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(B) False

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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;

$R=\dfrac{abc}{4S}$â€¦â€¦â€¦â€¦â€¦â€¦.(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 $\Delta ABC$ involving circumradius R as;

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{1}{2R}..............\left( 2 \right)$

As we have a formula of area with involvement of two sides and angle between them.

Let the area be represented by S.

$Area=S=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}ab\sin C=\dfrac{1}{2}ac\sin B........\left( 3 \right)$

Now, from equation (2) and (3), we can write an equation with respect to one angle as

$\dfrac{\sin A}{a}=\dfrac{1}{2R}\text{ and }S=\dfrac{1}{2}bc\sin A$

Substituting value of sin A from the relation $\dfrac{\sin A}{a}=\dfrac{1}{2R}\text{ to }S=\dfrac{1}{2}bc\sin A$, we get;

As $\sin A=\dfrac{a}{2R}$ from the first relation, now putting value of sin A in $S=\dfrac{1}{2}bc\sin A$, we get

$\begin{align}

& S=\dfrac{1}{2}bc\dfrac{a}{2R} \\

& S=\dfrac{abc}{4R} \\

\end{align}$

Transferring R to other side, we get;

$R=\dfrac{abc}{4S}$

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 heronâ€™s formula for proving i.e.

$\begin{align}

& S=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \\

& s=\dfrac{a+b+c}{2} \\

\end{align}$

Which will make the solution very complex.

One can go wrong while writing sine rule as;

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{2R}{1}$ which is wrong.

Correct equation of sine rule will be,

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{1}{2R}$.

Complete step-by-step answer:

Here, we have given R as a radius of circumcircle i.e. circumradius and need to prove the relation;

$R=\dfrac{abc}{4S}$â€¦â€¦â€¦â€¦â€¦â€¦.(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 $\Delta ABC$ involving circumradius R as;

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{1}{2R}..............\left( 2 \right)$

As we have a formula of area with involvement of two sides and angle between them.

Let the area be represented by S.

$Area=S=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}ab\sin C=\dfrac{1}{2}ac\sin B........\left( 3 \right)$

Now, from equation (2) and (3), we can write an equation with respect to one angle as

$\dfrac{\sin A}{a}=\dfrac{1}{2R}\text{ and }S=\dfrac{1}{2}bc\sin A$

Substituting value of sin A from the relation $\dfrac{\sin A}{a}=\dfrac{1}{2R}\text{ to }S=\dfrac{1}{2}bc\sin A$, we get;

As $\sin A=\dfrac{a}{2R}$ from the first relation, now putting value of sin A in $S=\dfrac{1}{2}bc\sin A$, we get

$\begin{align}

& S=\dfrac{1}{2}bc\dfrac{a}{2R} \\

& S=\dfrac{abc}{4R} \\

\end{align}$

Transferring R to other side, we get;

$R=\dfrac{abc}{4S}$

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 heronâ€™s formula for proving i.e.

$\begin{align}

& S=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \\

& s=\dfrac{a+b+c}{2} \\

\end{align}$

Which will make the solution very complex.

One can go wrong while writing sine rule as;

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{2R}{1}$ which is wrong.

Correct equation of sine rule will be,

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=\dfrac{1}{2R}$.

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