Mathematics - Properties of Triangle Question with Solution | TestHub

MathematicsProperties of Triangler1,r–R formula basedMedium2 minQB

MathematicsMediumsingle choice

In a triangle ABC, if $Δ = r$ , then

Statement I: $a^{2} + b^{2} + c^{2} + 2 ab c < 2$ .

Statement II: As $s = 1$ , we get $1 - a, 1 - b, 1 - c > 0$ .

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Answer:

Solution:

Since $Δ = rs, Δ = r$ gives $s = 1$ , so $a + b + c = 2$ . Then $1 - a = s - a > 0$ , and similarly $1 - b > 0, 1 - c > 0$ . So $(1 - a) (1 - b) (1 - c) > 0$ , which gives $ab + b c +$ $c a - ab c > 1$ . Hence $a^{2} + b^{2} + c^{2} + 2 ab c = 4 - 2 (ab + b c + c a - ab c) < 2$ . Both statements are true. Statement II gives exactly this positivity, so it correctly explains Statement I.

Stream:JEESubject:MathematicsTopic:Properties of TriangleSubtopic:r1,r–R formula based

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