Question

The length if two parallel chords are 6 cm and 8 cm if the smaller code of is a distance of 4 cm from the centre what is the distance of the Other code from the centre

Answer 1

Answer:

AB=8cm

CD=6cm

AE=BE=4cm

OF=4cm

CF=FD=3cm

In triangle OCF

OV^2=OF^2+FC^2

=4^2+3^2

=16+9

=25

OC=5cm

OC=OA=5cm. (radius)

In triangle AOE

OE^2=AO^2-AE^2

=5^2-4^2

=25-16

=9

OE=3cm

Therefor distance of chord AB (8cm) is 3cm

Step-by-step explanation: