Question : A chord of length 40 cm is drawn in a circle having a diameter of 50 cm. What is the
a minimum distance of another parallel chord of length 30 cm in the same circle from a 40 cm long chord?
Option 1: 10 cm
Option 2: 15 cm
Option 3: 5 cm
Option 4: 20 cm
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Correct Answer: 5 cm
Solution :
Let C be the centre of the circle and AB and CD be the shorter and longer chords respectively. Since the minimum distance is required, the chords must be on the same side of the centre of the circle.
AB = 30 cm and CD = 40 cm
OF is the perpendicular distance between O and AB.
Since AB and CD are parallel, OF is also perpendicular to CD at E.
Radius of the circle = $\frac{50}{2}$ = 25 cm
OC = OD = OA = OB = 25 cm = Radii
According to the concept,
AF = FB = $\frac{30}{2}$ = 15 cm
CE = ED = $\frac{40}{2}$ = 20 cm
$\triangle$OEC is a right-angle triangle at E.
So, OE = $\sqrt{25^2−20^2}$ = 15 cm
Likewise, OF = $\sqrt{25^2−15^2}$ = 20 cm
Distance between AB and CD, EF = 20 – 15 = 5 cm
∴ The minimum distance is 5 cm.
Hence, the correct answer is 5 cm.
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