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JEE MainMathematicsEasyProperties of triangle

In ΔABC\Delta ABC, if r=1, R=3 and S=5, then the value of
a2+b2+c2a^{2} + b^{2} + c^{2} is
ΔABC\Delta ABC r=1, R=3 rFkk S=5] rc a2+b2+c2a^{2} + b^{2} + c^{2}dk
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Show answer and solution
r=ΔS⇒Δ=5r = \frac{\Delta}{S} \Rightarrow \Delta = 5
Δ=abc4R⇒abc=60\Delta = \frac{abc}{4R} \Rightarrow abc = 60
Δ2=s(s−a)(s−b)(s−c)\Delta^{2} = s(s - a)(s - b)(s - c)
5=(5−a)(5−b)(5−c)5 = (5 - a)(5 - b)(5 - c)
=125−25(a+b+c)+5(ab+bc+ca)−abc= 125 - 25(a + b + c) + 5(ab + bc + ca) - abc
a2+b2+c2=24a^{2} + b^{2} + c^{2} = 24

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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