# Find Area and Perimeter of inscribed triangle

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?

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 $4\sin 30^{\circ}$.

Yes Angelo Mario. if you draw the triangle with these angles you could find the answer.(The longest side of the triangle goes through the center of the circle.) the answer will be 13.856

The condition on the three arcs' $1:2:3$ ratio, together with the fact that the arcs add up to $360^\circ$, means that the arcs are $60^\circ$, $120^\circ$, and $180^\circ$.

The angles in triangle $ABC$ are inscribed angles, each being exactly half of the corresponding arc; so the angles are $30^\circ$, $60^\circ$, and $90^\circ$, i.e. $ABC$ is a right triangle. The hypotenuse of the $30$-$60$-$90$ triangle $ABC$ is the diameter of the given circle; so the hypotenuse is 8, the short leg is 4, and the long leg is $4\sqrt3$.

From this we find the perimeter, $12+4\sqrt3$, as well as the area $8\sqrt3$ of triangle $ABC$.

Hint:

If the three arcs are in $1:2:3$ ratio ( and the sum is $2 \pi$), than the first arc goes form $0$ to $\pi/3$, the second from $\pi/3$ to $\pi$ and the third from $\pi$ to $2\pi$.

Can you do from this? ( see the figure)

One can observe that the arc lengths can be shown as parts of 6, for 1+2+3 is equal to 6. Once this has been observed, one can see the triangle has a side that of which is a diameter, and another as a radius.

It may be easier to think of it as parts of a hexagon divided into 6 equal parts.

Because an angle on the circumference is equal to half of the angle in the center sub tended by the same arc, one can see it is a right triangle and use Pythagorean Theorem to get an area of 8*3^1/2