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?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
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 $ABC$.
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