Geometry Practice Test 2

Let OE = x
OD = y
AB = 6 cm BC = 2 cm
OC = √50 = radius.
By Pythagoras theorem,

(OA)2 = OD2+AD2 and OC2 = EC2+EO2
(√50)2 = y2+(6−x)2
(√50)2 = x2+(y+2)2
50 = y2+36+x2−12x and
50 = x2+y2+4+4y
Equating both the equations, we get:
- 12x + 36 = 4 + 4y
12x + 4y = 32
3x + y = 8
By hit and trial, there are only two possibilities i.e., x = 1 and y = 5
and x = 2 and y = 2 [but both x and y are not equal, so these values are rejected.]
Hence, x = 1 and y = 5
In ΔBOE,
(OB)2 = (OE)2+EB2
(Since EB = OD)
OB2 = OE2+OD2
OB2 = x2+y2
OB2 = 12+52 = 26