AB and CD are two parallel chords on the same side of the circle. AB = 6 cm, CD = 8 cm. The small chord is at a distance of 4 cm from the centre. At what distance from the centre is the other chord?
Answers
Answer:
Let O be the centre of the circle. AB and CD are two parallel chords on the same side of the centre, where AB=6 cm and CD=8 cm. Also, OB and OD are the radius of the circle, OE=4 cm. To find OF.
Since OE⊥AB, so AE=EB=3 cm. Similarly, BF=FD=4 cm.
Now, from right angled triangle OEB, we have\begin{gathered}OB^2=OE^2+EB^2\\\Rightarrow OB=\sqrt{OE^2+EB^2}=\sqrt{4^2+3^2}=5.\end{gathered}
OB
2
=OE
2
+EB
2
⇒OB=
OE
2
+EB
2
=
4
2
+3
2
=5.
So, OB=OD=5 cm.
Again, from right angled triangle OFD, we have
\begin{gathered}OD^2=OF^2+FD^2\\\Rightarrow OF=\sqrt{OD^2-FD^2}=\sqrt{5^2-4^2}=3.\end{gathered}
OD
2
=OF
2
+FD
2
⇒OF=
OD
2
−FD
2
=
5
2
−4
2
=3.
Thus, the distance of the other chord CD from the centre is 3 cm.
Step-by-step explanation:
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