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GATE Civil 2021 Set 2 | GA Question: 10

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In an equilateral triangle $\text{PQR}$, side $\text{PQ}$ is divided into four equal parts, side $\text{QR}$ is divided into six equal parts and side $\text{PR}$ is divided into eight equals parts. The length of each subdivided part in $\text{cm}$ is an integer. The minimum area of the triangle $\text{PQR}$ possible, in $\text{cm}^{2}$, is

  1. $18$
  2. $24$
  3. $48\sqrt{3}$
  4. $144 \sqrt{3}$
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Migrated from GO Civil 2 years ago by Arjun

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Let the side length of an equilateral triangle be $’x’\;\text{cm}.$

As mentioned in the question, the length of each subdivided part in $\text{cm}$ is an integer. So, the side length must be the LCM of $(4,6,8) \implies x = 24\;\text{cm}.$ 

Now, the area of an equilateral triangle of side $x\;\text{cm} = \dfrac{\sqrt{3}}{4}\; x^{2} = \dfrac{\sqrt{3}}{4} \;24^{2} = 144\;\sqrt{3}\;\text{cm}^{2}.$

So, the correct answer is $(D).$

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