If R and r are respectively the circumradius and in radius of triangle having sides 40 cm, 41 cm and 9 cm, then find the value of 2 (R + r).

For a triangle with sides \(a\), \(b\), and \(c\), the circumradius \(R\) and the inradius \(r\) are given by:

\[ R = \frac{abc}{4K} \]

and

\[ r = \frac{K}{s} \]

where \(K\) is the area of the triangle and \(s = \frac{a + b + c}{2}\) is the semi-perimeter of the triangle.

For the given triangle with sides \(a = 40\) cm, \(b = 41\) cm, and \(c = 9\) cm:

\[ s = \frac{40 + 41 + 9}{2} = \frac{90}{2} = 45 \text{ cm} \]

Using Heron’s formula, the area \(K\) of the triangle is:

\[ K = \sqrt{s(s – a)(s – b)(s – c)} \]
\[ K = \sqrt{45(45 – 40)(45 – 41)(45 – 9)} \]
\[ K = \sqrt{45 \times 5 \times 4 \times 36} \]
\[ K = \sqrt{32400} \]
\[ K = 180 \text{ cm}^2 \]

Now, we can find the circumradius \(R\):

\[ R = \frac{abc}{4K} \]
\[ R = \frac{40 \times 41 \times 9}{4 \times 180} \]
\[ R = \frac{14760}{720} \]
\[ R = 20.5 \text{ cm} \]

And the inradius \(r\):

\[ r = \frac{K}{s} \]
\[ r = \frac{180}{45} \]
\[ r = 4 \text{ cm} \]

Finally, the value of \(2(R + r)\) is:

\[ 2(R + r) = 2(20.5 + 4) \]
\[ 2(R + r) = 2 \times 24.5 \]
\[ 2(R + r) = 49 \]

Therefore, the value of \(2(R + r)\) is 49 cm.