Question

If R is the radius of circumcentre of $\mathrm{\Delta }ABC,$ then $R={\displaystyle \frac{abc}{4S}}$
(A) True
(B) False

Answer 1

Hint: Use area of triangle formula, where two sides of triangle and angle between them is given. Use sine rule related with circumradius to get the given relation.

Complete step-by-step answer:
Here, we have given R as a radius of circumcircle i.e. circumradius and need to prove the relation;
$R={\displaystyle \frac{abc}{4S}}$……………….(1)
Where (a, b, c) are sides of the triangle as denoted in the diagram.

Where O is the centre of the circle C, which is circumscribing the triangle ABC.
R = Circumradius of triangle ABC.
We can write sine rule in $\mathrm{\Delta }ABC$ involving circumradius R as;
${\displaystyle \frac{\sin A}{a}}={\displaystyle \frac{\sin B}{b}}={\displaystyle \frac{\sin C}{c}}={\displaystyle \frac{1}{2R}}..............\left(2\right)$
As we have a formula of area with involvement of two sides and angle between them.
Let the area be represented by S.
$Area=S={\displaystyle \frac{1}{2}}bc\sin A={\displaystyle \frac{1}{2}}ab\sin C={\displaystyle \frac{1}{2}}ac\sin B........\left(3\right)$
Now, from equation (2) and (3), we can write an equation with respect to one angle as
${\displaystyle \frac{\sin A}{a}}={\displaystyle \frac{1}{2R}}\text{ and }S={\displaystyle \frac{1}{2}}bc\sin A$
Substituting value of sin A from the relation ${\displaystyle \frac{\sin A}{a}}={\displaystyle \frac{1}{2R}}\text{ to }S={\displaystyle \frac{1}{2}}bc\sin A$, we get;
As $\sin A={\displaystyle \frac{a}{2R}}$ from the first relation, now putting value of sin A in $S={\displaystyle \frac{1}{2}}bc\sin A$, we get
$\begin{array}{rl}& S={\displaystyle \frac{1}{2}}bc{\displaystyle \frac{a}{2R}}\\ & S={\displaystyle \frac{abc}{4R}}\end{array}$
Transferring R to other side, we get;
$R={\displaystyle \frac{abc}{4S}}$
Hence, the relation given in the problem is true.
Note: One can go wrong with the formula of area of the triangle. One can apply heron’s formula for proving i.e.
$\begin{array}{rl}& S=\sqrt{s(s-a)(s-b)(s-c)}\\ & s={\displaystyle \frac{a+b+c}{2}}\end{array}$
Which will make the solution very complex.
One can go wrong while writing sine rule as;
${\displaystyle \frac{\sin A}{a}}={\displaystyle \frac{\sin B}{b}}={\displaystyle \frac{\sin C}{c}}={\displaystyle \frac{2R}{1}}$ which is wrong.
Correct equation of sine rule will be,
${\displaystyle \frac{\sin A}{a}}={\displaystyle \frac{\sin B}{b}}={\displaystyle \frac{\sin C}{c}}={\displaystyle \frac{1}{2R}}$.