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A circle is the collection of all points, which are equidistant from a fixed point in the plane.
Any line segment starting at the centre of the circle and joining anywhere on the border on the circle is a Radius of Circle.
Any line segment that’s both endpoints (starting and ending) anywhere on the border on the circle is known as Chord .
The longest chord of a circle and which passes through the centre of the Circle is a Diameter of the circle.
An arc is a smooth curve joining two points anywhere on the border on the circle . A continuous piece of a circle is Arc of the circle .
It is the length of the circle if we open and straightened out to make a line segment.
The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. It could be a major or minor segment.
The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector. It could be a major or minor sector.
Equal chords of a circle subtend equal angles at the centre.
If the angles subtend by two chords of a circle at the centre ( corresponding centers ) are equal, then the chords must be equal.
Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.
Solution :
Given : Two circles having centres O and O’ and having equal radius.
Let AB and CD be the equal chords of 2 congruent circles.
AB = CD
We want to prove that the angles subtended by the chords at the centers are equal, i.e.,
$\angle $ AOB = $\angle $ CO’D
In Δ AOB and Δ CO'D
We know that the radius of both the circles are equal. So ,
OA = O’C
OB = O’D
( Equal radius of the congruent circles )
AB = CD
( Given equal chords )
Since all three corresponding sides of Δ AOB and Δ CO'D are equal, the two triangles are congruent by the SSS congruence criterion.
Δ AOB ≅ Δ CO'D
By the property of Corresponding Parts of Congruent Triangles (CPCT), the corresponding angles of the congruent triangles must be equal. Therefore,
∴ $\angle $ AOB = $\angle $ CO’D
Hence Proved.
Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.
Solution :
Given that two circles are congruent with centers O and O' and having equal radius..
Let chords of congruent circles subtend equal angles at their centres.
$\angle $ AOB = $\angle $ CO’D
To Prove: AB and CD be the equal chords of 2 congruent circles.
AB = CD
In Δ AOB and Δ CO'D
We know that the radius of both the circles are equal. So ,
OA = O’C
( Equal radius of the congruent circles )
$\angle $ AOB = $\angle $ CO’D
( Chords subtend equal angles at centre )
OB = O’D
( Equal radius of the congruent circles )
Since two corresponding sides and the included angle of Δ AOB and Δ CO'D are equal, the two triangles are congruent by the SAS congruence criterion.
Δ AOB ≅ Δ CO'D
By the property of Corresponding Parts of Congruent Triangles (CPCT), the corresponding angles of the congruent triangles must be equal. Therefore,
∴ AB = CD
Hence Proved.
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