Since Z is an integer only one positive value(length can’t be negative) is possible. Thus, \(Z = 1\) and \(√Z = 1\) and \(Z^2 = 1\) as \(Z = 2\) would not make a triangle because other two side would be \(√Z = √2\) and \(Z^2 = 4\). Here \(Z + √Z < Z^2\) where sum of two smaller sides must be greater than the largest side.
Hence the triangle has each side equal to 1. So, it’s an equilateral triangle.
Therefore Area of equilateral triangle \(= \frac{√3}{4} * side^2\).
\(= \frac{√3}{4}\)
Also, using Hero’s formula
\(S = \frac{(a + b + c)}{2}\) where a, b and c are three sides of triangle.
\(S = \frac{3}{2}\)
Area of triangle \(= √(s(s-a)(s-b)(s-c))\)
\(= √(\frac{3}{2} * (\frac{3}{2}-1) * (\frac{3}{2}-1) * (\frac{3}{2}-1))\)
\(= √(\frac{3}{2} * (\frac{1}{2})^3)\)
\(= \frac{√3}{4}\)
Answer (D).
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Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019
GMATPREP2 610(Q44,V29) June 10, 2019
GMATPREPSoft1 680(Q48,V35) June 26, 2019