KVPY 2019 SA Exam Previous Papers

Question : 14 of 80

Marks: +1, -0

Let l > 0 be a real number, C denote a circle with circumference l, and T denote a triangle with perimeter l. Then

given any positive real number α, we can choose C and T as above such that the ratio ${\Area(C)}/{\Area (T)}$ is greater than α
given any positive real number α, we can choose C and T as above such that the ratio ${\Area(C)}/{\Area(T)}$ is less than α
given any C and T as above, the ratio ${\Area(C)}/{\Area(T)}$ is independent of C and T
there exist real numbers a and b such that for any circle C and triangle T as above, we must have a < ${\Area(C)}/{\Area(T)}$ < b

Solution:

Area

(C) = π(l/{2π})^2=l^2/{4π}

Area

(T)≤√3/4(l/3)^2=l^2/{12√3}⇒(A)

Hence

{\Area(c)}/{\Area( τ)}≥{3√3}/π

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