Advertisements
Advertisements
प्रश्न
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Advertisements
उत्तर

Let AB and CD be two parallel chords in a circle centered at O. Join OB and OD.
Distance of smaller chord AB from the centre of the circle = 4 cm
OM = 4 cm
MB = AB/2 = 6/2 = 3cm
In ΔOMB,
OM2 + MB2 = OB2
(4)2 + (3)2 = OB2
16 + 9 = OB2
OB2 = 25
`OB = sqrt25`
OB = 5cm
In ΔOND,
OD = OB = 5cm (Radii of the same circle)
ND = CD/2 = 8/2 = 4cm
ON2 + ND2 = OD2
ON2 + (4)2 = (5)2
ON2 = 25 - 16 = 9
ON = 3
Therefore, the distance of the bigger chord from the centre is 3 cm.
संबंधित प्रश्न
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
