A, B, C and D lie on a circle, centre O, radius 8 cm.
AB and CD are tangents to a circle, centre O, radius 4 cm.
ABCD is a rectangle.
( a ) Calculate the distance AE.
Since AB and CD are tangents to the circle with centre O and radius 4 cm, they are perpendicular to the radius at the point of contact. Therefore, triangles OAE and OBE are right-angled at A and B respectively.
Since AB and CD are tangents, AO and CO are also perpendicular to AB and CD respectively. Therefore, AO = CO = 4 cm.
Now, in triangle OAE, we have:
OE^2 + AE^2 = OA^2
OE^2 + AE^2 = 8^2
OE^2 + AE^2 = 64
In triangle OBE, we have:
OE^2 + BE^2 = OB^2
OE^2 + BE^2 = 8^2
OE^2 + BE^2 = 64
Since OE is common in both equations, we can subtract the two equations to get:
AE^2 - BE^2 = OA^2 - OB^2
AE^2 - BE^2 = 64 - 64
AE^2 - BE^2 = 0
As ABCD is a rectangle, AE = BE. Therefore, substituting AE = BE into the equation, we get:
AE^2 - AE^2 = 0
0 = 0
Therefore, AE = BE = 0.
Hence, the distance AE is 0.