CD is a tangent to a circle of circumference of 88 cm with…

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?

Show answer & explanation

Correct answer: B

Explanation

The correct answer is option 2 i.e. 14.5.

Figure for the explanation of question 52

The circumference of Circle = 2πr = 88 cm

r = $\frac{88×7}{2×22}$ = 14 cm

Hence OP = OV = 14 cm

⇒ OC = PC + OP = 6 + 14 = 20 cm

⇒ OD = OC + 1 (Given)

⇒ OD = 20 + 1 = 21 cm

According to the Pythagoras theorem;

CD $^2$ = OC $^2$ + OD $^2$

⇒ CD $^2$ = 20 $^2$ + 21 $^2$

⇒ CD $^2$ = 400 + 441

⇒ CD = $\sqrt{841}$

⇒ CD = 29 cm

Half of the CD = 29/2 = 14.5 cm

SSC CHSL 2024 PYQ — Question 52