Question

The radius of a circle is $2.5$ $cm.$ $AB$ and $CD$ are two parallel chords $2.7$ $cm$ apart. If $AB=4.8cm$ .then $CD$ is equal to

• A. $4.8$ $cm$
• B. $2.4$ $cm$
• C. $3$ $cm$
• D. $4$ $cm$

Answer 1

The correct option is C $3$ $cm$

Here, $AB\parallel CD$

$AB=4.8cm$ and $AM=MB=\frac{AB}{2}=2.4cm$

$OM$ and $ON$ are perpendicular to $AB$ and $CD$

In $△AMO,$

$\Rightarrow$ $(OA{)}^2=(AM{)}^2+(OM{)}^2$ [By Pythagoras theorem ]

$\Rightarrow$ $(2.5{)}^2=(2.4{)}^2+(OM{)}^2$

$\Rightarrow$ $6.25=5.76+(OM{)}^2$

$\Rightarrow$ $0.49=(OM{)}^2$

$\therefore$ $OM=0.7cm$

$\Rightarrow$ $ON=MN-OM$

$\Rightarrow$ $ON=2.7-0.7$

$\Rightarrow$ $ON=2cm$

In $CNO,$

$\Rightarrow$ $(OC{)}^2=(ON{)}^2+(CN{)}^2$

$\Rightarrow$ $(2.5{)}^2=(2{)}^2+x^2$

$\Rightarrow$ $x^2=6.25-4$

$\Rightarrow$ $x^2=2.25$

$\therefore$ $x=1.5$

$\Rightarrow$ $CD=(1.5+1.5)cm$

$\therefore$ $CD=3cm$