Question

Pq and rs are two parallel chord of a circle whose centre is o nd radius is 10dm. If pq = 16cm abd rs = 12 cm , find the distnce between pq and rs, they lie

Answer 1

Distance between two parallel chords

Answer:

Distance between PQ and RS when radius of circle is 10dm that is 100 cm is 199.5cm.

Distance between PQ and RS when radius of circle is taken as  10cm is 14cm.

Explanation:

Consider attached figure

Given that PQ = 16 cm , RS = 12 cm also PQ II RS.

O is center of circle having radius 10dm = 10 × 10 cm = 100 cm

⇒ OP = OQ = OR = RS = 100 cm ( radius of circle )

Now construct a Perpendicular from O on chord A .

⇒ PA = AQ = PQ/2 = 16/ 2 = 8 cm   [ Since perpendicular from center on chord bisects the chord ]

Now extend Line AO meeting SR at B.

As PQ is parallel to SR , so line BA which is perpendicular to PQ will also be perpendicular to SR   [ As line perpendicular to one of the two parallel lines is perpendicular to other line also ]

so OB that OA extended is perpendicular from center O On RS which means OB bisects chord RS and SB = BR = Rs/2 = 12/2 = 6 cm.

We need to determine length of AB which is distance between chord PQ and RS.

AB = OA + OB

CALCULATION FOR RADIUS AS 10DM that is 100 CM

lets first calculate OA

In right angled triangle AOP

PA² + AO² = OP²    [ Pythagoras theorem ]

⇒ 8² + AO² = 100²

⇒AO² = 100² - 8²

⇒AO² = 10000-64 = 9936 cm

⇒AO = √9936 = 99.679 cm

Now calculating OB

In right angled triangle SOB

OB² + BS² = SO²    [ Pythagoras theorem ]

⇒OB² + 6² = 100²

⇒OB² =100² - 6²

⇒OB² =10000 - 36

⇒OB² = 9964

⇒OB = √9964 = 99.819

so AB = OA + OB =  99.679cm + 99.819 cm = 199.49 ≈ 199.5 cm

So according to question our answer is distance between PQ and RS when radius of circle is 10dm that is 100 cm is 199.5cm.

But some how answer is not so satisfactory . so considering one more scenarion in which radius is not 10 dm but it is 10 cm

CALCULATION FOR RADIUS AS 10CM

lets first calculate OA

In right angled triangle AOP

PA² + AO² = OP²    [ Pythagoras theorem ]

⇒ 8² + AO² = 10²

⇒AO² = 10² - 8²

⇒AO² = 100-64 = 36 cm

⇒AO = √36 = 6 cm

Now calculating OB

In right angled triangle SOB

OB² + BS² = SO²    [ Pythagoras theorem ]

⇒OB² + 6² = 10²

⇒OB² =10² - 6²

⇒OB² =100- 36

⇒OB² =64

⇒OB = √64 = 8 cm

so AB = OA + OB = 6cm + 8 cm = 14 cm

So according to question our answer is distance between PQ and RS when radius of circle is taken as  10cm  is 14cm.

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#BAL.