Advertisements
Advertisements
प्रश्न
Two chords AB and CD of a circle are parallel and a line L is the perpendicular bisector of AB. Show that L bisects CD.
उत्तर
We know that the perpendicular bisector of any chord of a circle always passes through the centre of the circle.
Since, L is the perpendicular bisector of AB.
Therefore, L passes through the centre of the circle.
But L ⊥ AB and AB || CD
⇒ L ⊥ CD
Thus, L ⊥ CD and passes through the centre of the circle.
So, L is perpendicular bisector of CD.
APPEARS IN
संबंधित प्रश्न
A chord of length 24 cm is at a distance of 5 cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance of 12 cm from the centre.
A chord CD of a circle whose centre is O, is bisected at P by a diameter AB.
Given OA = OB = 15 cm and OP = 9 cm. calculate the length of:
(i) CD (ii) AD (iii) CB
Two circle with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
The given figure shows two circles with centres A and B; and radii 5 cm and 3 cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
In following figure , AB , a chord of the circle is of length 18 cm. It is perpendicularly bisected at M by PQ.
From a point P outside a circle, with centre O. tangents PA and PB are drawn as following fig., Prove that ∠ AOP = ∠ BOP and OP is the perpendicular bisector of AB.
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
OP is the ⊥ bisector of chord AB.
A chord of length 6 cm is drawn in a circle of radius 5 cm.
Calculate its distance from the center of the circle.
M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O.
Prove that: (i) ∠BMN = ∠DNM
(ii) ∠AMN = ∠CNM
Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 13 cm.