Question

Two parallel chords AB and CD are 3.9cm apart and lie on the opposite sides of the centre of a circle.if AB=1.4cm and CD=4cm,find the radius of the circle

Answer 1

copy little bit from photo.
cf=fd=1/2 cd same reason
cf = fd = 1/2 × 4
cf = fd = 2
let oe be x
oe=x
ef=eo+of
3.9 = x+of
of=3.9-x
ao = co radii of same circle .......1
In triangle AEO m angle AEO = 90 .....2
In triangle OFC m angle OFC = 90 ......3
from 1 2 and 3
${ae}^{2} + {eo}^{2} = {of}^{2} + {fc}^{2} \\ {0.7 }^{2} + {x}^{2} = (3.9 - {x}^{2}) + {2}^{2} \\ 0.49 + {x}^{2} = {x}^{2} - 7.8x + 15.21 + 4 \\ 7.8x = 18.72 \\ x = \frac{18.72}{7.8 } \\ x = 2.4cm$
ao^2=ae^2+oe^2
ao^2=
${0.7}^{2} + {2.4 }^{2} \\ 0.49 + 5.76 \\ 6.25 \\ \sqrt{6.25 } \\ 2.5$
radius of circle is 2.5 cm.

Answer 2

Step-by-step explanation:

1/2 ofcd

=0.7

1/2 of PQ 1.95