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Question

In the given figure, $$BDC$$ 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 dawn 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.

971335_02032a88518b4460992592b291f08d8b.png


Solution

(i)

AB, BC and AC are tangents to the circle at E, D and F.

$$BD = 30 cm,DC = 7 cm,\angle BAC = 90$$

From the theorem stated,

$$BE = BD = 30 cm$$

Also $$FC = DC = 7 cm$$

Let $$AE = AF = x$$ …. (1)

Then $$AB = BE + AE = (30 + x)$$

$$AC = AF + FC = (7 + x)$$

$$BC = BD + DC = 30 + 7 = 37 cm$$

Consider right trianlge ABC, 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 x cannot be negative]

$$\therefore  AF = 5 cm $$[From (1)]

Therefore$$ AB =30 +x = 30 + 5 = 35 cm$$


(ii)

$$AC = 7 + x = 7 + 5 = 12 cm$$

Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.

Join point O, F; points O, D and points O, E.

From the figure,

$$\dfrac{1}{2} \times  AC \times  AB = \dfrac{1}{2} \times  AB \times  OE+ \dfrac{1}{2} \times  BC \times  OD + \dfrac{1}{2} \times  AC \times  OC$$

$$ AC \times  AB = AB \times  OE + BC \times  OD + AC \times  OC$$

$$12 \times  35 = 35 \times  r + 37 \times  r + 12 \times  r$$

$$ 420 = 84 r$$

$$\therefore r = 5$$

Thus the radius of the circle is 5 cm.

Mathematics

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