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A) \( \Large y^{4}=432x^{2} \) |

B) \( \Large y^{4}=216x^{2} \) |

C) \( \Large y^{2}=432x^{2} \) |

D) None of these |

Correct Answer:

A) \( \Large y^{4}=432x^{2} \) |

Description for Correct answer:

Area of equilateral triangle

\( \Large \frac{\sqrt{3} a^{2}}{4} \)=x ....(i)

and perimeter = 3a = y => a = \( \Large \frac{y}{3} \) ...(ii)

Now, putting the value of a from Eq. (ii) in Eq. (i), we get

\( \Large \frac{\sqrt{3}\left(\frac{y}{3}\right)^{2}}{4} \)=x => x=\( \Large \frac{\sqrt{3}\times y^{2}}{9\times 4} \)

=> x=\( \Large \frac{y^{2}}{3\sqrt{3}\times 4} \) => x=\( \Large \frac{y^{2}}{12\sqrt{3}} \)

=> \( \Large 12\sqrt{3}x=y^{2} \)

On squaring both sides, we get

\( \Large y^{4}=432x^{2} \)

Area of equilateral triangle

\( \Large \frac{\sqrt{3} a^{2}}{4} \)=x ....(i)

and perimeter = 3a = y => a = \( \Large \frac{y}{3} \) ...(ii)

Now, putting the value of a from Eq. (ii) in Eq. (i), we get

\( \Large \frac{\sqrt{3}\left(\frac{y}{3}\right)^{2}}{4} \)=x => x=\( \Large \frac{\sqrt{3}\times y^{2}}{9\times 4} \)

=> x=\( \Large \frac{y^{2}}{3\sqrt{3}\times 4} \) => x=\( \Large \frac{y^{2}}{12\sqrt{3}} \)

=> \( \Large 12\sqrt{3}x=y^{2} \)

On squaring both sides, we get

\( \Large y^{4}=432x^{2} \)

Part of solved Area and perimeter questions and answers : >> Elementary Mathematics >> Area and perimeter

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