How do you find the area of the largest possible circle inside a triangle???

Question

Cont. When you only have the measurments of the sides of the triangle???
Ex.
Side 1 - 240 km
Side 2 - 330 km
Side 3 - 270 km

Answer 1

Area of a triangle = rs
where r = radius of incircle and s = semiperimeter (=(a + b+ c)/2)

Also area = √(s(s-a)(s-b)(s-c))

Therefore rs = √(s(s-a)(s-b)(s-c))
so r = √(s(s-a)(s-b)(s-c))/s

Area of circle = πr²
= (√(s(s-a)(s-b)(s-c))/s)²
= π(s-a)(s-b)(s-c))/s
For this triangle a = 240, b = 330, c = 270, so s = 420
Thus required area = π (420-240)(420 - 330)(420 - 270)/420
= π*180*110*150/420
=4950π/7 km²

Answer 2

Imagine putting a small circle inside the triangle and inflating it to make it as big as possible. What will it look like? First, the circle will touch all 3 sides of course, or else it could be bigger. It will touch each side in 1 point, which means it is tangent to all 3 sides, which means the distance from the center to each side is the same - call it r.

Now compute the area of the circle. By drawing the lines from the vertices to the center of the circle you split the triangle into 3 triangles, where the area is 1/2side1*r + 1/2side2*r + 1/2side3*r = 1/2r(perimeter).

You can also compute the area of a triangle from the sides. Do you know the formula? Something like sqrt(s(s - a)(s - b)(s - c)), where s is the semiperimeter (a + b + c)/2. Solve this for the given triangle and then set it equal to the equation derived above for the area in terms of r, and solve.

Answer 3

area of the circle = pie*square(inradius)
inradius =area/semiperimeter
area can be found by heros formula......... and semiperimeter is half the perimeter

Answer 4

See http://courses.ncssm.edu/goebel/statecon/Topics/triangl/TRIANGL.htm