Question

The diagram shows six equal circles inscribed in equilateral triangle ABC. The circles touch exter- nally among themselves and also touch the sides of the triangle. If the radius of each circle is R, area of the triangle is A B с (a) (6+π✓3)R (c) R2 (12+7✓3) (b) 9 R2 (d) R2 (9+6✓3). R2 is r squared.​

Answer 1

Answer:

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Step-by-step explanation:

ur answer is there in images of 3 pictures

Answer 2

The radius of each circle is the triangle is $R^{2}(7\sqrt{3}+12)$.

Step-by-step explanation:

$ABC$ is an equilateral triangle, each angle is $60^{0}$

$BP$ bisects angle $ABC$ such that angle $PBQ$ is $30^{0}$

consider the triangle $PBQ$

$tan30=\frac{PQ}{BQ}$

$\frac{1}{\sqrt{3} }= \frac{R}{BQ}$

$BQ=\sqrt{3}R$

$sc=BQ=\sqrt{3}R$

$QS=R+2R+R$

$QS=4R$

$BC=BQ+QS+SC$

$=\sqrt{3} R+4R+\sqrt{3}R$

$=R(4+2\sqrt{3})$

$=\frac{\sqrt{3[R(4+2\sqrt{3)]^{2}} } }{4}$

$=\frac{\sqrt{3}[R(4+2\sqrt{3)}]^{2} }{4}$

$=\frac{\sqrt{3} R^{2}(16+16\sqrt{3}+12) }{4}$

$=\sqrt{3}R^{2}(7+4\sqrt{3})$

$=R^{2}(7\sqrt{3}+12)$.