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.
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. %)
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.
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.
Letting the height of triangle BQC = h, and AB=BC=CD=AD = k and drawing a line L P parallel to BC through the midpoints M,N of BQ and QC with L on AB and P on CD, [ALPD] = 9 + 4 and (1/4) [BQC] = [LMB] +[PCN] So k.k which is [ABCD] is [ ALPD] + 4 [LMB] +4 [PCN] (Almost there but it is Christmas so I will be back later)
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.
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. %)
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.
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.
An Elegant Solution : )
Letting the height of triangle BQC = h, and AB=BC=CD=AD = k
and drawing a line L P parallel to BC through the midpoints M,N of BQ and QC with L on AB and P on CD, [ALPD] = 9 + 4 and (1/4) [BQC] = [LMB] +[PCN]
So k.k which is [ABCD] is [ ALPD] + 4 [LMB] +4 [PCN]
(Almost there but it is Christmas so I will be back later)
(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).
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.