Question

2 AJ Complete and Write the following activities: 1) P R Solution :- In AOQP and AORP ZOQP - 90° Hypotenuse OP = Hypotenuse OP Side OQ IN [Tangent theorem] [Common side ] [Radii of the same circle] [Hypotenuse side test) (C. s. c. t. ] AORP Seg PR​

Answer 1

Answer:

Draw a line segment, from Centre O to external point P { i.e. P is the intersecting point of both the tangents}

Now ∆POR and ∆POQ.

In order to prove they have the same length, we will first prove that both triangles are similar.

We know that the tangents make a right angle with a radius of the circle.

Here, OR and OQ is the radius of the circle

So, ∠OQP = ∠ORP = 90°

Now, it is clear that both the triangles ∆POQ and ∆POR are right-angled triangles and a common hypotenuse OP in them.

Proof

Now proving the similarity between triangles ∆POQ and ∆POR

Here,

∠PQO = ∠PRO = 90°

Common hypotenuse OP between them.

And OQ = OR [Radius of circle].

So, by the R.H.S. rule of similarity

∆POQ ~ ∆POR

Hence, both the triangles are similar to each other.

Therefore,

OP/OP = PQ/PR = OQ/OR

PQ/PR = 1 {since OP/OP = 1} ;

Hence, PQ = PR;

Hence, Proved that The lengths of tangents drawn from an external point to a circle are equal.