Can you find area of the Yellow shaded triangle? | (Cyclic Quadrilateral) | #math #maths | #geometry

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Can you find area of the Yellow shaded triangle? | (Cyclic Quadrilateral) | #math #maths | #geometry

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[drJavi]

This one was a lot of fun to solve. Thanks for sharing.

[michaelkouzmin]

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.

[jmlfa]

This one is worth searching (a lot) ! Highly intricate - Thanks

[崔允豪]

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.

[allanflippin]

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.

[aljawad]

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.

[jimlocke]

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.

[prossvay]

ADC=PBC=60°
Connect A to C
In∆ ADC
Law of cosines

AC^2=6^2+9^2-2(6)(9)(1/2)
so AC=3√7
ABC=180°-60°=120°
In∆ ABC
Law of cosines


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


So a=7
So AP=14 and BP=14-6=8
So yellow square units.❤❤❤ Thanks sir.

[SumitVerma-lgqh]

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 👀

[זאבגלברד]

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...

[Birol]

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

[quigonkenny]

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

[LuisdeBritoCamacho]

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.

[zdrastvutye]

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
30
40 100
50 ngl1=a12*a21:ngl2=a22*a11
60 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
70
80
90 xl=zx/ngl:yl=zy/ngl:print
100 gosub
110 if xm2=xm1 or ym2=ym1 then 210
120 gosub 160:goto 200
130
140
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
220 240
230 den geradenschnittpunkt berechnen ***
240
250
260
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<masy then mass=masx else mass=masy
360 goto 380
370 xbu=x*mass:ybu=y*mass:return
380 for a=0 to 1:gcola+11:x=x(a, 0):y=y(a, 0):gosub 370:xba=xbu:yba=ybu:for b=1 to 4:ib=b:if ib=4 then ib=0
390 x=x(a, ib):y=y(a, ib):gosub 370:xbn=xbu:ybn=ybu:goto 410
400 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
410 gosub 400:next b:next a:gcol 8:x=xm1:y=ym1:gosub 370:circle xbu, ybu, r1*mass
420 print "run in bbcbasic sdl and hit ctrl tab to copy from the results window"
premath-can you find area of the yellow shaded triangle
x=3.93241866y=2.35288833
1.86264511E-10 1.51408722E-10
7.038971725.72176471
x=15.0350819y=5.47232229
die gesuchte flaeche ist=-10.3923048
run in bbcbasic sdl and hit ctrl tab to copy from the results window
>

[marioalb]

Cosine rule, twice:

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 √ )

[giuseppemalaguti]

6√3....non lo controllo più..con il teorema del coseno risulta del triangolo giallo conosco tutti gli angoli e un lato...percio calcolo l'area che risulta 6√3

[unknownidentity]

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