The distance between two parallel chords of length 6 cm each in circle of diameter 10 cm is

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According to question,Given :

AB = CD = 8 cmr = 5 cm\( \Large \therefore In \triangle OMB \)\( \Large OB^{2}=OM^{2}+MB^{2} \)\( \Large r^{2}OM^{2}+(4)^{2} \)\( \Large (5)^{2}=OM^{2}+16 \)\( \Large 25-16=OM^{2} \)\( \Large OM^{2}=25-16 \)\( \Large OM^{2}=9 \)\( \Large OM=3 \)\( \Large \therefore MN = 2 \times OM \)

\( \Large MN = 2 \times 3=6cm \)

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Home » Aptitude » Plane Geometry » Question

The distance between two parallel chords of length 8 cm each in a circle of diameter 10 cm is

As per the given in question , we draw a figure circle

Given , Diameter = 10 cm AB = CD OP = OQ From ∆ OAP,

OP = √OA² - AP²

OP = √5² - 4²
OP = √25 - 16
OP = √9 = 3 cm.
∴ QP = 2 × OP = 6 cm.

Answer (Detailed Solution Below)

Option 1 : 6 cm

Free

100 Qs. 100 Marks 120 Mins

Given:

The length of the parallel chords is 8 cm and the diameter of the circle is 10 cm

Concept:

A line passing through the center perpendicularly bisect the chord into two equal parts.

Equal Chords are equal distance from the center.

Calculation:

Construct a circle with center 'O' and draw parallel chord AB and CD, EF is the perpendicular bisector of the chords AB and CD.

AB = CD = 8 cm, R = 10/2 = 5cm

We have to find the length of EF.

In ΔBEO, ∠E is 90°, OB = 5 cm, BE = 8/2 = 4 cm,

By Pythagoras theorem

EO = (OB2 - BE2)1/2

⇒ (52 - 42)1/2

⇒ (25 - 16)1/2

⇒ (9)1/2

⇒ 3 cm

Now, The length of the EF is

⇒ 2 × OE

⇒ 2 × 3

⇒ 6 cm

∴ The required distance between the chords is 6 cm.

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