Question

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

Answer 1

Hint: 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 $\cos (\dfrac {A}{2})$, $\cos (\dfrac {B}{2})$ and $\cos (\dfrac {C}{2})$. In this first we will write down the values of in-radius, circum-radius in the form of formula and

Formula used: * \[R = \dfrac{{abc}}{{4\Delta }}\]
* \[r = \dfrac{\Delta }{S}\]
* \[\cos \left( {\dfrac {A}{2}} \right) = \sqrt {\dfrac{{S(S - a)}}{{bc}}} \];\[\cos \left( {\dfrac {B}{2}} \right) = \sqrt {\dfrac{{S(S - b)}}{{ac}}} \];\[\cos \left( {\dfrac {C}{2}} \right) = \sqrt {\dfrac{{S(S - c)}}{{ab}}} \]
Here S is the semi perimeter; \[\Delta \] is the area of triangle; a, b, c are sides of triangle; \[\cos \left( {\dfrac {A}{2}} \right);\cos \left( {\dfrac {B}{2}} \right);\cos \left( {\dfrac {C}{2}} \right)\] 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 $S = 4\operatorname{R} r\cos (\dfrac {A}{2})\cos (\dfrac {B}{2})\cos (\dfrac {C}{2})$. Here $R$ is circum-radius, $r$ is in-radius, $S$ is semi perimeter. By taking the R.H.S and substituting the values in the expression of $R$, $r$ etc. as \[R = \dfrac{{abc}}{{4\Delta }}\];\[r = \dfrac{\Delta }{S}\];\[\cos \left( {\dfrac {A}{2}} \right) = \sqrt {\dfrac{{S(S - a)}}{{bc}}} \];\[\cos \left( {\dfrac {B}{2}} \right) = \sqrt {\dfrac{{S(S - b)}}{{ac}}} \];\[\cos \left( {\dfrac {C}{2}} \right) = \sqrt {\dfrac{{S(S - c)}}{{ab}}} \]
Step2: On substituting the values we get
R.H.S:
$ \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:
$ \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 ${S^3}$ into ${S^2} \times S$ and taking the square root of ${S^3}$ and ${a^2}{b^2}{c^2}$
$ \Rightarrow \dfrac{{abc}}{S} \times \dfrac{S}{{abc}}\sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
$ \Rightarrow \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - c} \right)} $
This expression is equal to the $\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.