Given:
The radius of a circle = 50 cm
Distance of chord AB from the center of the circle = 14 cm
Formula used:
In a circle, the perpendicular bisector of a chord passes through the center of the circle.
If a perpendicular is drawn from the center of a circle to a chord, it bisects the chord.
In a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
Calculation:
Let's solve the question:
First, let's draw a rough diagram of the circle and chord AB:
Let O be the center of the circle, and let M be the midpoint of the chord AB. Draw OM, which is the perpendicular bisector of AB.
By the property of perpendicular bisectors, OM passes through the center O of the circle.

Since OM is perpendicular to AB, we can use the Pythagorean theorem to find the length of AM (or BM), which is half of the chord AB:
AM2+OM2=OA2 AM2+142=502 AM2=502−142 AM=BM=√(502−142) AM=BM=√(2500−196) AM=BM=√2304 AM=BM=48
Therefore, the length of the chord AB is:
AB = 2 × AM
= 2 × 48
= 96 cm
The diameter of the circle is twice the radius, so it is:
diameter = 2 × radius
= 2 × 50
= 100 cm
Therefore, the sum of the chord AB and the diameter of the circle is:
AB + diameter = 96 + 100
= 196 cm
Hence, the sum of chord AB and the diameter of this circle is 196 cm.