Q4. - Geometry

Question

$\overline{JT}$ is tangent to $\odot$$O$ at $T$. Complete.

If $JK=9$ and $KO=8$, then $JT=?$

Step-by-Step Solution

Answer

The value of $JT=3\sqrt{17}$.

Step 1. Given information.

It is given that, $JK=9,KO=8$ in the figure,

$\begin{array}{l}JO=JK+KO\\ =17\end{array}$

Let, $JT=x$

Step 2. Concept used.

When a secant segment and a tangent segments are drawn to a circle from an external point then the product of the secant segment and its external segment is equal to the square of the tangent segment.

Therefore, from the given figure,

$\begin{array}{c}J{T}^2=JK\times JO\\ x^2=9\times 17\\ x^2=153\\ x=3\sqrt{17}\end{array}$

Step 3. Neglect negative sign.

Since, length is positive

 So, neglecting the negative value. 

Therefore, the value of $JT=3\sqrt{17}$.