Question

8. In a circle with centre P, radius is 13 cm.
The distance between two parallel chords AB and CD is 17cm. If AB=24cm, then find CD.​

Answer 1

Answer:

Here, O is the centre of the given circle of radius 13 cm. AB and CD are two parallel chords, such that AB = 24 cm and CD = 10 cm.

Join OA and OC.  

Since ON⊥AB and 0M ⊥CD.  

∴ M, O and N are collinear and M, N are mid-points of CD and AB.  

Now, in rt. ∠ed ∆ANO, we have

$ON^2$ = 02 $AN^{2}$  

= $13^{2}$ – $12^{2}$  = 169 – 144 = 25

ON = √25 = 5cm  

Similarly, in it. ∠ed ∆CMO, we have  

$OM^{2}$ = $OC^{2}$ – $CM^{2}$  

= $13^{2}$ – $5^{2}$

= 169 – 25

= 144  

OM = √144 = 12cm  

Hence, distance between the two chords  NM = NO + OM  = 5 + 12  = 17 cm

Answer 2

Step-by-step explanation:

Q :-

8. In a circle with centre P, radius is 13 cm.

The distance between two parallel chords AB and CD is 17cm. If AB=24cm, then find CD.

A :-

Let the diameter be AOB = 26 cm(r = 13 cm), 2 parallel chords be CD = 10 cm and EF = 24 cm.

Let the diameter be AOB = 26 cm(r = 13 cm), 2 parallel chords be CD = 10 cm and EF = 24 cm.From centre O to the chords draw a perpendicular and a radius, the chord will be half the length ie CD/2 = 5 cm and EF/2 = 12 cm.

Let the diameter be AOB = 26 cm(r = 13 cm), 2 parallel chords be CD = 10 cm and EF = 24 cm.From centre O to the chords draw a perpendicular and a radius, the chord will be half the length ie CD/2 = 5 cm and EF/2 = 12 cm.Using Pythagoras’ Theorem, OC² = (CD/2)² + perpendicular² and OF² = (EF/2)² + perpendicular² => 13² = 5² + perpendicular² and 13² = 12² + perpendicular²

13² = 5² + perpendicular² => perpendicular = sqrt(169 - 25) = 12 cm.

=> perpendicular = sqrt(169 - 25) = 12 cm.13^2 = 12^2 + perpendicular^2 => perpendicular = sqrt(169 - 144) = 5 cm.

=> perpendicular = sqrt(169 - 25) = 12 cm.13^2 = 12^2 + perpendicular^2 => perpendicular = sqrt(169 - 144) = 5 cm.Distance between chords = 12 + 5 = 17 cm.