Quadrilateral \(ABCD\) is inscribed in a circle with segment \(AC\) a diameter of the circle. If \(m\angle DAC = 30^\circ\) and \(m\angle BAC = 45^\circ\), the ratio of the area of \(ABCD\) to the area of the circle can be expressed as a common fraction in simplest radical form in terms of \(\pi\) as \(\frac{a+\sqrt{b}}{c\pi}\), where a, b, and c are positive integers. What is the value of a+b+c ?
+128408 We can choose any convenient value for the diameter.....let this = 2...so the radius = 1
Angles ADC and ABC will be right angles
So in right triangle ADC angle DAC = 30° and angle DCA = 60°
So DC = 1 AD = sqrt (3)
So area of triangle ADC = (1/2) (1) (sqrt (3) = sqrt (3)/2
And triangle ABD will be isosceles right with AB = BC = 2/sqrt (2) = sqrt (2)
So the area of triangle ABD = (1/2) [sqrt (2)]^2 = 1
So [ ABCD ] = 1 + sqrt (3) / 2 = [ 2 + sqrt (3) ] /2
And the area of the circle = pi 1^2 = pi
So [ABCD ] / area of circle = (2 + sqrt (3) ) / ( 2 * pi ) = (2 + sqrt (3) / (2pi)
So a + b + c = 2 + 3 + 2 = 7
Here's a pic :
