Question

two parallel chord 10cm and 24cm long are drawn on the same side of the centre of a circle of radius 13cm find the distance between the chords

Answer 1

Drop a perpendicular from the centre and mar the intersecting point E
We know that the perpendicular from the centre bisects the chord.

=> CE = 1/2 CD

=> CE = 1/2 × 24

=> CE = 12 cm

Now, ∆ODC is a right triangle

=> OD² + DC² = OC²

=> OD² + 12² = 13²

=> OD² = 13² - 12²

=> OD² = (13 - 12)(13 + 12)

=> OD² = 25

=> OD = √25

=> OD = 5 cm

Now again, extend the perpendicular such that it bisects AB at F

=> AF = 1/2 AB

=> AF = 5 cm

=> OF² + AF² = OA²

=> OF² + 5² = 13²

=> OF² = 13² - 5²

=> OF² = (13 - 5)(13 + 5)

=> OF² = (8)(18)

=> OF² = 144

=> OF = 12 cm

So distance between them

= OF - OE

= 12 - 5

= 7 cm

Answer 2

Answer:

7cm

Step-by-step explanation:

Drop a perpendicular from the centre and mar the intersecting point E

We know that the perpendicular from the centre bisects the chord.

=> CE = 1/2 CD

=> CE = 1/2 × 24

=> CE = 12 cm

Now, ∆ODC is a right triangle

=> OD² + DC² = OC²

=> OD² + 12² = 13²

=> OD² = 13² - 12²

=> OD² = (13 - 12)(13 + 12)

=> OD² = 25

=> OD = √25

=> OD = 5 cm

Now again, extend the perpendicular such that it bisects AB at F

=> AF = 1/2 AB

=> AF = 5 cm

=> OF² + AF² = OA²

=> OF² + 5² = 13²

=> OF² = 13² - 5²

=> OF² = (13 - 5)(13 + 5)

=> OF² = (8)(18)

=> OF² = 144

=> OF = 12 cm

So distance between them

= OF - OE

= 12 - 5

= 7 cm