Question

Let AB and CD be parallel chords of a circle, with centre O; M is the mid point of AB and N is the midpoint of CD. Prove that O, N, M are collinear. If MN$=3$cm., AB$=4$cm., CD$=10$cm., find the radius of the circles.

Answer 1

$AB\parallel CD$ (chords)

$M$ is midpoint $AM=BM$

$N$ is midpoint $CN=DN$

$MN=3cmAB=4cmCD=10cm$

Since, $N$ is midpoint of $CD$, Line passing through $N$ and Center of circle is perpendicular bisector of chord $CD$. $O$ and $N$ lies on a line.

Similarly,

$M$ is midpoint of $AB$ and line passing through center of circle and $M$ is the perpendicular bisector of $AB$. $O$ and $M$ lies on a straight line.

Hence, $O,M,N$ are collinear.

In $△OCD$

$(ON{)}^2=(OC{)}^2-\left(CN{)}^2\right(ON)=\sqrt{R^2-5^2}(1)$

In $△OAB$

$(OM{)}^2=(OA{)}^2-\left(AM{)}^2\right(OM)=\sqrt{R^2-2^2}(2)$

Subtracting $\left(1\right)$ from $\left(2\right)$

$OM-ON=\sqrt{R^2-2^2}-\sqrt{R^2-5^2}NM=\sqrt{R^2-2^2}-\sqrt{R^2-5^2}3+\sqrt{R^2-5^2}=\sqrt{R^2-2^2}9+R^2-5^2+6\sqrt{R^2-5^2}=R^2-2^{2}6\sqrt{R^2-5^2}=25-9-4=12\sqrt{R^2-5^2}=2R=\sqrt{25-4}=\sqrt{21}cm$