• 4.12-masala.
  • Dasturlash tili




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    4.11-masala. A, B, C sonlar mos ravishda uchta kesmaning uzunliklarini ifodalaydi. Agar kesmalar uchburchakning tomonlarini ifodalasa, uchburchakning yuzi S, uchburchakka tashqi va ichki chizilgan aylanalarning radiuslari r1 va r2 larni toping.

    Yechish. Agar
    p a b c
    2
    deb belgilash kiritsak, uchburchakning mavjud bo’lish

    sharti p  ( p a)  ( p b)  ( p c)  0 ko‘rinishda yoziladi. Uchburchakning yuzi

    s  , tashqi aylananing radiusi
    r a b c , ichki aylananing




    s
    radiusi esa r2  formula yordamida aniqlanadi.
    p
    1 4  s

    import math A=float(input("Uchburchakning A tomonini kiriting\nA=")) B=float(input("Uchburchakning B tomonini kiriting\nB=")) C=float(input("Uchburchakning C tomonini kiriting\nC="))
    if (A+B)>C and (A+C)>B and (B+C)>A: p=(A+B+C)/2
    S=math.sqrt(p*(p-A)*(p-B)*(p-C)) r1=(A*B*C)/4*S
    r2=S*p
    print("Uchburchakning yuzi=",S) print("Uchburchakka tashqi chizilgan
    aylananing radiusi=",r1) print("Uchburchakka ichki chizilgan
    aylananing radiusi=",r2) else:
    print("Berilgan sonlar bilan uchburchak yasab bo'lmaydi!")
    4.12-masala. Ax+B=0 tenglamani yeching.
    Yechish. Ma’lumki, a x b  0 tenglamaning yechimi quyidagicha aniqlanadi:
    1). A=0, b=0 bo’lsa, tenglama cheksiz ko‘p yechimga ega; 2). A=0, b 0 bo‘lsa, tenglama yechimga ega emas;

    3). A 0, bo‘lsa, tenglama
    x   b

    a
    yagona yechimga ega;









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