Question

Two chords AB and CD of length 5cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of it centre.If the distance between AB and CD is 6cm,find the radius of the circle.​

Answer 1

Figure:-

Given:-

• Two chords AB and CD of length 5cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of it centre.
• the distance between AB and CD is 6cm.

To find:-

• find the radius of the circle.

Solutions:-

• OM ┻ AB
• ON ┻ CD
• OB and OD are join.

BM = AB/2 = 5/2 (perpendicular from the centre bisects the chord)

ND = CD/2 = 11/2

Let ON be x.

Therefore,

OM be 6 - x.

In ∆MOB,

OM² + MB² = OB²

(6 - x)² + (5/2)² = OB²

36 + x² - 12x + 25/4 = OB² ............(i).

In ∆NOD,

ON² + ND² = OD²

x² + (11/2)² = OD² ............(ii).

OB = OD (Radii of the same circle)

Therefore,

from Eq (i). and (ii).

36 + x² - 12x + 25/4 = x² + 121/4

12x = 36 + 25/4 - 121/4

12x = (144 + 25 - 121)/4

12x = 48/4

12x = 12

x = 1

from Eq (ii).

(1)² + (121/4) = OD²

OD² = 1 + 121/4

OD² = (121 + 4)/4

OD² = 125/4

OD = √125/4

OD = 5/2

Hence, the radius of the circle is 5/2 cm.