Question

If $AB$ is a chord and $P$ is a point on $AB$ such that $AP=8$ cm, $PB=5$ cm and $P$ is $3$ cm from the centre of the circle, find the radius.

Answer 1

Solution:

Given that:

$AP=8cm,PB=5cm$ and $OP=3cm$

To find:

$OA=OB=?$

Solution:

$MB=\frac{AB}{2}=\frac{13}{2}=6.5cm$

$MP=MB-PB=6.5cm-5cm=1.5cm$

In $△OMP$

$O{P}^2=O{M}^2+M{P}^2$ (By pythagorus theorem)

or, $O{M}^2=3^2-(1.5{)}^2=9-2.25=6.75c{m}^2$

Now, In $△OMA$

$O{A}^2=O{M}^2+A{M}^2$ (By pythagorus theorem)

or, $O{A}^2=6.75c{m}^2+(6.5cm{)}^2$

or, $O{A}^2=6.75c{m}^2+42.25c{m}^2$

or, $O{A}^2=49c{m}^2$

or, $OA=7cm$

Hence, radius of the circle $=7cm.$