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

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

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New post 18 Sep 2019, 21:42
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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 18 Sep 2019, 22:04
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Let Z=1
If a side Z=1, then √Z=1 and Z^2=1....which is nothing but equilateral triangle.

Area of equilateral triangle ABC= √3/4 * side^2 = √3/4 * 1^2 = √3/4

Imo answer is option D

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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 18 Sep 2019, 22:16
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Ah an Excellent question.

The catch here is that, for no values of Z except 1,a triangle is possible.

Lets take the case of Z=2, the sides will be 2 , root2 and 4. Now third side is longer than the sum of other two sides, which is not possible.

Hence Z=1. Triangle is equilateral. And area is root3/4.

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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 18 Sep 2019, 23:13
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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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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 18 Sep 2019, 23:39
Z=1
So triangle will be equilateral
Answer is √3/4 D

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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 18 Sep 2019, 23:58
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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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 19 Sep 2019, 05:24
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
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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t  [#permalink]

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New post 19 Sep 2019, 06:19
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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Re: Triangle ABC has sides Z, √Z and Z^2, where Z is an integer. What is t   [#permalink] 19 Sep 2019, 06:19
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