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.

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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