Can't believe I was that close to getting the answer. Even thought I found the ratio, did the ratio variable, and found the similarity of the 2 right triangles, but didn't pursue it by trying to spread the ratio variable for the other variables, thinking it can't get me anywhere, it can't get me to the answer I'm looking for, but once I saw you pursuing it, I immediately went back, pursued it, and got the answer straight away. Fascinating how, sometimes, pursuing seemingly pointless things can get you shocking results.
Let's note c the side length of the square, a = AP and b = PQ. Let's also note H the orthogonal projection of Q (or P) on (BC). We have: 2.(area of APQB) = a.(c + b) = 18 and 2.(area of DPQC) = (c -a). (c + b) = 8, so a/(c - a) = 18/8 = 9/4, which gives that a = (9/13).c In triangle BQP: QH^2 = HB.HC = a.(c -a) = ((9/13).c).((4/13).c) = (36/169).(c^2), so QH = (6/13).c. As QH = c - b we have then c - b = (6/13).c and b = (7/13).c. 2.(area of APQB) was a.(c + b), so it is ((9/13).c).((20/13).c) = (180/169).(c^2) So we get that (180/169).(c^2) = 18, which gives that c^2 = 169/10. Finally the area of triangle BCQ is c^2 - 9 - 4 = (169/10) - 13 = 39/10.
Razón entre áreas =s²=4/9---> Razón de semejanza =s=2/3---> Si BQ=b---> QC=2b/3. El cuadrado ABCD se puede dividir en cuatro triángulos BCQ, cuyas hipotenusas son los cuatro lados de ABCD, y un cuadrado central de lado BQ-QC=b/3 ---> Área BCQ =T =b*(2b/3)/2 =b²/3---> Área del cuadrado central =b²/9 =T/3 ---> 3T+(T/3) =10T/3 =9+4=13 ---> T=39/10 = Área BQC. Gracias y saludos.
An Elegant Solution : ) The only difference in our solutions was that of finding the height by using h = sqrt(9a * 4a). I suppose my subconscious wanted to make use of that semicircle : )
I solved it this way. I drew perpendiculars to the sides of the square passing through the point Q. They divide the square into 4 rectangles whose areas we can calculate. The two rectangles at the bottom are divided by the sides of the triangle BCQ each into two congruent triangles and let a be the area of the triangles at the bottom left and b be the area of those at the bottom right. Therefore: the area of the lower left rectangle is 2a the area of the lower right rectangle is 2b the area of the upper left rectangle is 9 - a the area of the upper right rectangle is 4 - b It is easy to demonstrate that the cross product of the areas of the rectangles is equal, therefore we can write: (9 - a)*2b = (4 - b)*2a from which: b = 4/9a this tells us that the area of the triangle QMC is 4/9 the area of the triangle QMB, therefore the constant of proportionality K = √4/9=2/3 Since QMC and QMB are similar it follows that given BM = x QM = 2/3x MC = 4/9x BC = x + 4/9x = 13/9x PQ = 13/9x - 2/3x = 7/9x Area trapezoidAPQB=(13/9x + 7/9x)*x*1/2 = 9 x = 9/10√10 BQC Area = 13/9*(9/10√10)*2/3*(9/10√10)*1/2 =39/10
9:03 - а вот об этом я и не вспомнил! Можно ж высоту было быстро из подобия найти: 4a/h=h/9a⇔h²=4a/9a=4/9⇔h=2/3. Тогда 169a²=13+13a*(2/3)÷2=13(1+a/3)⇔13a²=1+a/3⇔13a²-a/3-1=0. D=1/9+4*13=49+1/9, a=(1/9+√(49+1/9)). √(49+1/9)=√(442/9)=2√110½/3. Чтобы получить площадь, надо разделить ещё на 3, итого 2√110½/9.
PQ=b, AB=BC=CD=AD=a, PD=h, AP=a-h. By theorem about altitude of right triangle and formulas of square of trapezoid (a+b)(a-h)/2=9, (a+b)h/2=4, (a-b)^2=(a-h)h. By deciding of system of equations a^2=16,9 - square of square and 3,9 - square of right triangle.
Start with the point Q lying on a circle with diameter B (0, 0) and C(x, 0). Give point Q parametric equations (x/2 + x/2 * cos theta, x/2 * sin theta). Calculating the areas gives 2 equations with unknowns x and theta: [x^2*(4 - sin theta)*(1 + cos theta) = 9*8, x^2*(4 - sin theta)*(1 - cos theta) = 4*8]. Quick calculation gives cos theta = 5/13, sin theta = 12/13 and x^2 = 13^2/10. Solution is 13^2/10 - 9 - 4 = 39/10.
Имеем систему: c²=a²+b², c²=9+4+ab/2, ch/2=ab/2, a²=(9x)²+h², b²=(4x)²+h², 9x+4x=c. Попробуем её упростить. c=13x. 169x²=81x²+16x²+2h²⇔2h²=(169-81-16)x²⇔h²=(84½-40½-8)x²=(44-8)x²=(40-4)x²=36x²⇔h=6x, т. е. наша искомая площадь S=6*13x=78x. При этом мы знаем, что общая площадь 9x*9x=81x², значит 81x²-78x=13⇔81x²-78x-13=0. D=6084+4*81*13=10296, √D=√10296=4√643½. x=(78±4√643½)/162, оставляем положительный x=(78+4√643½)/162. Множим на 78: S=78(78+4√643½)/162=(6084+312√643½)/162=37+90/162+52√643½/27=37+45/81+1²⁵/₂₇√643½. В общем, как-то так... %)
Good method, but I will correct some mistakes for you, my friend. The area of the right triangle is s=(6x*13x)/2=39x², and the equation becomes (13x)²-39x²=13, and from it x²=1/10, and the area of the triangle is s=39x²=39/10.
@@ناصريناصر-س4ب Когда я решаю ночью - становлюсь очень невнимательным. Я догадывался, что решаю правильно, просто где-то посчитал неверно. Хорошо, что деньги считать легче, иначе меня бы обсчитывали продавцы. %)
Letting the height of triangle BQC = h, and AB=BC=CD=AD = k and drawing a line L R parallel to BC through the midpoints M,N of BQ and QC with L on AB and R on CD, [ALRD] = 9 + 4 and (1/4) [BQC] = [LMB] +[RCN] So k.k which is [ABCD] is [ ALRD] + 4 [LMB] +4 [RCN] (Almost there but it is Christmas so I will be back later) AP =9/13 k PD = 4/13 k 9/13 k( k-h/2) =9 so ( k. k -k.h/2 ) = 13 QB.QB = h.h + (81/169) k.k QC .QC= h.h +( 16/169)k.k so h.h + 81/169 k.k + h.h + 16/169 k.k = BC.BC = k.k three uses of Pythagoras' theorem 2 h.h = k.k (1 - (81+16)/169) and QB.QB.QC.QC = Four times the square of the shaded area. = k.h.k.h = (h.h + (81/169)k.k) (h.h +(16/169)k.k) = h^4 + k.h.k.h (97/169) + k^4( 16.81/(169.169) h^4 - ((169-97)/169 )k.h.k.h + (16.81/(169.169) k^4 =0 Solving a quadratic h^2 = ( 72/169 )k.k .... +/- sqrt (b^2-4ac))/2a etc and still getting only h^2 in terms of k^2is not way to go about this
Another attempt Also OQ = (1/2) k where O is the midoint of BC because BQC is a right-angle . 1/2- 4/13 = 5/26 9/13- 1/2 = 5/26 height is perpendicular to base so QOS is a small right-angled triangle with sides h, (5/26)k and (1/2)k so h is 12/26 k in a (5,12,13) triangle h = ( 6/13) k PQ = 7/13 k 13 =(7/13 k+ 3/13k).k 13 is ( 10/13)k.k k.k is 16 + 9/10 k is (1 +3/10)sqrt(10) [BQC] has an area (1/2).base. height)( k/2)(6k/13) =(3/13)[ABCD] =(169/10)(3/13) = 39/10 [BQC] = 3 + 9/10
Can't believe I was that close to getting the answer.
Even thought I found the ratio, did the ratio variable, and found the similarity of the 2 right triangles, but didn't pursue it by trying to spread the ratio variable for the other variables, thinking it can't get me anywhere, it can't get me to the answer I'm looking for, but once I saw you pursuing it, I immediately went back, pursued it, and got the answer straight away.
Fascinating how, sometimes, pursuing seemingly pointless things can get you shocking results.
Let's note c the side length of the square, a = AP and b = PQ. Let's also note H the orthogonal projection of Q (or P) on (BC).
We have: 2.(area of APQB) = a.(c + b) = 18 and 2.(area of DPQC) = (c -a). (c + b) = 8, so a/(c - a) = 18/8 = 9/4, which gives that a = (9/13).c
In triangle BQP: QH^2 = HB.HC = a.(c -a) = ((9/13).c).((4/13).c) = (36/169).(c^2), so QH = (6/13).c.
As QH = c - b we have then c - b = (6/13).c and b = (7/13).c.
2.(area of APQB) was a.(c + b), so it is ((9/13).c).((20/13).c) = (180/169).(c^2)
So we get that (180/169).(c^2) = 18, which gives that c^2 = 169/10.
Finally the area of triangle BCQ is c^2 - 9 - 4 = (169/10) - 13 = 39/10.
Razón entre áreas =s²=4/9---> Razón de semejanza =s=2/3---> Si BQ=b---> QC=2b/3.
El cuadrado ABCD se puede dividir en cuatro triángulos BCQ, cuyas hipotenusas son los cuatro lados de ABCD, y un cuadrado central de lado BQ-QC=b/3 ---> Área BCQ =T =b*(2b/3)/2 =b²/3---> Área del cuadrado central =b²/9 =T/3 ---> 3T+(T/3) =10T/3 =9+4=13 ---> T=39/10 = Área BQC.
Gracias y saludos.
An Elegant Solution : )
The only difference in our solutions was that of finding the height by using h = sqrt(9a * 4a).
I suppose my subconscious wanted to make use of that semicircle : )
I solved it this way. I drew perpendiculars to the sides of the square passing through the point Q. They divide the square into 4 rectangles whose areas we can calculate. The two rectangles at the bottom are divided by the sides of the triangle BCQ each into two congruent triangles and let a be the area of the triangles at the bottom left and b be the area of those at the bottom right. Therefore:
the area of the lower left rectangle is 2a
the area of the lower right rectangle is 2b
the area of the upper left rectangle is 9 - a
the area of the upper right rectangle is 4 - b
It is easy to demonstrate that the cross product of the areas of the rectangles is equal, therefore we can write:
(9 - a)*2b = (4 - b)*2a
from which:
b = 4/9a
this tells us that the area of the triangle QMC is 4/9 the area of the triangle QMB, therefore the constant of proportionality K = √4/9=2/3
Since QMC and QMB are similar it follows that given BM = x
QM = 2/3x
MC = 4/9x
BC = x + 4/9x = 13/9x
PQ = 13/9x - 2/3x = 7/9x
Area trapezoidAPQB=(13/9x + 7/9x)*x*1/2 = 9
x = 9/10√10
BQC Area = 13/9*(9/10√10)*2/3*(9/10√10)*1/2 =39/10
AP=9k
DP=4k (k> 0)
PQ=ak
台形PQCDの面積
(13+a)k×4k/2=4
ak=2/k-13k
PQを延長した直線とBCとの
交点をRとすると、
QR=26k-2/k
ここで、BCを直径とする
外接円の円周上にQがあるので
、方べきの定理より
(26k-2/k)^2=4k×9k
26k-2/k=6k
k=1/√(10)
△QBCの面積=
169k^2-4-9
=39/10
9:03 - а вот об этом я и не вспомнил! Можно ж высоту было быстро из подобия найти: 4a/h=h/9a⇔h²=4a/9a=4/9⇔h=2/3. Тогда 169a²=13+13a*(2/3)÷2=13(1+a/3)⇔13a²=1+a/3⇔13a²-a/3-1=0. D=1/9+4*13=49+1/9, a=(1/9+√(49+1/9)). √(49+1/9)=√(442/9)=2√110½/3. Чтобы получить площадь, надо разделить ещё на 3, итого 2√110½/9.
PQ=b, AB=BC=CD=AD=a, PD=h, AP=a-h. By theorem about altitude of right triangle and formulas of square of trapezoid (a+b)(a-h)/2=9, (a+b)h/2=4, (a-b)^2=(a-h)h. By deciding of system of equations a^2=16,9 - square of square and 3,9 - square of right triangle.
Start with the point Q lying on a circle with diameter B (0, 0) and C(x, 0). Give point Q parametric equations (x/2 + x/2 * cos theta, x/2 * sin theta). Calculating the areas gives 2 equations with unknowns x and theta: [x^2*(4 - sin theta)*(1 + cos theta) = 9*8, x^2*(4 - sin theta)*(1 - cos theta) = 4*8]. Quick calculation gives cos theta = 5/13, sin theta = 12/13 and x^2 = 13^2/10. Solution is 13^2/10 - 9 - 4 = 39/10.
Имеем систему: c²=a²+b², c²=9+4+ab/2, ch/2=ab/2, a²=(9x)²+h², b²=(4x)²+h², 9x+4x=c. Попробуем её упростить. c=13x. 169x²=81x²+16x²+2h²⇔2h²=(169-81-16)x²⇔h²=(84½-40½-8)x²=(44-8)x²=(40-4)x²=36x²⇔h=6x, т. е. наша искомая площадь S=6*13x=78x. При этом мы знаем, что общая площадь 9x*9x=81x², значит 81x²-78x=13⇔81x²-78x-13=0. D=6084+4*81*13=10296, √D=√10296=4√643½. x=(78±4√643½)/162, оставляем положительный x=(78+4√643½)/162. Множим на 78: S=78(78+4√643½)/162=(6084+312√643½)/162=37+90/162+52√643½/27=37+45/81+1²⁵/₂₇√643½. В общем, как-то так... %)
Good method, but I will correct some mistakes for you, my friend. The area of the right triangle is s=(6x*13x)/2=39x², and the equation becomes (13x)²-39x²=13, and from it x²=1/10, and the area of the triangle is s=39x²=39/10.
@@ناصريناصر-س4ب Когда я решаю ночью - становлюсь очень невнимательным. Я догадывался, что решаю правильно, просто где-то посчитал неверно. Хорошо, что деньги считать легче, иначе меня бы обсчитывали продавцы. %)
No problem again you have to sleep well so you can get some rest and think well
@@ناصريناصر-س4ب спасибо. У меня ночные процедуры - болею, но постараюсь пораньше с ними заканчивать - тут согласен.
Letting the height of triangle BQC = h, and AB=BC=CD=AD = k
and drawing a line L R parallel to BC through the midpoints M,N of BQ and QC with L on AB and R on CD, [ALRD] = 9 + 4 and (1/4) [BQC] = [LMB] +[RCN]
So k.k which is [ABCD] is [ ALRD] + 4 [LMB] +4 [RCN]
(Almost there but it is Christmas so I will be back later)
AP =9/13 k PD = 4/13 k
9/13 k( k-h/2) =9 so ( k. k -k.h/2 ) = 13
QB.QB = h.h + (81/169) k.k QC .QC= h.h +( 16/169)k.k
so h.h + 81/169 k.k + h.h + 16/169 k.k = BC.BC = k.k three uses of Pythagoras' theorem
2 h.h = k.k (1 - (81+16)/169) and
QB.QB.QC.QC = Four times the square of the shaded area. = k.h.k.h
= (h.h + (81/169)k.k) (h.h +(16/169)k.k) = h^4 + k.h.k.h (97/169) + k^4( 16.81/(169.169)
h^4 - ((169-97)/169 )k.h.k.h + (16.81/(169.169) k^4 =0 Solving a quadratic h^2 = ( 72/169 )k.k .... +/- sqrt (b^2-4ac))/2a etc
and still getting only h^2 in terms of k^2is not way to go about this
Another attempt
Also OQ = (1/2) k where O is the midoint of BC because BQC is a right-angle . 1/2- 4/13 = 5/26 9/13- 1/2 = 5/26
height is perpendicular to base so QOS is a small right-angled triangle with sides h, (5/26)k and (1/2)k
so h is 12/26 k in a (5,12,13) triangle h = ( 6/13) k PQ = 7/13 k
13 =(7/13 k+ 3/13k).k 13 is ( 10/13)k.k k.k is 16 + 9/10 k is (1 +3/10)sqrt(10)
[BQC] has an area (1/2).base. height)( k/2)(6k/13) =(3/13)[ABCD] =(169/10)(3/13) = 39/10
[BQC] = 3 + 9/10
(9)^2 (4)^2={81+16}=97{90°A90°B+90°C+90°D}=360°ABCD/97=3.69ABCD 3.6^3^2 12^33^1 2^1^3^1 23 (ABCD ➖ 3ABCD+2).
Muito bela resolution!!
Amazing
B=9/4;9/√10 a=13/4;13/√10 S∆=0,5a*(a-c)=a²-13=-2 7/16×;3,9✓
21.125-13=8.125
V zadaní sa neuvádza, že ide o štvorec
Nine hours later it does state that ABCD is a square. Happy Christmas.