In Fig. 10.61, BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the circle and meet when produced at A making BAC a right angle triangle. Calculate (i) AF (ii) radius of the circle.
Answers
(i) AF = 5 cm
(ii) radius of the circle r = 5 cm
From given diagram, we have,
BC = BD + DC = 30 + 7 =37 cm
BD = BE = 30 cm
CD = CF = 7 cm
AE = AF
From Pythagoras theorem, we have,
BC^2 = AB^2 + AC^2
(BD + DC)^2 = (AE + EB)^2 + (AF + FC)^2
(30 + 7)^2 = (AF + 30)^2 + (AF + 7)^2
(37)^2 = AF^2 + 30^2 + 60AF + AF^2 + 7^2 + 14AF
37^2 - 30^2 - 7^2 = 2AF^2 + 74AF
420 = 2AF^2 + 74AF
AF^2 + 37AF - 210 = 0
AF = 5 cm ..............(i)
Let "O" be the center of the circle.
Area of Δ ABC = Area of Δ BOA + Area of Δ BOC + Area of Δ AOC
1/2 × AC × AB = 1/2 × AB × OE + 1/2 × BC × OD + 1/2 × AC × OF
(AF + FC) × (AE + EB) = AB × OE + BC × OD + AC × OF
(5 + 7) × (5 + 30) = AB × r + BC × r + AC × r
(12) × (35) = AB × r + BC × r + AC × r
(12) × (35) = 35 × r + 37 × r + 12× r
420 = r (35 + 37 + 12)
420 = r (84)
r = 5 cm .............(ii)
ANSWER:
BD = 30 cm and DC = 7 cm and ∠BAC = 90°
- BE = BD = 30 cm
- FC = DC = 7 cm
Let AE = AF = x → ------ (1)
Then AB = BE + AE = (30 + x)
AC = AF + FC = (7 + x)
BC = BD + DC = 30 + 7 = 37 cm
by Pythagoras theorem we have,
BC^2 = AB^2 + AC^2
⇒ (37)^2 = (30 + x)^2 + (7 + x)^2
⇒ 1369 = 900 + 60x + x^2 + 49 + 14x + x^2
⇒ 2x^2 + 74x + 949 – 1369 = 0
⇒ 2x^2+ 74x – 420 = 0
⇒ x^2 + 37x – 210 = 0
⇒ x^2 + 42x – 5x – 210 = 0
⇒ x (x + 42) – 5 (x + 42) = 0
⇒ (x – 5) (x + 42) = 0
⇒ (x – 5) = 0 or (x + 42) = 0
⇒ x = 5 or x = – 42
⇒ x = 5 [Since length cannot be negative]
∴ AF = 5 cm
- r = AF (since AE= AF, tangents) (since in a square, all the sides are equal)
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