 6√3 as answer 👀
Perfect
My way of solution is ▶
Lets consider the triangles: ΔADC and ΔCBA
∠ ADC= 60°
∠CBA= 180°-60°
∠CBA= 120° (according to the rule of Cyclic Quadrilateral)
if we apply the cosine theorem for the [AC]
for the ΔADC:
⇒
AC²= 6²+9²-2*6*9*cos(60°)
AC²= 36+81-108*0,5
AC²= 63
the cosine theorem for the ΔCBA:
BC= x
AC²= x²+6²-2*x*6*cos(120°)
cos(120°)= -0,5
AC²= x²+36+12x*0,5
AC²= x²+36+6x
⇒
x²+36+6x= 63
x²+6x-27=0
Δ= 36-4*1*(-27)
Δ= 144
√Δ= 12
x= (-6+12)/2
x= 3 length units
∠PBC= 180°-120°
∠PBC= 60°
ΔCPB ~ ΔAPD
BC/AD= CP/PA= BP/DP
CP= x
PB= y
⇒
3/6= x/(6+y)= y/(9+x)
⇒
1/2= x/(6+y)
2x= 6+y
y= 2x-6
1/2= y/(9+x)
2y= 9+x
2*(2x-6)= 9+x
4x-12= 9+x
3x= 12+9
3x= 21
x= 7
y= 2*7-6
y= 8
Ayellow= 1/2*BC*PB*sin(60°)
Ayellow= 1/2*3*8*√3/2
Ayellow= 6√3 square units
As ABCD is a cyclic quadrilateral, the opposite angles sum to 180°. Therefore ∠ABC = 180°-60° = 120°.
Let BC = x. Draw AC. By the law of cosines:
Triangle ∆CDA:
AC² = DA² + CD² - 2(DA)CDcos(60°)
AC² = 6² + 9² - 2(6)(9)(1/2)
AC² = 36 + 81 - 54 = 63
AC = √63 = 3√7
Triangle ∆ABC:
AC² = AB² + BC² - 2(AB)BCcos(120°)
63 = 6² + x² - 2(6)(x)(-1/2)
63 = 36 + x² + 6x
x² + 6x - 27 = 0
(x-3)(x+9) = 0
x = 3 | x = -9 ❌
As ∠ABC = 120°, ∠CBP = 180°-120° = 60°. As ∠CBP = ∠CDA and ∠P is common, ∆APD and ∆BPC are similar triangles.
Triangle ∆BPC:
PC/CB = AP/DA
PC/3 = (6+BP)/6
PC = (6+BP)/2 = BP/2 + 3
BP/CB = PD/DA
BP/3 = (PC+9)/6
BP = (PC+9)/2 = PC/2 + 9/2
BP = (BP/2+3)/2 + 9/2
BP = BP/4 + 3/2 + 9/2
3BP/4 = 6
BP = 6(4/3) = 8
Aᴛ = CB(BP)sin(60°)/2
Aᴛ = 3(8)(√3/4) = 6√3 ≈ 10.39 sq units
There is the theorm that m(m+6)=n(n+9) plus cosine theorm in triangle ADP construct two equations with two unknowns. Also a way to solve...
I used simply means of Geometry.
01) Let Point D has the Coordinates (0 ; 0)
02) Point P has Coordinates (16 ; 0)
03) Pont D has Coordinates (9 ; 0)
04) Area of Triangle [ADP] = 16 * 5,196 / 2 ; A = 41,568 Sq un. Height (h) of Triangle ADP = AA' = h = 6 * sin (60º) = 6 * (sqrt(3) / 2) ; h = 3*sqrt(3) lin un ; h ~ 5,1962 lin un.
05) So : Segment CP = (16 - 9) = 7 lin un
06) Vertical Distance between B and Line DP is equal to 2,969 lin un
07) Triangle [BCP] Area = (2,969 * 7) / 2 ; A = 20,783 / 2 ; A = 10,3915 sq un
Thus,
OUR ANSWER IS : Yellow Area is equal to approx. 10,4 Square Units.
NOTE: I think that should be more Geometry and less Algebra in the Process of Solving these kind of Problems. To find the Center of the Circle I used a simple Method, drawing two Straight Lines Perpendicular and at the Middle Point of Line AD and Line DC, and I found its Intersect Point with Coordinates of C (4,5 ; sqrt(3)/2) and Radius equal to sqrt(21). 4,5^2 + (sqrt(3)/2)^2 = 20,25 + 0,75 = 21. R = sqrt(21).
Then I got all the Coordinates and Distances I need to solve the given Problem.
Greetings from The Islamic State Caliphate in Cordoba.
Excellent!
Thanks for sharing ❤️
1. DE = 6 이 되도록 CD 위에 점 E를 설정합니다
2. △ADE 는 정삼각형입니다
3. △ABE 는 이등변삼각형입니다
4. △BCE 도 이등변삼각형이 됨을 파악합니다.
(∠BEC 와 ∠EBC 크기가 왜 같을까요?)
5. CE =3 이므로 BC = 3 임을 얻습니다.
6. 이후의 풀이는 영상의 진행과 동일합니다.
Translation by Translator
1. Set point E above the CD to DE = 6
2. △ADE is an equilateral triangle
3. △ABE is an isosceles triangle
4. Understand that the △BCE is also an isosceles triangle.
(Why is the ∠ BEC and ∠ EBC the same size?)
5. Get BC = 3 because CE = 3.
6. The subsequent solution is the same as the progress of the video.
Cosine rule, twice:
x²+6²-2.6.x.cos120°=9²+6²-2.6.9.cos60°
x²+6x=27
x = 3cm
Similarity of triangles:
n/3 = (m+6)/6
n = (m+6)/2
Secant-secant theorem:
m(m+6)=n(n+9)
Replacing:
m(m+6)=(m+6)/2 . [(m+6)/2 + 9]
m = (m+6)/4 + 9/2 = m/4 + 6
m = 8 cm
Area of triangle:
A = ½.b.h = ½ m. x. sin60°
A = ½ 8.(3.sin60°)
A = 10,39 cm² ( Solved √ )
there is only one solution. the area has been calculated using the gauss area formula:
10 print "premath-can you find area of the yellow shaded triangle":dim x(1,3),y(1,3)
20 l1=9:l2=6:sw=l2/30:x1=0:y1=0:x2=l1*cos(rad(20)):y2=l1*sin(rad(20))
30 x3=l2*cos(rad(80)):y3=l2*sin(rad(80)):a11=2*(x2-x1):a12=2*(y2-y1):a13=x2^2+y2^2-x1^2-y1^2
40 a21=2*(x3-x2):a22=2*(y3-y2):a23=x3^2+y3^2-x2^2-y2^2:goto 100
50 ngl1=a12*a21:ngl2=a22*a11
60 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
70 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2
80 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2
90 xl=zx/ngl:yl=zy/ngl:print "x=";xl;"y=";yl:r=sqr((x1-xl)^2+(y1-yl)^2):return
100 gosub 50:xm1=xl:ym1=yl:r1=r:xm2=x3:ym2=y3:r2=l2:xs=-l1
110 if xm2=xm1 or ym2=ym1 then 210
120 gosub 160:goto 200
130 ys=(r1^2-xm1^2-ym1^2-r2^2+xm2^2+ym2^2-2*xs*(xm2-xm1))/2/(ym2-ym1)
140 dg=((xs-xm2)^2+(ys-ym2)^2)/r2^2:dg=dg-1:return
150 if xm1=xm2 then else 150
160 gosub 130
170 dg1=dg:xs1=xs:xs=xs+sw:gosub 130:xs2=xs:if dg1*dg>0 then 170
180 xs=(xs1+xs2)/2:gosub 130:if dg1*dg>0 then xs1=xs else xs2=xs
190 if abs(dg)>1E-10 then 180 else return
200 print xs,ys:xs=xs+sw:gosub 160:print xs,ys:goto 240
210 if xm2=xm1 then 230
220 xs=xm1:ys=(r1^2-ym1^2-r2^2+ym2)^2/2/(ym2-ym1):goto 240
230 ys=ym1:xs=(r1^2-xm1^2-r2^2+xm2)^2/2/(xm2-xm1):rem den geradenschnittpunkt berechnen ***
240 xg11=0:yg11=0:xg12=x2:yg12=y2:xg21=x3:yg21=y3:xg22=xs:yg22=ys
250 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12)
260 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22)
270 a13=a131+a132:a23=a231+a232
280 gosub 50:x(0,0)=x2:y(0,0)=y2:x(0,1)=xl:y(0,1)=yl:x(0,3)=xs:y(0,3)=ys:x(0,2)=(x(0,1)+x(0,3))/2
290 y(0,2)=(y(0,1)+y(0,3))/2:ar=0
300 for a=0 to 3:iam=a-1:if a=0 then iam=3
310 iap=a+1:if a=3 then iap=0
320 da=x(0,a)*(y(0,iam)-y(0,iap))/2:ar=ar+da:next a
330 print "die gesuchte flaeche ist=";ar
340 x(1,0)=x1:y(1,0)=y1:x(1,1)=x2:y(1,1)=y2:x(1,2)=xs:y(1,2)=ys:x(1,3)=x3:y(1,3)=y3
350 masx=1200/(xl-x3):masy=850/(yl-y1):if masx
Let's find the area:
.
..
...
....
.....
The sum of opposite angles in a cyclic quadrilateral is 180°. So we obtain:
∠ADC + ∠ABC = 180° ⇒ ∠ABC = 180° − ∠ADC = 180° − 60° = 120°
Now we apply the law of cosines to the triangles ACD and ABC:
AC² = AD² + CD² − 2*AD*CD*cos(∠ADC)
AC² = 6² + 9² − 2*6*9*cos(60°)
AC² = 6² + 9² − 2*6*9*(1/2)
AC² = 36 + 81 − 54 = 63
⇒ AC = 3√7
AC² = AB² + BC² − 2*AB*BC*cos(∠ABC)
63 = 6² + BC² − 2*6*BC*cos(120°)
63 = 6² + BC² − 2*6*BC*(−1/2)
63 = 36 + BC² + 6*BC
0 = BC² + 6*BC − 27
BC = −3 ± √(3² + 27) = −3 ± √(9 + 27) = −3 ± √36 = −3 ± 6
Since BC>0, the only useful solution is BC=−3+6=3. Now we are able to apply the theorem of Ptolemy:
AC*BD = AB*CD + AD*BC
(3√7)*BD = 6*9 + 6*3 = 54 + 18 = 72
⇒ BD = 72/(3√7) = 24/√7
Let's have a look at the triangles ACP and BDP. Both triangles share the angle BPC. Since A and D are both located on the circle and on the same side of the chord BC, we know that the angles ∠BAC(=∠PAC) and ∠BDC(=∠BDP) are identical. Therefore these two triangles are similar and we can conclude:
AC/BD = AP/DP = CP/BP
AC/BD = (AB + BP)/(CD + CP) = CP/BP
(3√7)/(24/√7) = (6 + PB)/(9 + CP) = CP/BP
7/8 = (6 + PB)/(9 + CP) = CP/BP
7/8 = CP/BP ⇒ CP = (7/8)*BP
7/8 = (6 + BP)/(9 + CP)
7*(9 + CP) = 8*(6 + BP)
63 + 7*CP = 48 + 8*BP
63 + (49/8)*BP = 48 + 8*BP
15 = (64/8)*BP − (49/8)*BP
15 = (15/8)*BP
⇒ BP = 8
⇒ CP = 7
Now we are able to calculate the area of the yellow triangle by applying the formula of Heron:
BC = 3
BP = 8
CP = 7
s = (3 + 8 + 7)/2 = 18/2 = 9
A = √[s*(s − BC)*(s − BP)*(s − CP)] = √(9*6*1*2) = = √(3²*2²*3) = 6√3
Best regards from Germany
Excellent!
Thanks for sharing ❤️
When I look for a good explaination, I'm jumping to your solution every time. Best regards from Belgium.
It is a very difficult puzzle.😢
6√3....non lo controllo più..con il teorema del coseno risulta BC=3..DAB=arccos(-1/7)...quindi del triangolo giallo conosco tutti gli angoli e un lato...percio calcolo l'area che risulta 6√3
Excellent!
Thanks for sharing ❤️
That was haed for me 😂