Впрямоугольный треугольник с катетами 3 см и 4 см вписан квадрат, имеющий с треугольником общий прямой угол. вычислите периметр и площадь квадрата.

Abdilda Tazhibayev was born in 1909 in Akmeshit (now Kyzylorda). Medium

one hundred

The Kipchak tribe Tory clan is descended from Bessary's grandfather. [1] [2] His father, Tazhibai, died in 1915 and was left in the care of his mother, Aimankul. Due to his condition, six-year-old Abdilda was taken to a suburb of Karaozek. Then the future poet lived until the age of thirteen with his mother Aimankul and uncle Daldabay

grows in upbringing. Young Abdilda's poems and literature

His mother and aunt also aroused his curiosity

will be the grandfather. Aimankul is fluent in Arabic and Persian

He was literate, open-minded, and cheerful.

He writes letters to Abdilda and teaches her literature. Abdilda's mother Aimankul, who later became a well-known folk poet and published a collection of poems, at the request of his father Daldabay, recited to the villagers the stories of the East and various Kazakh poems. Passionate, silent, silent

One of his listeners was Abdil. Later, in a small group, Abdilda would recite the stories and epics to his guests and villagers.

a)по т.менелая 

(сд: да)•(ае: ек)•(вк: вс)=1 

(, откуда получим вк: ек=1, следовательно, вк=ек.   

2)проверим ∆ авс по т.пифагора.

ab² > ас²+вс² => угол с тупой.

по ф.герона s(abc)-√(14•7•6•1)=14√3 

вd- медиана и делит треугольник на два равных по площади. 

s(bcd)=7√3

по другой формуле 

s (abc)=ac•bc•sinc: 2

14√3=8•7•sinc: 2 => sinc=√3/2 => тупой угол с=120° 

(можно подтвердить по  т.косинусов – получим cos c= -1/2)

из вершины а проведем высоту ан. 

высота тупоугольного треугольника, проведенная из острого угла, находится вне треугольника и пересекается с продолжением стороны, к которой проведена. 

угол асн =180°-120°=60°( смежный углу асв)

нс=ас•cos60°=4

ah=ac•sin60°=4√3

примем вк=ек=х

в ∆ анк

ак=ае+ек=7+х

кс=вс-вк=7-х

ак²=ан²+кн² =(7+х)²=(4√3)²+(4+7-х)²

49+14х+х²=121-22х+х²+48=>

вк=х=10/3

ск=7-10/3=11/3 

ak=7+10/3=31/3

площадь  dекс найдем из разности площадей ∆ аск и ∆ аеd. 

s(akc)=ah•ck: 2=(4√3•11/3): 2=22/√3 =>  

sincak=22/√3: (8•31/6)=11√3/62 

s(aed)=ad•ae•sina: 2=77√3/31

s(cdek)= (ед. площади)