Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t

Let Z,1
p=1+1+1=3
According to Heron's formula, we have
P=√p/2*(p/2-1)*(p/2-1)*(p/2-1)=
√3/2*1/2*1/2*1/2=√3/16= √3 /4
Option D

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Z=1
So triangle will be equilateral
Answer is √3/4 D

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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).

if we check the sum of sides property

z+z^2 > sqrt(z) for many values
also sqrt(z)+z^2 >z for many values
but z+sqrt(z)>z^2 only if z=1

so area is (sqrt(3)*z)/4 ...... OA: D

According to given sides triangle ABC is possible only when all sides are equal I.e when z=√z=z^2, it happens when z=1 , for any other integer third side rule does not satisfy, so since all sides are equal area of equilateral triangle is√3/4 a^2 = √3/4

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