Verified answer

Answer:

The perpendicular bisector of the common chord passes through the centres of both circles.As the circles intersect at two points, we can construct the above figure.

Consider AB as the common chord and O and O’ as the centers of the circles

O’A = 5 cm

OA = 3 cm

OO’ = 4 cm [Distance between centres is 4 cm]

As the radius of bigger circle is more than the distance between two centers, we know that the center of the smaller circle lies inside the bigger circle

The perpendicular bisector of AB is OO’

OA = OB = 3 cm

As O is the midpoint of AB

AB = 3 cm + 3 cm = 6 cm

Length of common chord is 6 cm

It is clear that common chord is the diameter of the smaller circle

Step-by-step explanation:

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Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Solution:

The perpendicular bisector of the common chord passes through the centers of both circles.

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Given that the circles intersect at two points, so we can draw the above figure. Let AB be the common chord. Let O and O’ be the centers of the circles, respectively.

O’A = 5 cm, OA = 3 cm

OO’ = 4 cm [Given distance between the centres is 4cm]

Since the radius of the bigger circle is more than the distance between the 2 centers, we can say that the center of the smaller circle lies inside, the bigger circle itself.

OO’ is the perpendicular bisector of AB.

So, OA = OB = 3 cm

AB = 3 cm + 3 cm = 6 cm [Since, O is the mid point of AB]

The length of the common chord is 6 cm.

It is also evident that the common chord is the diameter of the smaller circle.

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