Advertisements
Advertisements
Question
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
Solution
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.
RELATED QUESTIONS
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
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.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
