Original and impressive ! Here are my methods : P,Q,T,V are on the (semi)circle of center S and radius R. angle(RPV)=c ; angle(RVP)=d c+60°+d=180° then d=120°-c area(PQS)=16 PS*cos(c)*PS*sin(c)=16 1/2*R^2*sin(2*c)=16 R^2*sin(2*c)=32 area(VTS)=30 SV*cos(d)*SV*sin(d)=30 1/2*R^2*sin(2*d)=30 R^2*sin(2*d)=60 R^2*sin(240°-2*c)=60 R^2*(sin(240°)*cos(2*c)-cos(240°)*sin(2*c))=60 R^2*(-sqrt(3)/2*cos(2*c)-(-1/2)*sin(2*c))=60 R^2*(-sqrt(3)*cos(2*c)+sin(2*c))=2*60 -sqrt(3)*R^2*cos(2*c)+R^2*sin(2*c)=120 -sqrt(3)*R^2*cos(2*c)+32=120 sqrt(3)*R^2*cos(2*c)=-88 R^2*cos(2*c)=-88/sqrt(3) (R^2*cos(2*c))^2+(R^2*sin(2*c))^2=(-88/sqrt(3))^2+32^2 R^4*((cos(2*c))^2+(sin(2*c))^2)=88^2/3+32^2 R^4=8^2*(121/3+16) R^4=8^2*169/3 R^4=(8*13)^2/3 R^2=104/sqrt(3) Method 1 : we will notice that QST is equilateral and RQT similar to PRV with ratio=1/2 angle(QST)=180°-angle(PSQ)-angle(VST) angle(PSQ)=180°-2*angle(RPV)=180°-2*c angle(VST)=180°-2*angle(RVP)=180°-2*d angle(QST)=180°-(180°-2*c)-(180°-2*d) angle(QST)=2*(c+d)-180° angle(QST)=2*120°-180° angle(QST)=60° But QST is an isosceles triangle then : angle(SQT)=angle(STQ)=(180°-angle(QST))/2=(180°-60°)/2=60° Then, QST is an equilateral triangle. QT=QS=ST=R angle(RQT)=180°-angle(SQT)-angle(PQS) angle(RQT)=180°-60°-c angle(RQT)=120°-c angle(RQT)=d angle(RTQ)=180°-angle(STQ)-angle(STV) angle(RQT)=180°-60°-d angle(RQT)=120°-d angle(RQT)=c Triangle RQT is similar to triangle PRV : same angles : c, d, 60° QT=R=PV/2. Then : QR=RV/2 and RT=PR/2 and : Area(RQT)=(1/2)^2*Area(PRV) Area(RQT)=1/4*Area(PRV) 4*Area(RQT)=Area(PRV) 4*Area(RQT)=Area(RQT)+Area(QST)+Area(PQS)+Area(STV) 3*Area(RQT)=Area(QST)+Area(PQS)+Area(STV) Area(RQT)=1/3*Area(QST)+1/3*(Area(PQS)+Area(STV)) Area(RQT)=1/3*Area(QST)+1/3*(16+30) Area(QRTS)=Area(RQT)+Area(QST) Area(QRTS)=4/3*Area(QST)+1/3*(16+30) Area(QRTS)=4/3*sqrt(3)/4*R^2+46/3 Area(QRTS)=1/sqrt(3)*104/sqrt(3)+46/3 Area(QRTS)=104/3+46/3 Area(QRTS)=50 u² Method 2 : we will calculate tan(c) and tan(d) tan(2*c)=(R^2*sin(2*c))/(R^2*cos(2*c)) tan(2*c)=32/(-88/sqrt(3)) tan(2*c)=-4*sqrt(3)/11 With formula : tan(2*c)=2*tan(c)/(1-(tan(c))^2), we obtain the following second degree equation : 2*sqrt(3)*T^2-11*T-2*sqrt(3)=0 (with T=tan(c)) T=tan(c)=2*sqrt(3) (because we can see that tan(c)>0 then the other solution T=tan(c)=-sqrt(3)/6 is impossible) tan(d)=tan(120°-c) tan(d)=(tan(120°)-tan(c))/(1+tan(120°)*tan(c)) tan(d)=(-sqrt(3)-2*sqrt(3))/(1-sqrt(3)*2*sqrt(3)) tan(d)=-3*sqrt(3)/(-5) tan(d)=3*sqrt(3)/5 Area(PRV)=1/2*Base*Height Base=PV=2*R Height=Base*tan(c)*tan(d)/(tan(c)+tan(d)) : i don't explain this formula i know Area(PRV)=2*R^2*tan(c)*tan(d)/(tan(c)+tan(d)) Area(PRV)=2*104/sqrt(3)*2*sqrt(3)*3*sqrt(3)/5/(2*sqrt(3)+3*sqrt(3)/5) Area(PRV)=96 Area(PRV)=Area(QRTS)+Area(PQS)+Area(STV) 96=Area(QRTS)+16+30 Area(QRTS)=50 u²
Supposing that isosceles triangle TSV is a right triangle: Side of isosceles right triangle: ½s² = 30cm² ---> s= √60=2√15cm Base of larger triangle: b = 2s = 4√15 cm Angle RPV: α = 180°-60°-45°= 75° Sine rule: c/sin75°= b/sin60° --> c=17,279 cm Area of triangle RPV: A₁= ½.b.c.sin45° = 94,641 cm² Quadrilateral shaded area: A = A₁-30-16 = 48,64 cm² (which is an approximate result)
It's a pity. In the drawing of this video (which seems to be done with right proportions), triangle TSV seems to be an isosceles right-angled triangle. But it's false. tan(angle(SVT))=3*sqrt(3)/5=1,039 on average (and not equal to 1) angle(SVT)=angle(STV)=46,1° on average (and not 45°) angle(VST)=2*arctan(5/(3*sqrt(3)))=87,8° on average (and not 90°) Then, the answer is A=50 u² (and not 48,64 cm²)
@@marioalb9726 I computed the correct value for the angle that you assumed was 90°, and followed your method. I've posted my calculations as a separate comment.
It's a pity. In the drawing of this video (which seems to be done with right proportions), triangle TSV seems to be an isosceles right-angled triangle. But it's false. tan(angle(SVT))=3*sqrt(3)/5=1,039 on average (and not equal to 1) angle(SVT)=angle(STV)=46,1° on average (and not 45°) angle(VST)=2*arctan(5/(3*sqrt(3)))=87,8° on average (and not 90°) Then, the answer is A=50 u² (and not 49,64 cm²)
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Let's assume that
This is marioalb9726's method, done with the correct value of
{16°p+30°TV+60°R}={106°PTVR+74°S}=180°PTVRS 360°/180°PTVRS=2PTVRS (PTVRS ➖ 2PTVRS+2).
Original and impressive !
Here are my methods :
P,Q,T,V are on the (semi)circle of center S and radius R.
angle(RPV)=c ; angle(RVP)=d
c+60°+d=180° then d=120°-c
area(PQS)=16
PS*cos(c)*PS*sin(c)=16
1/2*R^2*sin(2*c)=16
R^2*sin(2*c)=32
area(VTS)=30
SV*cos(d)*SV*sin(d)=30
1/2*R^2*sin(2*d)=30
R^2*sin(2*d)=60
R^2*sin(240°-2*c)=60
R^2*(sin(240°)*cos(2*c)-cos(240°)*sin(2*c))=60
R^2*(-sqrt(3)/2*cos(2*c)-(-1/2)*sin(2*c))=60
R^2*(-sqrt(3)*cos(2*c)+sin(2*c))=2*60
-sqrt(3)*R^2*cos(2*c)+R^2*sin(2*c)=120
-sqrt(3)*R^2*cos(2*c)+32=120
sqrt(3)*R^2*cos(2*c)=-88
R^2*cos(2*c)=-88/sqrt(3)
(R^2*cos(2*c))^2+(R^2*sin(2*c))^2=(-88/sqrt(3))^2+32^2
R^4*((cos(2*c))^2+(sin(2*c))^2)=88^2/3+32^2
R^4=8^2*(121/3+16)
R^4=8^2*169/3
R^4=(8*13)^2/3
R^2=104/sqrt(3)
Method 1 : we will notice that QST is equilateral and RQT similar to PRV with ratio=1/2
angle(QST)=180°-angle(PSQ)-angle(VST)
angle(PSQ)=180°-2*angle(RPV)=180°-2*c
angle(VST)=180°-2*angle(RVP)=180°-2*d
angle(QST)=180°-(180°-2*c)-(180°-2*d)
angle(QST)=2*(c+d)-180°
angle(QST)=2*120°-180°
angle(QST)=60°
But QST is an isosceles triangle then :
angle(SQT)=angle(STQ)=(180°-angle(QST))/2=(180°-60°)/2=60°
Then, QST is an equilateral triangle.
QT=QS=ST=R
angle(RQT)=180°-angle(SQT)-angle(PQS)
angle(RQT)=180°-60°-c
angle(RQT)=120°-c
angle(RQT)=d
angle(RTQ)=180°-angle(STQ)-angle(STV)
angle(RQT)=180°-60°-d
angle(RQT)=120°-d
angle(RQT)=c
Triangle RQT is similar to triangle PRV : same angles : c, d, 60°
QT=R=PV/2.
Then : QR=RV/2 and RT=PR/2 and :
Area(RQT)=(1/2)^2*Area(PRV)
Area(RQT)=1/4*Area(PRV)
4*Area(RQT)=Area(PRV)
4*Area(RQT)=Area(RQT)+Area(QST)+Area(PQS)+Area(STV)
3*Area(RQT)=Area(QST)+Area(PQS)+Area(STV)
Area(RQT)=1/3*Area(QST)+1/3*(Area(PQS)+Area(STV))
Area(RQT)=1/3*Area(QST)+1/3*(16+30)
Area(QRTS)=Area(RQT)+Area(QST)
Area(QRTS)=4/3*Area(QST)+1/3*(16+30)
Area(QRTS)=4/3*sqrt(3)/4*R^2+46/3
Area(QRTS)=1/sqrt(3)*104/sqrt(3)+46/3
Area(QRTS)=104/3+46/3
Area(QRTS)=50 u²
Method 2 : we will calculate tan(c) and tan(d)
tan(2*c)=(R^2*sin(2*c))/(R^2*cos(2*c))
tan(2*c)=32/(-88/sqrt(3))
tan(2*c)=-4*sqrt(3)/11
With formula : tan(2*c)=2*tan(c)/(1-(tan(c))^2), we obtain the following second degree equation :
2*sqrt(3)*T^2-11*T-2*sqrt(3)=0 (with T=tan(c))
T=tan(c)=2*sqrt(3) (because we can see that tan(c)>0 then the other solution T=tan(c)=-sqrt(3)/6 is impossible)
tan(d)=tan(120°-c)
tan(d)=(tan(120°)-tan(c))/(1+tan(120°)*tan(c))
tan(d)=(-sqrt(3)-2*sqrt(3))/(1-sqrt(3)*2*sqrt(3))
tan(d)=-3*sqrt(3)/(-5)
tan(d)=3*sqrt(3)/5
Area(PRV)=1/2*Base*Height
Base=PV=2*R
Height=Base*tan(c)*tan(d)/(tan(c)+tan(d)) : i don't explain this formula i know
Area(PRV)=2*R^2*tan(c)*tan(d)/(tan(c)+tan(d))
Area(PRV)=2*104/sqrt(3)*2*sqrt(3)*3*sqrt(3)/5/(2*sqrt(3)+3*sqrt(3)/5)
Area(PRV)=96
Area(PRV)=Area(QRTS)+Area(PQS)+Area(STV)
96=Area(QRTS)+16+30
Area(QRTS)=50 u²
Supposing that isosceles triangle TSV is a right triangle:
Side of isosceles right triangle:
½s² = 30cm² ---> s= √60=2√15cm
Base of larger triangle:
b = 2s = 4√15 cm
Angle RPV:
α = 180°-60°-45°= 75°
Sine rule:
c/sin75°= b/sin60° --> c=17,279 cm
Area of triangle RPV:
A₁= ½.b.c.sin45° = 94,641 cm²
Quadrilateral shaded area:
A = A₁-30-16 = 48,64 cm²
(which is an approximate result)
It's a pity.
In the drawing of this video (which seems to be done with right proportions), triangle TSV seems to be an isosceles right-angled triangle.
But it's false.
tan(angle(SVT))=3*sqrt(3)/5=1,039 on average (and not equal to 1)
angle(SVT)=angle(STV)=46,1° on average (and not 45°)
angle(VST)=2*arctan(5/(3*sqrt(3)))=87,8° on average (and not 90°)
Then, the answer is A=50 u² (and not 48,64 cm²)
@@matthieudutriaux
Yes, I wrote "Supposing, it was an approximate result"
I couldn't solve this
You found that
@@marioalb9726 I computed the correct value for the angle that you assumed was 90°, and followed your method. I've posted my calculations as a separate comment.
Supposing that isosceles triangle TSV is a right triangle:
Side of Isosceles right triangle:
½s² = 30cm² ---> s= √60cm
Internal angles:
α = 180°-60°-45°= 75° (∠RPV)
β = 180°-(2*75°)= 30° (∠QSP)
δ = 90°-β = 60° , so ΔQST is equilateral
γ = 180°-α-δ = 45° (∠RQT)
Sine rule, triangle QRT:
a/sin75°=s/sin60° --> a=8,6395 cm
Triangle RQT:
A₁= ½.a.s.sin45° = 23,66cm²
Equilateral triangle QST:
A₂ = √3/4 s² = 25,98 cm²
Quadrilateral shaded area:
A = A₁+A₂ = 49,64 cm² ( Which is an approximate result )
It's a pity.
In the drawing of this video (which seems to be done with right proportions), triangle TSV seems to be an isosceles right-angled triangle.
But it's false.
tan(angle(SVT))=3*sqrt(3)/5=1,039 on average (and not equal to 1)
angle(SVT)=angle(STV)=46,1° on average (and not 45°)
angle(VST)=2*arctan(5/(3*sqrt(3)))=87,8° on average (and not 90°)
Then, the answer is A=50 u² (and not 49,64 cm²)
Yes, I wrote "an approximate result, supposing that ...." )
I Couldn't solve it, yet