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Question: In a triangle ABC, \[r{r_1} + {r_2}{r_3} = \] \(A.\) bc \(B.\) ac \(C.\) ab \(D.\) (a+b+c)a+...

In a triangle ABC, rr1+r2r3=r{r_1} + {r_2}{r_3} =
A.A. bc
B.B. ac
C.C. ab
D.D. (a+b+c)a+bc

Explanation

Solution

This question is based on the concept of circles associated with triangles. For that we have to understand the relation between in-radius (the radius of the inscribed circle of a triangle is called the in-radius). It is denoted by r. Also, circles which touch the side of the triangle by outside are called escribed circles. There can be three escribed circles of a triangle forming on all the three sides of triangles. Their radius is denoted by r1,r2,r3{r_1},{r_2},{r_3} respectively.

Complete step-by-step answer :
r=△sr = \dfrac{\vartriangle }{s}, where △→\vartriangle \to area of triangle [inradius]
r1=△s−a{r_1} = \dfrac{\vartriangle }{{s - a}} [exradius]
r2=△s−b{r_2} = \dfrac{\vartriangle }{{s - b}} [exradius]
r3=△s−c{r_3} = \dfrac{\vartriangle }{{s - c}} [exradius]
s→s \to semi-perimeter of triangle ( we can calculate s with the help of formula s=a+b+c2s = \dfrac{{a + b + c}}{2})
Now from the question, we have
r=△sr = \dfrac{\vartriangle }{s}, r1=△s−a{r_1} = \dfrac{\vartriangle }{{s - a}}, r2=△s−c{r_2} = \dfrac{\vartriangle }{{s - c}}, r3=△s−c{r_3} = \dfrac{\vartriangle }{{s - c}}
Substituting these value in
rr1+r2r3r{r_1} + {r_2}{r_3}, we get
rr1+r2r3=△s.△s−a+△s−b.△s−cr{r_1} + {r_2}{r_3} = \dfrac{\vartriangle }{s}.\dfrac{\vartriangle }{{s - a}} + \dfrac{\vartriangle }{{s - b}}.\dfrac{\vartriangle }{{s - c}}
= △2[(s−b)(s−c)+s(s−a)]s(s−a)(s−b)(s−c)\dfrac{{{\vartriangle ^2}[(s - b)(s - c) + s(s - a)]}}{{s(s - a)(s - b)(s - c)}}
= △2[2s2−s(a+b+c)+bc]△2\dfrac{{{\vartriangle ^2}[2{s^2} - s(a + b + c) + bc]}}{{{\vartriangle ^2}}}
= 2s2−2s2+bc=bc2{s^2} - 2{s^2} + bc = bc
Hence the correct option for the above answer is option A.A.

Note :Incircle of a triangle is the circle, which touches all three sides of a triangle. It is the largest circle contained in the triangle; it touches the three sides. The center of the incircle is a triangle center called the triangle’s incenter. The incenter can be found as the intersection of the three internal angle bisectors. Any line through a triangle that splits both the triangle’s area and its perimeter in half goes through the triangle’s incenter. There are either one, two, or three of these for any given triangle. The incircle radius is no greater than one-ninth the sum of the altitudes. All regular polygons have incircles tangent to all sides, but not all polygons do; those that do are tangential polygons. Some common in-radius formula-
r=△sr = \dfrac{\vartriangle }{s}
r=(s−a)tan⁡A2=(s−b)tan⁡B2=(s−c)tan⁡C2r = (s - a)\tan \dfrac{A}{2} = (s - b)\tan \dfrac{B}{2} = (s - c)\tan \dfrac{C}{2} [in terms of side angle, and semi-perimeter]
In terms of side and all half angles-
r=asin⁡B2sin⁡C2cos⁡A2r = \dfrac{{a\sin \dfrac{B}{2}\sin \dfrac{C}{2}}}{{\cos \dfrac{A}{2}}}
r=bsin⁡C2sin⁡A2cos⁡B2r = \dfrac{{b\sin \dfrac{C}{2}\sin \dfrac{A}{2}}}{{\cos \dfrac{B}{2}}}
r=csin⁡A2sin⁡B2cos⁡C2r = \dfrac{{c\sin \dfrac{A}{2}\sin \dfrac{B}{2}}}{{\cos \dfrac{C}{2}}}
There is one more concept of excircle and excenter of a triangle. It can be explained as the circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle’s sides.