Question

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

Answer 1

Answer:

Solution:

Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.

As we know, AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6− x.

Consider ΔMOB,

OB2 = OM2 + MB2

Or,

OB2 = 36 + x2 – 12x + 25/4 ……(1)

Consider ΔNOD,

OD2 = ON2 + ND2

Or,

OD2 = x2+121/4 ……….(2)

We know, OB = OD (radii)

From eq. (1) and eq. (2) we have;

36 + x2 -12x + 25/4 = x2 + 121/4

12x = 36 + 25/4 – 121/4

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

12x = 48/4 = 12

x = 1

Now, from eq. (2) we have,

OD2 = 11 + (121/4)

Or OD = (5/2) × √5

Hope it will be helpful :)

Answer 2

Answer:

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