(A) In a right-angled triangle $ABC$ with $\angle A = 90^{\circ}$,the inradius $r$ is given by $r = \frac{b+c-a}{2}$.The circumradius $R$ is given by

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(A) In a right-angled triangle $ABC$ with $\angle A = 90^{\circ}$,the inradius $r$ is given by $r = \frac{b+c-a}{2}$.The circumradius $R$ is given by $R = \frac{a}{2}$.Therefore,$r+R = \frac{b+c-a}{2} + \frac{a}{2} = \frac{b+c}{2}$.Multiplying by $2$,we get $2(r+R) = b+c$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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In $\triangle ABC$,if $\angle A = 90^{\circ}$,then $2(r+R) = $

AP EAMCET 2018 All AP EAMCET

A
$b+c$
B
$a+b$
C
$a+c$
D
$a+b+c$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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In $\triangle ABC$,if $\angle A = 90^{\circ}$,then $2(r+R) = $

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