Question

In figure 3.80, if $PQ=6$, $QR=10$, $PS=8$ find $TS$.

Answer 1

$\Rightarrow$ Here, we have, $PS=8,PQ=6$ and $QR=10$

$\therefore$ $PR=10+6=16$

We know that if the chord of the circle intersect externally then the products of the segments of the chords are equal.

$\therefore$ $PS\times PT=PQ\times PR$

$\Rightarrow$ $8\times PT=6\times 16$

$\therefore$ $PT=12$

Now, $PT=PS+TS$

$\Rightarrow$ $TS=PT-PS$

$\therefore$ $TS=12-8$

$\therefore$ $TS=4$