Highschool Level Algebra Problem | ab=100, bc=200, ca=300, find a+b+c

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[@nejla76]

AB+BC+CA=300+200+100 (10A+B)+(10B+C)+(10C+A)=600 11A+11B+11C=600 11(A+B+C)=600 A+B+C=600/11

[@avinashmehrunkar5041]

We can solve it in this way also (a × b) ÷( b × c) =100 ÷ 200 a ÷ c = 1/2 or a = c ÷ 2 Put this value of 'a' in equation c × a = 300 c × c ÷ 2 = 300 c^2 = 600 c = sq root of 600 i.e (10×sq root of 6 = c) c × a = 300 so 10 sq root 6 × a = 300 a = 300 ÷ (10 sq root 6) a = 30 ÷ sq root 6 or (30 × sq root 6) ÷ 6 This gives a = 5 × sq root 6 Putting this value of 'a' in a × b = 100 we get (5 × sq root 6 ) × b = 100 b = 100 ÷ ( 5 × sq root 6) b = 20 ÷ sq root 6 or 20 × sq root 6 ÷ ( 6) i.e b =10 × sq root 6 ÷ (3) So (a + b + c ) = 5 × sq root 6 + 10 sq root 6 ÷ (3) + 10 sq root 6 Solving we get ( 15 × sq root 6 + 10 × sq root 6 + 30 × sq root 6 ) ÷ ( 3) i.e (55 × sq root 6) ÷( 3) (a +b + c) = (55 sq root 6)÷(3)

[@avinashmehrunkar5041]

Thanks 👍

[@mrunalikalukhe8159]

AB+BC+AC=100+200+300= 600 We can write it as =(A+B+C)(A+B+C)=600 =2(A+B+C)=600 =(A+B+C)=600÷2 =(A+B+C)=300

[@SmokalotOPott]

That’s is not correct, ab+ac+bc is not 2(a+b+c).