Question

Help me with this question
Explain it nicely.

Find for
$x$

Answer 1

Solution :

_____________________________________________________________


Given :

In the given that,

The radius of circle is 13mm,

The lengths of the two chords,

RS = 24mm,

PQ = 10mm,.

_____________________________________________________________

To Find :

The value of x,

which is the perpendicular distance between the two parallel chords,

_____________________________________________________________

As we know that,

The mid-point of the chord is perpendicular (makes 90°) to the center of the circle (O),.

So, let M be the mid-point of the chord RS,. (RM = MS = $\frac{24}{2}$ = 12mm,)

So, we can say that, ∠OMR = ∠OMS = 90°,

In Δ OMR,

⇒ OM² + MR² = OR² (By pythagoras theorem),

⇒ OM² + 12² = 13² (Radius of circle = 13mm )

⇒ OM² = 13² - 12²

⇒ OM² = 169 - 144

⇒ OM² = 25

⇒ OM = 5mm,.

________________________

Let N be the mid point of chord PQ,

Hence,

⇒ NP = NQ = $\frac{PQ}{2} = \frac{10}{2}$ = 5mm,

∠ ONP  = ∠ ONQ = 90°

Hence,

In Δ ONP

⇒ ON² + NP² = OP² (By pythagoras theorem),

⇒ ON² + 5² = 13² (Radius o circle = 13mm,)

⇒ ON² = 13² - 5²

⇒ ON² = 169 - 25

⇒ ON² = 144

⇒ ON = 12mm,

__________________

The distance between the chords = ON - OM = 12 - 5 = 7mm,.

_____________________________________________________________

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Answer 2

Answer:

here is ur ans.......

Given :

In the given that,

The radius of circle is 13mm,

The lengths of the two chords,

RS = 24mm,

PQ = 10mm,.

_____________________________________________________________

To Find :

The value of x,

which is the perpendicular distance between the two parallel chords,

_____________________________________________________________

As we know that,

The mid-point of the chord is perpendicular (makes 90°) to the center of the circle (O),.

So, let M be the mid-point of the chord RS,. (RM = MS = 242\frac{24}{2}

2

24

= 12mm,)

So, we can say that, ∠OMR = ∠OMS = 90°,

In Δ OMR,

⇒ OM² + MR² = OR² (By pythagoras theorem),

⇒ OM² + 12² = 13² (Radius of circle = 13mm )

⇒ OM² = 13² - 12²

⇒ OM² = 169 - 144

⇒ OM² = 25

⇒ OM = 5mm,.

________________________

Let N be the mid point of chord PQ,

Hence,

⇒ NP = NQ = PQ2=102\frac{PQ}{2} = \frac{10}{2}

2

PQ

=

2

10

= 5mm,

∠ ONP = ∠ ONQ = 90°

Hence,

In Δ ONP

⇒ ON² + NP² = OP² (By pythagoras theorem),

⇒ ON² + 5² = 13² (Radius o circle = 13mm,)

⇒ ON² = 13² - 5²

⇒ ON² = 169 - 25

⇒ ON² = 144

⇒ ON = 12mm,

__________________

The distance between the chords = ON - OM = 12 - 5 = 7mm,..........