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Practice each question, then open the answer or worked solution.
Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then AB = CD.
True
Justification:
Given that AB and AC are chords that are at a distance of 4 cm from the centre of a circle.
Since, chords that are equidistant from the centre of a circle are equal in length,
We have, AB = CD.
Two chords AB and AC, of a circle with centre O are on the opposite sides of OA. Then ∠OAB = ∠OAC.
False
Justification:
Let AB and AC be the chord of the circle with centre O on the opposite side of OA.
Consider the triangles AOC and AOB:
AO = AO (Common side in both triangles)
OB = OC (Both OB and OC are radius of the circle)
But we can’t show that either the third side of both triangles is equal or any angle is equal. Therefore ΔAOB is not congruent to ΔAOC.
∴ ∠OAB ≠ ∠OAC.
Two congruent circles with centres O and O’ intersect at two points A and B. Then ∠AOB = ∠AO’B.
True
Justification:
Equal chords of congruent circles subtend equal angles at the respective centre.
Hence, the given statement is true.
Through three collinear points a circle can be drawn.
False
Justification:
A circle through two points cannot pass through a point which is collinear to these two points.
Hence, the given statement is false.
A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
True
Justification:
According to the question,
Radius of circle = 3 cm
Diameter of circle = 2 × r
= 2 × 3 cm
= 6 cm
Now,
From the question we have,
AB = 6 cm
So, the given statement is true because AB will be the diameter