Properties of triangles

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${In a △ABC, if r=1,R=3,s=5, then the value of a}^{2} + b^{2} + c^{2} i s$

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$\begin{array}{l} Given r = 1, s = 5 △ = r s = 5 \\ Also s = 5 \Rightarrow a + b + c = 10 \\ a b c = 4 R △ \\ a b c = 4 \times 3 \times 5 = 60 \\ now, △^{2} = s (s - a) (s - b) (s - c) \\ 25 = 5 (5 - a) (5 - b) (5 - c) \\ 5 = 125 - 25 (a + b + c) + 5 (a b + b c + c a) - a b c \\ 5 = 125 - 25 x 10 + 5 (a b + b c + c a) - 60 \\ a b + b c + c a = 38 \\ hence, a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (38) = 24 \end{array}$

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Properties of triangles

Remember concepts with our Masterclasses.

In a △ABC, if r=1,R=3,s=5, then the value of a2+b2+c2 is

Given r=1,s=5 △=rs=5Also s=5 ⇒ a+b+c=10​abc=4R△​abc=4×3×5=60​now, △2=ss−as−bs−c​25=55−a5−b5−c​5=125−25a+b+c+5ab+bc+ca−abc​5=125−25x10+5ab+bc+ca−60​ab+bc+ca=38​hence, a2+b2+c2=a+b+c2−238=24

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${In a △ABC, if r=1,R=3,s=5, then the value of a}^{2} + b^{2} + c^{2} i s$