Question

The sides of a triangle are 25, 39, 40, the diameter of circumscribed circle is:
(a) $\dfrac{{133}}{3}$
(b) $\dfrac{{125}}{3}$
(c) 42
(d) 41 (e) 40

Answer 1

Hint: we should know the formula to find the radius of the circle circumscribing a triangle at sides a, b, c, i.e,
\[R = \dfrac{{abc}}{{\sqrt {\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)} }}\]

Complete step by step solution:
First we need to find the radius of the circle circumscribing the triangle.
We know that a, b, c is 25, 39, 40 respectively on the sides of the triangle.
Using the above formulae, we get
     $R = \dfrac{{25 \times 39 \times 40}}{{\sqrt {\left( {25 + 39 + 40} \right)\left( {39 + 40 - 25} \right).\left( {39 + 25 - 40} \right)\left( {25 + 40 - 39} \right)} }}$
     $ = \dfrac{{25 \times 39 \times 40}}{{\sqrt {\left( {104} \right) \times 54 \times 24 \times 26} }}$
     $ = \dfrac{{25 \times 39 \times 40}}{{\sqrt {8 \times 13 \times 2 \times 9 \times 3 \times 8 \times 3 \times 13 \times 2} }}$
     $ = \dfrac{{25 \times 39 \times 40}}{{8 \times 13 \times 2 \times 9}} = \dfrac{{125}}{6}$
Now diameter is twice the radius
$Therefore, $ $D = 2R = \dfrac{{2 \times 125}}{6} = 125$
$Therefore, $ Diameter $ = \dfrac{{125}}{3}$



Note: The above formulae can be derived using the solution of triangle and heron’s formula. It will be helpful if you know the formulae beforehand.
Proof:
By using extended sine rule i.e.,
$\dfrac{a}{{{\text{Sin}}{\text{A}}}} = \dfrac{b}{{{\text{SinB}}}} = \dfrac{c}{{Sinc}} = 2R$
And using formulae from solution of triangle area of triangle $\left( \Delta \right)$ is
$\Delta = \dfrac{1}{2}ab\;Sinc$
 $Therefore, \dfrac{c}{{{\text{Sinc}}}} = 2R$
${\text{Sinc}} = \dfrac{c}{{2R}}$
Substituting it to area equation we get
$\Delta = \dfrac{1}{2}ab\dfrac{c}{{2R}} = \dfrac{{abc}}{{4R}}$
Now using hess's formulae
\[\Delta = \sqrt {s\left( {s - a} \right)\left( {s - 6} \right)\left( {s - 6} \right)} \] where $s = \dfrac{{a + b + c}}{2}$
equating both $\Delta $ values we get
\[\dfrac{{abc}}{{4R}} = \sqrt {\left( {\dfrac{{a + b + c}}{2}} \right)\left( {\dfrac{{a + b + c - a}}{2}} \right)\left( {\dfrac{{a + b + c - b}}{2}} \right)\left( {\dfrac{{a + b + c - c}}{2}} \right)} \]
$R = \dfrac{{abc}}{{4 \times \dfrac{{\sqrt {\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {a + c - b} \right)\left( {a + b - c} \right)} }}{4}}}$
\[R = \dfrac{{abc}}{{\sqrt {\left( {a + b + c} \right)\left( {b + c - a} \right)\left( {a + c - b} \right)\left( {a + b - c} \right)} }}\]