Question

PQ is the diameter of a circle and PR is a chord, if PQ = 34cm and

PR = 30cm. Find the distance of PR from the center of the circle​

Answer 1

★Given:-

• PQ is the diameter of a circle.
• PR is a chord
• PQ = 34cm
• PR = 30cm

★To find:-

• The distance of PR from the center of the circle

★Solution:-

From the diagram,

OX is the perpendicular distance from centre O to the chord PR

∴ PX=XR

→PX+XR=PR

As [PX = XR],

⇒PX + PX = PR

⇒2PX=PR

⇒2PX=30

⇒PX= 30/2

⇒PX=15

Also,

→PO+OQ=PQ

As [PO = OQ],

⇒ PO+PO = PQ

⇒2PO=PQ

⇒2PO=34

⇒PO=17

We know,

• ΔPOX is a right angled triangle.

By pythagoras theorem,

✦(Hypotenuse)²=(perpendicular)²+(base)²

Putting values,

⇒ (PO)²=(OX)²+(PX)²

⇒ 17² = (OX)²+15²

⇒ (OX)² = 17²-15²

⇒ OX = √64

⇒ OX = 8cm

Hence,

The distance of PR from center of circle is 8cm.

______________

Answer 2

Given : The diameter of the circle PQ (D) = 34 cm

             ⇒ Radius (r) = 17 cm

           Chord PR = 30 cm

Let O be the centre of the circle

OS be the perpendicular bisector of the chord PR  &  

PS  = SR = 30/2 = 15 cm

OS is the distance of the chord from the centre (d)

Now ΔOPS is a right angled triangle with right angle at S

so, from Pythogaras theorem

OP² = PS² + OS²

⇒ r² = 15² + d²

⇒ 17² = 15² + d²

⇒ d² = 64

∴ d = 8 cm