Question

In the figure given below, $RA=SA=4cm$ and $QA=3cm$. If $PQ$ is the diameter, then the radius is

• A. $5$ cm
• B. $6$ cm
• C. $7$ cm
• D. $\frac{25}{6}$ cm

Answer 1

Answer: (D)

$\frac{25}{6}$ cm

We can apply Intersecting chord theorem.

Chord $RS=RA+SA=8cm$

The intersecting chords theorem tells us that $AP\times QA=RA\times SA$

let $l=RS$ and $h=QA$

$\Rightarrow RA+QA=\frac{1}{2}l$

$\Rightarrow \frac{1}{2}l\times \frac{1}{2}l=h\times AP$

i.e $AP=\frac{1}{4h}{l}^2$.The diameter $PQ=AP+QA$ and

$QA+AP=h+\frac{l^2}{4h}=\frac{4h^2+l^2}{4h}$

The radius is one half of this ,i.e .

$OP=\frac{4h^2+l^2}{8h}$

$=\frac{4\times 3^2+8^2}{8\times 3}$

$=\frac{25}{6}$

$\therefore$ radius $OP=\frac{25}{6}$cm