Question

निम्नलिखित प्रश्नों में से किसी एक प्रश्न का उत्तर दीजिए।
(a) दी हुई आकृति में वृत्त की त्रिज्या 5 सेंटीमीटर है। जीवायें AB और
CD समान्तर हैं और उनके बीच की दूरी 7 सेंटीमीटर है। यदि A
CD = 6 सेंटीमीटर है तो AB ज्ञात किजिए।
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Answer 1

Given:

The radius of the circle is 5 cm

Chords AB and CD are parallel and the distance between them is 7 cm

CD = 6 cm

To find:

The length of AB

Solution:

We know,

A line passing through the centre of the circle and perpendicular to the chords of the circle must bisect the chord.

Since  

OP ⊥ AB and PQ passes through the centre

∴  $\bold{PA = \frac{1}{2} AB}$ ....... (i)

and  

OQ ⊥ CD and PQ passes through the centre

∴  $QC = \frac{1}{2} CD$

$\implies QC = \frac{1}{2}\times 6$  

$\implies \bold{QC = 3 \:cm}$ ....... (ii)

Considering Δ COQ and applying the Pythagoras theorem, we get

$OC^2 = QC^2 + OQ^2$

substituting the value of OC = radius = 5 cm and the value of QC from (ii)

$\implies 5^2 = 3^2 + OQ^2$

$\implies 25 = 9 + OQ^2$

$\implies OQ = \sqrt{25 - 9}$

 $\implies \bold{OQ = 4\: cm}$

 

 The distance between the chords AB and CD is given as,

PQ = OP + OQ = 7 cm

∴ OP = PQ - OQ = 7 - 4 = 3 cm

Considering Δ AOP and applying the Pythagoras theorem, we get

$OA^2 = PA^2 + OP^2$

on substituting the value of OA = radius = 5 cm and OP = 3 cm  

$\implies 5^2 = PA^2 + 3^2$

$\implies 25 = PA^2 + 9$

$\implies PA = \sqrt{25 - 9}$

 $\implies \bold{PA = 4\: cm}$ ...... (iii)

On substituting from (iii) in (i), we get  

$4 = \frac{1}{2} AB$

$\implies AB = 4 \times 2$

$\implies \bold{AB = 8 \:cm}$

 

Thus,  the length of AB is → 8 cm.

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Also View:

A circle is radius 5cm find the length of the chord of distance 3cm from its centre

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