-
N is the foot of the perpendicular from a point P of a circle with radius 7 cm, on a diameter AB of the circle. If the length of the chord PB is 12 cm, the distance of the point N from the point B is
-
-
6 5 cm. 7 -
12 2 cm. 7 -
3 5 cm. 7 -
10 2 cm. 7
-
Correct Option: D
As per the given in question , we draw a figure a circle,
AB = 14 cm, PB = 12 cm
∠APB = 90° (angle in the semi circle)
∴ AP = √14² - 12²
AP = √(14 + 12)(14 - 12)
AP = √26 × 2 = √52
Let AN = y ⇒ NB = 14 – y
In ∆ APN, PN² = AP² – AN² = 52 – y² ...(i)
In ∆ PNB, PN² = PB² – NB² = 12² – (14 – y)² ...(ii)
From (i) and (ii)
52 – y² = 144 – (14 – y)²
52 – y² = 144 – (196 + y² – 28 y)
⇒ 52 – y² = 144 –196 – y² + 28 y
52 = – 52 + 28 y
| ⇒ 28 y = 104 ⇒ y = | = | ||
| 28 | 7 |
NB = 14 – y
| NB = 14 – | = | = 10 | cm. | |||
| 7 | 7 | 7 |
