I have a circle which has a triangle inscribed in it.
The circle radius R = 4
The triangle ABC vertices divide circle into 3 arcs in 1:2:3 ratio
Find the perimeter and area of triangle.
Can you guys help me with this one?
[Math] Find Area and Perimeter of inscribed triangle
areaplane-geometrytriangles
Related Solutions
Study the above diagram which describes the problem.
From the calculations of the arc lengths, we have:
$$3 = rA, \qquad4 = rB, \qquad 5 = rC$$ $$3 + 4 + 5 = rA + rB + rC = r(A + B + C)$$ $$r = \frac{12}{A + B + C} = \frac{6}{\pi}$$
We also know that the angles $A,B,C$ are in proportions corresponding to the respective arc length proportions. So, $$A = \frac{6\pi}{12}, \qquad B = \frac{8\pi}{12}, \qquad C = \frac{10\pi}{12}$$
I believe it is easy to compute the area of the inscribed triangle from here: It is given by
$$\frac{1}{2}r^2\sin A + \frac{1}{2}r^2\sin B + \frac{1}{2}r^2\sin C$$
Let a triangle be inscribed in a unit circle, and let $A$ and $B$ mark two vertices. Let $\theta$ be half the length of the arc connecting $A$ and $B$, and let $\ell$ be the length of the chord, you have that from elementary trigonometry
$$ \ell / 2 = \sin (\theta) $$
Now let $\theta_1, \theta_2, \theta_3$ be half the lengths of the three arcs demarcated by the vertices of the triangle, from the above we have that the perimeter is equal to
$$ 2 \left[ \sin (\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] $$
and we know that
$$ \theta_1 + \theta_2 + \theta_3 = \pi $$
while
$$ \theta_1, \theta_2, \theta_3 \in [0,\pi] $$
Now from the fact that $\sin$ is a concave function on $[0,\pi]$, we have that the perimeter is equal to
$$ 6 * \frac13 \left[ \sin(\theta_1) + \sin(\theta_2) + \sin(\theta_3) \right] \leq 6 * \sin (\pi / 3) $$
by Jensen's inequality, with the optimum (being the case of the equality) holding when $\theta_1 = \theta_2 = \theta_3$.
Best Answer
Hint:
The ratio between the angles of the triangle is $1:2:3$, so the angles are $30^{\circ}$, $60^{\circ}$ and $90^{\circ}$. Now, the hypotenuse is also a diameter of the circle (why?) then the sides of the triangle are $4$, $4\cos 30^{\circ}$ and $4\sin 30^{\circ}$.