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  • The lengths of the sides of a triangle are <img alt="10+x^{2}, 10+x^{2}$ and $20-2 x^{2}" src="https://entrancecorner.oncodecogs.com/gif.latex?10&plus;x%5E%7B2%7D%2C%2010&plus;x%5E%7B2%7D%2

The lengths of the sides of a triangle are 10+x^{2}, 10+x^{2}$ and $20-2 x^{2}. If for x=\mathrm{k}, the area of the triangle is maximum, then 3 \mathrm{k}^{2} is equal to:

Option: 1

5


Option: 2

8


Option: 3

10


Option: 4

12


Answers (1)

best_answer

Semiperimeter s of the triangle

\mathrm{=\frac{10+x^{2}+10+x^{2}+20-2 x^{2}}{2}=20 }

\mathrm{\text { Area of } \Delta=\sqrt{S(S-a)(S-b)(S-c)}}

\mathrm{\Rightarrow \Delta^{2}=20\left(10-x^{2}\right)\left(10-x^{2}\right)\left(2 x^{2}\right)}

\mathrm{\Rightarrow \Delta=2 \sqrt{10} x\left(10-x^{2}\right)=2 \sqrt{10}\left(10 x-x^{3}\right) }

\mathrm{\Delta^{\prime}=2 \sqrt{10}\left(10-3 x^{2}\right)=0 \Rightarrow x=\pm \sqrt{\frac{10}{3}}, x=\sqrt{\frac{10}{3}} } will be point of maxima so,  \mathrm{3 x^{2}=3 \times \frac{10}{3}=10 }

Hence the correct answer is option 3

Posted by

Deependra Verma

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