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Q.

In a circle of radius 17 cm, two parallel chords of the length 30 cm and 16 cm respectively are drawn on the opposite sides of the centre. Then, which of the following is the distance between them?

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a

7cm

b

12cm

c

23cm

d

32cm

answer is C.

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Detailed Solution

Consider the given question,
In a circle of radius 17 cm two parallel chords of the length 30 cm and 16 cm respectively are drawn on the opposite sides of the centre.
Construct a diagram according to the given data.
Join OP and OR.
Draw a perpendicular ON from O to RS at N.
Since

PQ ∥ RS,

therefore OM will be perpendicular to PQ at M.
Question Image

Question Image

From figure,

PQ = 30 cm

RS = 16 cm

OR = OP = 17 cm

PQ ∥ RS .

Here,

∠ ONR = 90^{o} = ∠ OMP

(corresponding angles)

PM = \frac{1}{2} PQ

[Perpendicular from the centre bisects the chord]

\Rightarrow PM = \frac{1}{2} \times 30 cm

\Rightarrow PM = 15 cm

We know

ΔOPM

is a right angled triangle.
Pythagoras theorem:

{hypotenuse}^{2} = {opposite}^{2} + {adjacent}^{2}

By Pythagoras theorem, we have,

{OP}^{2} = {OM}^{2} + {PM}^{2}

\Rightarrow OM = \sqrt{{OP}^{2} - {PM}^{2}}

\Rightarrow OM = \sqrt{17^{2} - 15^{2}} cm

\Rightarrow OM = \sqrt{64} cm

\Rightarrow OM = 8 cm

Again,

RN = \frac{1}{2} RS

[Perpendicular from the centre bisects the chord]

\Rightarrow RN = \frac{1}{2} \times 16 cm

\Rightarrow RN = 8 cm

Here,

ΔORN

is a right triangle.
By Pythagoras theorem, we have,

{OR}^{2} = {ON}^{2} + {RN}^{2}

\Rightarrow {ON}^{2} = {OR}^{2} - {RN}^{2}

\Rightarrow ON = \sqrt{{OR}^{2} - {RN}^{2}}

\Rightarrow ON = \sqrt{17^{2} - 8^{2}} cm

\Rightarrow ON = \sqrt{225} cm

\Rightarrow ON = 15 cm

Now, the distance between PQ and RS is MN.

MN = ON + OM

\Rightarrow MN = (15 + 8) cm

\Rightarrow MN = 23 cm

Hence, the distance between PQ and RS is 23 cm.
Therefore, option 3 is correct.

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