Question

in the given figure O is the centre of circle of radius 5cm. OP perpendicular CD,AB parallel to CD
AB = 6 cm and CD= 8cm. Determine PQ

Answer 1

Answer:

Step-by-step explanation:

AP = 3 {radius when perpendicular to chord bisects it }

QC = 4 {radius when perpendicular to chord bisects it }

Applying pythagoras theorem 

In OAP

OP=4

In OCQ

OQ=3 

OP=PQ+OQ

4=PQ+3

therefore,

PQ = 1 .

Hope this help u ,

Sonjeevon kumar

Answer 2

Given:

The radius of the circle=5cm

OP is perpendicular to CD

AB is parallel to CD

AB=6cm, CD=8cm

To find:

The length of the PQ

Solution:

The length of the line PQ= 1cm.

We can find the length by following the steps given below-

We know that OP is perpendicular to CD.

Since CD is parallel to AB, OP is also perpendicular to AB.

This implies that OP bisects both AB and CD.

We have CD=8cm and AB=6cm.

So, CQ=CD/2=8/2=4cm

AP=AB/2=6/2=3cm

Now we will join OC and OA to form two right-angled triangles, ∆OQC and ∆OPA.

In ∆OQC,

OC is the radius of the circle=5cm

CQ=4cm

By using the Pythagoras theorem,

OC²=OQ²+CQ²

On putting the values,

5²=OQ²+4²

25-16=OQ²

9=OQ²

OQ=3cm

In ∆OPA,

OA is the radius of the circle=5cm

AP=3cm

By using the Pythagoras theorem,

OA²=OP²+AP²

On putting the values,

5²=OP²+3²

25-9=OP²

16=OP²

OP=4cm

Now, to obtain PQ, we will subtract OP and OQ.

PQ=OP-OQ

We will put the values,

PQ=4-3=1cm

PQ=1cm

Therefore, the length of the line PQ is 1cm.