Please be very detailed in your response and show clearly how you get from 1 ste
ID: 2943811 • Letter: P
Question
Please be very detailed in your response and show clearly how you get from 1 step to the other.....Thank you!!Calculate the radius of a circle circumscribed about and isosceles triangle with altitude 40 and base 60?
Explanation / Answer
The perpendicular bisector of a chord passes through the center of a circle. Let your isosceles triangle be formed by the point (0,0), (60,0), and (30,40). The perpendicular bisector of the chord defined by the base of your triangle passes through the midpoint of the base at (30,0) and has the equation x = 30. The perpendicular bisector of the chord defined by the line segment formed by the points (60,0) and (30,40) passes through the midpoint of the line segment at the point (45,20), i.e., (60+30)/2 and (0+40)/2. Since the slope of the line segment is -4/3, i.e., (0-40)/(60-30), the slope of the perpendicular bisector is 3/4. From the point-slope equation, we can find the equation of the perpendicular bisector... y-20 = 3/4(x-45) ==> y = (3/4)x - (3/4)45+20 ==> y = (3/4)x - 55/4 If we solve y = (3/4)x - 55/4 and x = 30, we can find the center of the circle... y = (3/4)30 - 55/4 ==> y = 90/4 - 55/4 ==> y = 35/4 Center of the circle is (30, 35/4); therefore, the radius is... Sqrt[(30-60)^2 + (35/4-0)^2] = 125/4 Let's check the other 2 points... Sqrt[(30-30)^2 + (40-35/4)^2] = 125/4 Sqrt[(30-0)^2 + (35/4-0)^2] = 125/4