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In the given diagram, an incircle DEF is circumscribed by the right angled triangle in which AF = 6 cm and EC = 15 cm. Then find the difference between CD and BD.
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- 1 cm.
- 3 cm.
- 4 cm.
- 5 cm.
- 1 cm.
Correct Option: A
As per the given figure in question , 
OE = OF (In–radius)
BF = BD (tangents from external point B)
and EC = DC = 15 cm.
Let BF = y cm. AB = (6 + y) cm.
AC = 6 + 15 = 21 cm.
From ∆ABC ,
BC² = AB² + AC²
⇒ (y + 15)² = (6 + y)² + (21)²
(∵ BC = BD + CD)
⇒ y² + 30y + 225 = 36 + y² + 12y + 441
⇒ 30y – 12y = 441 + 36 – 225
⇒ 18y = 252
| ⇒ y = | = 14 ⇒ BD = 14 cm. | |
| 18 |
∴ Required difference = CD – BD = 15 – 14 = 1 cm.
