prototypevenom wrote:
Simon wrote:
Note that both triangles OPQ and OXY are both isosceles triangles (OP=OQ=OX=OY=radius of the circle) => the height drawn from O to either bases cut the chord PQ and XY in equal halt => by Pythagorean theorem: radius^2 = (5.9)^2+(PQ/2)^2 = (5.8)^2+(XY/2)^2. Since (5.9)^2>(5.8)^2 => (PQ/2)^2<(XY/2)^2 => PQ<XY (PQ and XY are positive of course) => B is answer.
how do we infer that the perpendicular from center of circle to PQ or XY cuts it into half?
There's a nice/important triangle property that says:
The altitude of an isosceles (or an equilateral) triangle bisects the base
(screenshot from my course)
Since OX and OY are both radii of the same circle, we can conclude that OX = OY.
So, following the above
property, we know that the altitude cuts XY in half (i.e., bisects XY)
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep