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If I_{1}, I_{2}, I_{3} are the centers of escribed circles of \Delta ABC, the the area of the triangle I_{1}, I_{2}, I_{3} is

 

 

 

Option: 1

1


Option: 2

-1


Option: 3

\frac{abc}{2r}


Option: 4

-\frac{abc}{2r}


Answers (1)

best_answer

Given that,   

I_{1}, I_{2}, I_{3}  are the centers of escribed circles of \Delta ABC

Area=\frac{I_{1}I_{2}\times I_{2}I_{3}\times I_{1}I_{3}}{4R_{

Applying the formula to get,

Area=\frac{\left ( 4R\ cos\frac{A}{2} \right )\times \left ( 4R\ cos\frac{B}{2} \right )\times \left ( 4R\ cos\frac{C}{2} \right )}{2\times 4R}

Area=8R^{2}cos\frac{a}{2}\ cos\frac{b}{2}\ cos\frac{c}{2}

Area=\frac{8R^{2}\ sin\ A\ sin\ B\ sin\ C}{8\ sin\ \frac{A}{2}\ sin\ \frac{B}{2}\ sin\ \frac{C}{2}}

Area=\frac{R^{2}\ abc}{8R^{3}\ sin\ \frac{A}{2}\ sin\ \frac{B}{2}\ sin\ \frac{C}{2}}

Area=\frac{abc}{2\left (4R\ sin\ \frac{A}{2}\ sin\ \frac{B}{2}\ sin\ \frac{C}{2} \right )}

Area=\frac{abc}{2r}

 

 

Posted by

Sanket Gandhi

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