One geometry problem

[Auree123]

Q. OAB is an inscribed triangle in a circle with two vertices A, B on circle and another at center O. radius of the circle is 5. Angle <AOB = x; 90<x<180.

A. The perimeter of triangle AOB

B. 17

the OA. is A

Pls explain.

[sweetu1995]

The angle X is greater than 90.

For simplification, assume that the angle X is 90 and calculate the perimeter.

Side AB = 5sqrt2

Hence Perimeter = 10 + 5sqrt2 which is > 17

With X = 90, perimeter is > 17

If X > 90, length of AB will increase and hence the perimeter will be definitely > 17

[sweetu1995]

One more thing, a traingle is inscribed if all the 3 vertices lie on the circumference. The traingle you mentioned is not an inscribed traingle I guess

[Riyad Mohammad]

i have done in the same way as swwetu1995

[deichgraf]

The angle X is greater than 90.

For simplification, assume that the angle X is 90 and calculate the perimeter.

Side AB = 5sqrt2

Hence Perimeter = 10 + 5sqrt2 which is > 17

With X = 90, perimeter is > 17

If X > 90, length of AB will increase and hence the perimeter will be definitely > 17

Yep, think so to. x < 90 < 180 can be simplified to x > 90 because we know that there can't be an angle greater than 180 in a triangle, since the sum of all angles is already 180. The triangle we talk about is isosceles, since r=5 gives us two vertices. Since x is always greater than 90 the sum of all other to angles is at most less than 90. Therefore x is always the biggest angle... and we know that a vertice opposite to the biggest angle is always the biggest one in a triangle. The absolute value does not matter, but it is at least 2*sqrt5...