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Question: The given statement is \(S = 4rR\cos (\dfrac {A}{2})\cos (\dfrac {B}{2})\cos (\dfrac {C}{2})\). Find...

The given statement is S=4rRcos⁡(A2)cos⁡(B2)cos⁡(C2)S = 4rR\cos (\dfrac {A}{2})\cos (\dfrac {B}{2})\cos (\dfrac {C}{2}). Find out by solving the equation whether it is true or false.
(A) True
(B) False

Explanation

Solution

We are given a statement and we have to find whether the statement is true or false. The statement given shows that the semi-perimeter of the triangle is equal to the product of 4 circum-radius and in-radius and cos⁡(A2)\cos (\dfrac {A}{2}), cos⁡(B2)\cos (\dfrac {B}{2}) and cos⁡(C2)\cos (\dfrac {C}{2}). In this first we will write down the values of in-radius, circum-radius in the form of formula and

Formula used: * R=abc4ΔR = \dfrac{{abc}}{{4\Delta }}

  • r=ΔSr = \dfrac{\Delta }{S}
  • cos⁡(A2)=S(S−a)bc\cos \left( {\dfrac {A}{2}} \right) = \sqrt {\dfrac{{S(S - a)}}{{bc}}} ;cos⁡(B2)=S(S−b)ac\cos \left( {\dfrac {B}{2}} \right) = \sqrt {\dfrac{{S(S - b)}}{{ac}}} ;cos⁡(C2)=S(S−c)ab\cos \left( {\dfrac {C}{2}} \right) = \sqrt {\dfrac{{S(S - c)}}{{ab}}}
    Here S is the semi perimeter; Δ\Delta is the area of triangle; a, b, c are sides of triangle; cos⁡(A2);cos⁡(B2);cos⁡(C2)\cos \left( {\dfrac {A}{2}} \right);\cos \left( {\dfrac {B}{2}} \right);\cos \left( {\dfrac {C}{2}} \right) are the angles. On solving the equations we will get the value and then we will decide whether the statement is true or false.

Complete step-by-step solution:
Step1: We are given S=4R⁡rcos⁡(A2)cos⁡(B2)cos⁡(C2)S = 4\operatorname{R} r\cos (\dfrac {A}{2})\cos (\dfrac {B}{2})\cos (\dfrac {C}{2}). Here RR is circum-radius, rr is in-radius, SS is semi perimeter. By taking the R.H.S and substituting the values in the expression of RR, rr etc. as R=abc4ΔR = \dfrac{{abc}}{{4\Delta }};r=ΔSr = \dfrac{\Delta }{S};cos⁡(A2)=S(S−a)bc\cos \left( {\dfrac {A}{2}} \right) = \sqrt {\dfrac{{S(S - a)}}{{bc}}} ;cos⁡(B2)=S(S−b)ac\cos \left( {\dfrac {B}{2}} \right) = \sqrt {\dfrac{{S(S - b)}}{{ac}}} ;cos⁡(C2)=S(S−c)ab\cos \left( {\dfrac {C}{2}} \right) = \sqrt {\dfrac{{S(S - c)}}{{ab}}}
Step2: On substituting the values we get
R.H.S:
⇒4(abc4Δ)(ΔS)S(S−a)bcS(S−b)acS(S−c)ab\Rightarrow 4\left( {\dfrac{{abc}}{{4\Delta }}} \right)\left( {\dfrac{\Delta }{S}} \right)\sqrt {\dfrac{{S\left( {S - a} \right)}}{{bc}}} \sqrt {\dfrac{{S(S - b)}}{{ac}}} \sqrt {\dfrac{{S\left( {S - c} \right)}}{{ab}}}
On multiplying the values in the square root we get:
⇒abcSS3(S−a)(S−b)(S−c)a2b2c2\Rightarrow \dfrac{{abc}}{S}\sqrt {\dfrac{{{S^3}\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)}}{{{a^2}{b^2}{c^2}}}}
Splitting S3{S^3} into S2×S{S^2} \times S and taking the square root of S3{S^3} and a2b2c2{a^2}{b^2}{c^2}
⇒abcS×SabcS(S−a)(S−b)(S−c)\Rightarrow \dfrac{{abc}}{S} \times \dfrac{S}{{abc}}\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)}
⇒S(S−a)(S−b)(S−c)\Rightarrow \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)}
This expression is equal to the Δ\Delta i.e. area of the triangle. But in the statement it is equal to the semi perimeter. Hence the given statement is false.

Option (B) is the correct answer.

Note: In such types of questions no numerical calculations are required. It only requires the proper application of formula in such a question. Before solving such questions first revise all the formulas related to this topic it makes the solving easier. In doing the calculation while proving a formula be cautious as this type of calculation is quite tough as it involves square roots. Always remember that there is a difference between a semi-perimeter and area of a triangle. Semi-perimeter is the half the sum of all sides of a triangle while the area we calculate by heron’s formula using a semi-perimeter in it. So don’t get confused and solve it accordingly.