Solveeit Logo
Login

Question

Question: Prove that for a triangle \(ABC\) with sides \(a,b,c\) and circumradius and inradius respectively \(...

Prove that for a triangle ABCABC with sides a,b,ca,b,c and circumradius and inradius respectively RR and rr, area of the triangle S=4Rrcos⁡A2cos⁡B2cos⁡C2S = 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}.

Explanation

Solution

Area of a triangle is expressed using a different formula. Here we use equations using semi-perimeter, circumradius and inradius. Substituting these appropriately and making necessary simplifications we can prove the expression.

Formula used
For a triangle ABC with sides a,b,ca,b,c
Area of the triangle, S=s(s−a)(s−b)(s−c)S = \sqrt {s(s - a)(s - b)(s - c)}
where s=a+b+c2s = \dfrac{{a + b + c}}{2}, the semi-perimeter of the triangle.
Also, cos⁡A2=s(s−a)bc,cos⁡B2=s(s−b)ac,cos⁡C2=s(s−c)ab\cos \dfrac{A}{2} = \sqrt {\dfrac{{s(s - a)}}{{bc}}} ,\cos \dfrac{B}{2} = \sqrt {\dfrac{{s(s - b)}}{{ac}}} ,\cos \dfrac{C}{2} = \sqrt {\dfrac{{s(s - c)}}{{ab}}}
If the circumradius and inradius are represented by R,rR,r, then
Area of the triangle, S=abc4R=rsS = \dfrac{{abc}}{{4R}} = rs, where ss is the semi-perimeter.

Complete step-by-step answer:

Given that the triangle has sides a,b,ca,b,c and circumradius and inradius, R,rR,r respectively.
If ss is the semi-perimeter of a triangle, then s=a+b+c2s = \dfrac{{a + b + c}}{2}.
Also we have cos⁡A2=s(s−a)bc,cos⁡B2=s(s−b)ac,cos⁡C2=s(s−c)ab\cos \dfrac{A}{2} = \sqrt {\dfrac{{s(s - a)}}{{bc}}} ,\cos \dfrac{B}{2} = \sqrt {\dfrac{{s(s - b)}}{{ac}}} ,\cos \dfrac{C}{2} = \sqrt {\dfrac{{s(s - c)}}{{ab}}}
To prove the area of the triangle is the expression given, we can start with the RHS.
Substituting for cos⁡\cos terms in RHS we have,
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=4Rrs(s−a)bcs(s−b)acs(s−c)ab\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = 4Rr\sqrt {\dfrac{{s(s - a)}}{{bc}}} \sqrt {\dfrac{{s(s - b)}}{{ac}}} \sqrt {\dfrac{{s(s - c)}}{{ab}}}
Simplifying we get,
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=4Rrs3(s−a)(s−b)(s−c)a2b2c2\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = 4Rr\sqrt {\dfrac{{{s^3}(s - a)(s - b)(s - c)}}{{{a^2}{b^2}{c^2}}}}
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=4Rrsabcs(s−a)(s−b)(s−c)\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4Rrs}}{{abc}}\sqrt {s(s - a)(s - b)(s - c)}
Now since the expression inside the radical/root symbol is the area of a triangle, we can substitute it with SS.
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=4RrsabcS\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4Rrs}}{{abc}}S
Also, if the circumradius and inradius are represented by R,rR,r, then
Area of the triangle, S=abc4R=rsS = \dfrac{{abc}}{{4R}} = rs, where ss is the semi-perimeter.
So we get,
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=4Rabc×rs×S\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4R}}{{abc}} \times rs \times S
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=1S×S×S\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{1}{S} \times S \times S
Cancelling SS from numerator and denominator we have,
⇒4Rrcos⁡A2cos⁡B2cos⁡C2=S\Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = S
Hence we had proved the given expression.

Note: Here to substitute the terms easily we start from the RHS. Starting with LHS and getting RHS is somewhat impossible or will be a difficult task. So in these types of questions we have to observe and decide from where we have to start. After making necessary steps we can get into the solution.