Question

$O$ is the centre of the circle having radius $5$ cm. $AB$ and $AC$ are two chords such that $AB=AC=6$ cm. If OA meets $BC$ at $M$, then $OM$ is equal to

• A. $3.6$ cm
• B. $1.4$ cm
• C. $2$ cm
• D. $3$ cm

Answer 1

The correct option is B $1.4$ cm

$M$ is the mid point of $BC$
Thus, $BM=MC=x$ [Let]
$OA$ is the radius of the circle
Let $AM=y$
$OM=OA-AM$
$OM=5-y$ ----(i)
Now, $△ABM$ is a right angled triangle
$\therefore A{B}^2=A{M}^2+B{M}^2$
$=>6^2=y^2+x^2$
$=>x^2=36-y^2$ -----(ii)
Again in $△OMB$

$O{B}^2=O{M}^2+B{M}^2$
$=>5^2=(5-y{)}^2+x^2$ ....[Using (i)]
$=>25=25-10y+y^2+36-y^2$ ....[Using (ii)]
$=>10y=36$
$=>y=3.6$
Thus, $OM=(5-3.6)=1.4$cm