Triangle ABC

Calculate the sides of triangle ABC with area 1404 cm2 and if a: b: c = 12:7:18

Result

a =  87.69 cm
b =  51.15 cm
c =  131.53 cm

Solution:

 s1=12+7+182=18.5 cm S1=s1(s112)(s17)(s118)=26.295 cm2 k=1404S1=7.307  a=12k=87.69  cm  \ \\ s_1 = \dfrac{ 12 + 7 + 18}{2} = 18.5 \ cm \ \\ S_1 = \sqrt{ s_1(s_1-12)(s_1-7)(s_1-18)} = 26.295 \ cm^2 \ \\ k = \sqrt{ \dfrac{ 1404 }{ S_1} } = 7.307 \ \\ \ \\ a = 12 k = 87.69 \ \text{ cm }
b=7k=51.15  cm b = 7 k = 51.15 \ \text{ cm }
c=18k=131.53  cm c = 18 k = 131.53 \ \text{ cm }

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