CD is a tangent to a circle of circumference of 88 cm with its centre at O. OC cuts the circle at P while OD cuts circle at V such that ∠COD = 90°. Length of PC is 6 cm. What is half of CD (in cm) if OD exceeds OC by 1 cm?

Correct option is B

Given:
Circumference of the circle,  $$2\pi r=88\ cm.$$​​
Tangent of the circle CD
​$$\angle COD={90}^0.$$​​
PC = 6cm
OD = OC + 1cm
Concept and Theorem Used:
Tangent makes right Angle triangle with center of the circle.
Pythagoras theorem; $$H^2 = P^2 + B^2$$​​
Solution:

Circumference of the circle = 88 cm

​$$2\pi r=88\\2\times\frac{22}{7}\times r=88\\r=\frac{88\times7}{22\times2}\\r=14\ cm\\$$​​
Now,
PO = OV = r (radius of the circle)
PO = OV = 14 cm
OC = PO + PC
OC = 14 + 6
OC = 20 cm
if OD exceeds OC by 1 cm (Given)
OD = OC + 1
OD = 20 +1
OD = 21 cm
In ∆COD ,
​$$\angle COD={90}^0$$​​
Therefore, ∆COD is a Right-angle triangle.
By Pythagoras theorem;
​$$CD^2 = OD^2 + OC^2$$​​
CD² = 212 + 202
CD =$$\sqrt{441+400}$$​​
CD = $$\sqrt{841}$$​​
CD = 29 cm
Therefore, half of CD = $$\frac{1}{2}\times29=14.5\ cm$$​​