Question

Given the circle below with tangent \overline{NO} NO and secant \overline{QPO} QPO ​ . If QP=9QP=9 and QO=14QO=14, find the length of \overline{NO} NO . Round to the nearest tenth if necessary. Q

Answer 1

Given that that the lengths of the parts of the secant are, QO = 14 and QP = 9, we have that the tangent, NO = 17.9

How can the length of the tangent be found?

The given parameters are;

Length of side, QP = 9

Length of QO = 14

Required;

The length of;

$\overline{no}$

From the tangent and secant formula, we have;

${ \overline{no}}^{2} = { \overline{qo}} \times ( { \overline{qp} \: + { \overline{qo}}})$

Which gives;

NO^2 = 14 × (9 + 14) = 322

NO = √(322) = 17.9

• The length of the side NO = 17.9

Learn more about the tangent and secant theorem here;

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