At 15:20, Math Booster has 2 solutions and discards the larger solution, because point P must be outside square ABCD. Actually, the second solution is for the case where P lies inside the square and is the given distances away from points A, B, and C.
@guyhoghton399
4 ай бұрын
Let _θ = ∠ABP_ Using the cosine rule: In _ΔABP: 5² = x² + 3² - 6xcos(θ)_ ⇒ _6xcos(θ) = x² - 16_ S.B.S : _36x²cos²(θ) = x⁴ - 32x² + 256_ ... ① In _ΔBCP: 6² = x² + 3² - 6xcos(270° - θ)_ ⇒ _6xsin(θ) = -x² + 27_ S.B.S : _36x²sin²(θ) = x⁴ - 54x² + 729_ ... ② Adding ① and ②: _36x² = 2x⁴ - 86x² + 985_ ⇒ _2x⁴ - 122x² + 985 = 0_ ⇒ *area of **_ABCD = x² = ½(61 ± √1751)_*
@MarieAnne.
3 ай бұрын
s = side length of square. Using coordinates we get: A = (0,s), B = (s,s), C = (s,0), D = (0,0) P is point of intersection of circles centered at A, B, C with radii 5, 3, 6 respectively. These circles have equations: (1) x² + (y−s)² = 25 (2) (x−s)² + (y−s)² = 9 (3) (x−s)² + y² = 36 (1) − (2) gives x² − (x² − 2xs + s²) = 25 − 9 → x = (16+s²)/(2s) (3) − (2) gives y² − (y² − 2ys + s²) = 36 − 9 → y = (27+s²)/(2s) Plugging values of x and y into (2) we get ((16+s²)/(2s) − s)² + ((27+s²)/(2s) − s)² = 9 ((16−s²)/(2s))² + ((27−s²)/(2s))² = 9 (256−32s²+s⁴)/(4s²) + (729−54s²+s⁴)/(4s²) = 9 2s⁴ − 86s² + 985 = 36s² 2s⁴ − 122s² + 985 = 0 s² = (61+√1751)/2 ≈ 51.42 or s² = (61−√1751)/2 ≈ 9.58 Since s < 5, then s² < 25. Therefore: Area(ABCD) = s² = *(61−√1751)/2 ≈ 9.58*
@jimleahy3858
4 ай бұрын
Note that sin
@ChandanRoy-sr6yu
4 ай бұрын
Sir I have a math problem But I am not able to solve it can you solve it but how I can send the problem to you?
@tieshanhuang2466
4 ай бұрын
May an analytic geometry approach. assume coordinates as P(x, y), B(0,0), A(-h,0), and C(0, -h) . three unknown numbers x/y/h in three equations of the Pythagorean theorem. similar solution but easier to think about.
@joseluishablutzelaceijas928
4 ай бұрын
Thank you for the nice problem and its solution. My solution is different, thus maybe worth sharing here: I tried to exploit the observation that the angles BAP, BCP and APC add up to 90° by rotating the triangle BCP by 90° clockwise around B, i.e. such that C moves to A and P to another point, call it P'. One then has the triangle APP', whose sides are all known as the triangle PBP' is rectangular and isosceles, meaning that |PP'| = 3*sqrt(2) and whose angle PAP' is equal to the sum of the angles BCP and BAP. Applying the cosine rule, one has for triangle PAP' that (3*sqrt(2))^2 = 5^2 + 6^2 - 2*5*6*cos([angle PAP']), which means that cos([angle PAP']) = 43/60. Now one can come back to triangle APC and apply the cosine rule again, meaning that |AC|^2 = 5^2 + 6^2 - 2*5*6*cos([angle APC]) = 5^2 + 6^2 - 2*5*6*sin([angle BAP]+[angle BCP]) = 61 - sqrt(17*103), the second equation being the expression of the observation that the angles BAP, BCP and APC add up to 90° and the third equation resulting from sin([angle PAP']) = sqrt(1 - (cos([angle PAP']))^2) = sqrt(17*103)/60. The area of the square ADCB is then equal to |AC|^2/2 = (61 - sqrt(17*103))/2.
@michaeldoerr5810
4 ай бұрын
This definitely a problem that shows a subtle combination of algebra and geometric reasoning. Impossible means that it is easier than it looks. And lots of practice required in order to just comprehend how and why it is easier than it looks.
@zdrastvutye
4 ай бұрын
instigators trailer? i have written 4 equations with 4 unknown numbers and assumed that the common intersection point "P" is at 0;0: 10 print "math booster-impossible geometry problem-square given by 3 distances from 1 point" 20 l1=5:l2=3:l3=6:sw=l1^2/(l1+l2+l3)/10:xm1=sw:dim x(3),y(3):goto 60 30 ym1=sqr(l1^2-xm1^2):p=-(xm1+ym1):q=((xm1+ym1)^2-l3^2)/2 40 ym2=-p/2+sqr(abs(p*p/4-q)):xm2=xm1-ym2+ym1 50 dgu1=(xm2/l2)^2:dgu2=(ym1/l2)^2:dg=dgu1+dgu2-1:return 60 gosub 30 70 dg1=dg:xm11=xm1:xm1=xm1+sw:xm12=xm1:gosub 30:if dg1*dg>0 then 70 80 xm1=(xm11+xm12)/2:gosub 30:if dg1*dg>0 then xm11=xm1 else xm12=xm1 90 if abs(dg)>1E-10 then 80 100 print xm1;"%";ym1;"%";xm2;"%";ym2:la=xm1-xm2:print "laenge l=";la 110 x(0)=xm1:y(0)=ym1:x(1)=xm2:y(1)=ym1:x(2)=xm2:y(2)=ym2:x(3)=xm1:y(3)=ym2 120 xmin=(l1+l2+l3)/3:ymin=xmin:xmax=xmin:ymax=ymin 130 for a=0 to 3:if x(a)xmax then xmax=x(a) 150 if y(a)ymax then ymax=y(a) 170 next a:if xmin>0 then xmin=0 180 if ymin>0 then ymin=0 190 if xmax=xmin or ymax=ymin then else 230 200 if xmax=xmin then else 230 210 mass=850/(ymax-ymin):goto 240 220 mass=1200/(xmax-xmin):goto 240 230 masx=1200/(xmax-xmin):masy=850/(ymax-ymin):if masx run in bbc basic sdl and hit ctrl tab to copy from the results window
@ShreyasKote
4 ай бұрын
Nice video keep it up
@Arandomguy1yl
4 ай бұрын
You didn't even watch the full video to see if it's nice or not, it's literally uploaded 3 minutes ago and you commented 1 min ago and possibly watched for 2 mins
@ShreyasKote
4 ай бұрын
@@Arandomguy1ylI have done the answer in my head and that's why I watched it by speeding up by a factor of 10seconds
@imetroangola4943
4 ай бұрын
Uma questão bem desafiante! Parabéns!!!
@marcelowanderleycorreia8876
4 ай бұрын
Beautiful question!!!
@josecastro6666
4 ай бұрын
AREA SQUARE ABCD=(61-√1751)/2
@اقرء
4 ай бұрын
Thank for watching
@rabotaakk-nw9nm
4 ай бұрын
A≈9.578 (51.422) !!! 😁
@jimleahy3858
4 ай бұрын
Note that cos
@rabotaakk-nw9nm
4 ай бұрын
@@jimleahy3858🤔???
@harrymatabal8448
4 ай бұрын
If it's impossible then how you expect us to solve it. Are you from a mental home
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