Question

Please Solve Fast With Full Explanation. ​

Answer 1

Join OL. The triangle OLP will be right angled since the tangent PLQ is perpendicular to OL.

So by Pythagoras' theorem,

$\longrightarrow PL^2=OP^2-OL^2$

$\longrightarrow PL^2=(PT+OT)^2-OL^2$

Note that OT and OL are radii of the circle.

$\longrightarrow PL^2=\left(\dfrac{16}{3}+\dfrac{10}{3}\right)^2-\left(\dfrac{10}{3}\right)^2$

$\longrightarrow PL^2=\left(\dfrac{26}{3}\right)^2-\left(\dfrac{10}{3}\right)^2$

$\longrightarrow PL^2=\dfrac{676}{9}-\dfrac{100}{9}$

$\longrightarrow PL^2=\dfrac{576}{9}$

$\longrightarrow PL^2=64$

$\longrightarrow PL=8\ cm$

The triangle PQR is also right angled since PR, passing along OR, is perpendicular to QR.

So by Pythagoras' Theorem,

$\longrightarrow PQ^2=PR^2+QR^2$

$\longrightarrow (PL+QL)^2=(PT+RT)^2+QR^2$

Note that RT is a diameter of the circle.

And here $QL=QR$ because they're tangents to the circle drawn from the same point.

Therefore,

$\longrightarrow \left(8+QL\right)^2=\left(\dfrac{16}{3}+2\times\dfrac{10}{3}\right)^2+QL^2$

$\longrightarrow 64+QL^2+16QL=\left(\dfrac{16}{3}+\dfrac{20}{3}\right)^2+QL^2$

$\longrightarrow 64+16QL=144$

$\longrightarrow\underline{\underline{QL=5\ cm}}$

Hence 5 cm is the answer.