Question

AB is a chord of circle with center O. If OP perpendicularAB and extended OP meets the circle at C and AB=6cm,PC=2cm,then find the length of the radius.​

Answer 1

Given -

AB = 6 cm, OD = 4 cm and O is the center of the circle.

AB is a chord and OD is perpendicular to AB.

OA = Radius = ?

It is clear from the figure that AD = DB =1/2 of AD

=1/2 x 6

AD = 3 cm

Now. using Pythagoras Theorem,

(Hypotenuse)2 = (Base)? + (Perpendicular)

(OA)² = (AD)² + (OD)²

(OA)² = (3)²+ (4)²

(OA)² = 9 + 16

(OA)² = 25

taking square root,

OA = v25

OA = 5 cm

So, the radius of the circle is 5 cm.

Answer 2

Answer:

The radius is $3.25cm$.

Step-by-step explanation:

It is given that:

AB is a chord of circle with center O. If OP is perpendicular to AB and extended OP meets the circle at C.

$AB=6cm$

$PC=2cm$

We need to determine the radius.

If OP is perpendicular to AB, then

$AP=BP=3cm$

The theorem says that "the products of the lengths of the line segments on each chord are equal".

If $CD$ be diameter if circle, then

According to theorem

$AP*PB=PC*PD$

$3*3=2*PD$

$PD=4.5cm$

Diameter $CD=2+4.5=6.5cm$

Radius $=6.5/2=3.25cm$

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