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  • Let be a triangle with vertices  <img alt="A \equiv(6,2(\sqrt{3}+1)), B \equiv(4,2)" src="https://entra

Let A B C be a triangle with vertices  A \equiv(6,2(\sqrt{3}+1)), B \equiv(4,2) andC \equiv(8.2). Let R be the region consisting of all those points P inside \triangle A B C which satisfy  d(P, B C) \geq \max \{d(P . A B) ; d(P, A C)\}, where d(P, L)d(P, B C) \geq \max \{d(P . A B) ; d(P, A C)\}, \text{ where } d(P, L)denotes the distance of the point from the line L, then

 

Option: 1

 figure so formed is an equilateral triangle


Option: 2

figure obtained satisfying the above condition is a quadrilateral


Option: 3

 required area of quadrilateral is \frac{4 \sqrt{3}}{3} sq units


Option: 4

 required area of triangle is \sqrt{3} sq units


Answers (1)

best_answer

Clearly, we see that A B C is an equilateral triangle with side length 4. B D and C E are angle bisector of angle B and C respectively. Any point inside the triangle A E C is nearer to A C  than B C and point inside \triangle B D A is nearer to A B than B C, so any point inside the quadrilateral A E G D will satisfy the given condition.
Hence, required  area =2 \times area of \triangle E A G

\begin{aligned} & =2 \times \frac{1}{2} \times A E \times E G=\frac{1}{2} A B \times \frac{1}{3} C E \\ & =\frac{1}{6} \times 4 \times \sqrt{4^2-2^2}=\frac{4 \sqrt{3}}{3} \text { sq units } \end{aligned}

 

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vishal kumar

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