The radius of a circle inscribed within a triangle (also known as the inradius) can be determined using the triangle's area and side lengths. The key formula connecting these elements is:
$ A = rs $
The semi-perimeter ($s$) is half the sum of the lengths of the triangle's sides ($a$, $b$, $c$).
$ s = \frac{a+b+c}{2} $
Substitute the expression for the semi-perimeter ($s$) into the area formula ($A = rs$):
$ A = r \left( \frac{a+b+c}{2} \right) $
To find the radius ($r$), rearrange the equation:
$ r = \frac{A}{\left( \frac{a+b+c}{2} \right)} $
Simplify the expression:
$ r = \frac{2A}{a+b+c} $
Therefore, the radius of the inscribed circle in triangle ABC is given by the formula $\frac{2A}{a+b+c}$. This matches Option B.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.