Short Answer TypeIn the figure below, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to radius of the circumcircle where centre is O.
Long Answer TypeTwo chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Let the radius of the circle be r cm. Let OM = x cm.
Then ON = (6 - x) cm.
∵ OM ⊥ CD
∴ M is the mid-point of CD.
| The perpendicular from the centre of a circle to a chord bisects the chord
∵ ON ⊥ AB
∴ N is the mid-point of AB
| The perpendicular from the centre of a circle to a chord bisects the chord
In right triangle ONB,
OB2 = ON2 + NB2
| By Pythagoras Theorem
In right triangle OMD,
OD2 = OM2 + MD2
| By Pythagoras Theorem
From (1) and (2), we get
Putting x=1 in (2), we get
Hence, the radius of the circle is 
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?
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.
Short Answer TypeLong Answer TypeAC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
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