Question

If $AB$ is a chord of a circle with centre $O$ and $P$ is a point on $AB$ such that $BP=4PA$, $OP=5$ cm and the radius of the circle is $7$ cm, find the value of $(\sqrt{6}\times AB).$

Answer 1

Solution:

Given that:

$BP=4PA,OP=5cm$ and $OA=OB=7cm$

To find:

$\sqrt{6}\times AB=?$

Solution:

Let $PA=x,BP=4x$

$AB=PA+BP=x+4x=5x$

$BM=AM=\frac{AB}{2}=\frac{5x}{2}$

and $MP=\frac{5x}{2}-x=\frac{3x}{2}$

In $△OMB$

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

or, $O{M}^2=7^2-{\left(\frac{5x}{2}\right)}^2=49-\frac{25x^2}{4}$

In $△OMP$

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

or, $5^2={\left(\frac{3x}{2}\right)}^2+49-\frac{25x^2}{4}$

or, $25=\frac{9x^2}{4}+49-\frac{25x^2}{4}$

or, $4x^2=24$

or, $x^2=6$

or, $x=\sqrt{6}$

$AB=5x=5\sqrt{6}$

or, $\sqrt{6}\times AB=\sqrt{6}\times 5\sqrt{6}=30cm$