Question

A point P 15 cm from the centre of a arch. The radius of the circle is 5 cm, find the
length of the tangent drawn to the circle from the point P.{with diagram}
​

Answer 1

$\LARGE{ \underline{ \purple{ \sf{Required \: answer:}}}}$

The question can be solved by knowing the theoram that:

The tangent from an exterior point to the circle is perpendicular to its radius. So, the tangent, radius and the line joining the centre and the point is forming a right-angled triangle.

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Refer to the attachment...

Let the centre be O, exterior point be P and the point of tangency be Q. Then, ∆PQO is a right-angled triangle, right angled at Q.

By Pythagoras theoram,

⇛ PQ² + QO² = PO²

• PO = 15 cm
• QO = 5 cm

⇛ PQ² + 5² = 15²

⇛ PQ² = 15² - 5²

⇛ PQ² = 200

⇛ PQ = √200 cm

⇛ PQ = 10√2 cm or 14.14 cm

$\therefore$Thus, the required measure of the length of the tangent is 10√2 cm.

Answer 2

Step-by-step explanation:

Given : -

• A point P 15 cm from the centre of a arch.

• The radius of the circle is 5 cm

To Find : -

• find the length of the tangent drawn to the circle from the point P

Solution : -

Let OR is the radius and PR is the tangent..

OP = 15 cm

By P. G. T.

PR² = OR² + OP²

PR² = 5² + 15²

= 25 + 225

= 250

PR = √ 250

PR = 15.81 cm

Hence the tangent is 15.81 cm.

More information : -

• the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle.

• It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.