A triangle has sides of length 13,30 and 37. If the radius of the inscribed circle is pqp q (where p and q are coprime), then the value of qp+3qp+3 is
Answers
Answered by
3
Answer:
The seems to be a loss of detail in the expression for the inscribed circle in terms of p and q, but you can probably finish the problem off just fine once you know that the radius of the inscribed circle is 9/2.
Step-by-step explanation:
As usual, write s for the semiperimeter of the triangle. So s = (13+30+37)/2 = 40. Writing A for the area, Heron's Formula then gives us
A² = s(s-a)(s-b)(s-c) = 40 x 27 x 10 x 3 = 4 x 10 x 10 x 9 x 9
so A = 2 x 10 x 9 = 180.
Using the fact that A = rs, where r is the inradius, we have
r = A / s = 180 / 40 = 9 / 2.
Hopefully that gets you far enough to finish it off.
Similar questions