Two circles, each of radius 2 cm touch each other at P. AP and PB are ciameters of the two circes. M and N are the midpoints of AP and PB respectively. The perpendicular bisector of NB intersects the circle at C and D. What is the circumradius (in cm) of triangle ACD?
a) 26/7
b) 24/7 e) 4
c) 22/7
d)30/7
Answers
Given : Two circles, each of radius 2 cm touch each other at P. AP and PB are Diameters of the two circles. M and N are the midpoints of AP and PB respectively.
To Find : circumradius (in cm) of triangle ACD
a) 26/7
b) 24/7
c) 22/7
d)30/7
e) 4
Solution:
AP - Diameter => M is center of Circle
PB - Diameter => N is center of Circle
AM = MP = PN = NB = 2 cm
AB = 8 cm
and AMPNB is a straight line as circle touches at P
Let say E is the mid point of CD .
NE = EB = NB/2 = 2/2 = 1 cm
AE = AB - NE = 8 - 1 = 7 cm
NC = ND = 2 cm radius
=> CE² = NC² - NE² = 2² - 1² = 3
ACD is isosceles triangle
Hence circumcenter lies on AB
Let say AO = R cm = circumradius = OC = OD
OE = AE - AO = 7 - R cm
OE² + CE² = OC²
=> (7 - R)² + 3 = R²
=> 49 + R² - 14R + 3 = R²
=> 14R = 52
=> 7R = 26
=> R = 26/7
circumradius (in cm) of triangle ACD = 26/7 cm
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