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In the figure, \( B D C \) is a tangent to the given circle at point \( D \) such that \( B D=30 \mathrm{~cm} \) and \( C D=7 \mathrm{~cm} \). The other tangents \( B E \) and \( C F \) are drawn respectively from \( B \) and \( C \) to the circle and meet when produced at \( A \) making \( B A C \) a right angle triangle. Calculate \( A F \).
Given:
In the figure, \( B D C \) is a tangent to the given circle at point \( D \) such that \( B D=30 \mathrm{~cm} \) and \( C D=7 \mathrm{~cm} \).
The other tangents \( B E \) and \( C F \) are drawn respectively from \( B \) and \( C \) to the circle and meet when produced at \( A \) making \( B A C \) a right angle triangle.
To do:
We have to calculate \( A F \).
Solution:
$AB, BC$ and $AC$ are tangents to the circle at $E, D$ and $F$.
$BD = 30\ cm$ and $DC = 7\ cm$ and $\angle BAC = 90^o$ Tangents drawn from an exterior point to a circle are equal in length.
This implies,
$BE = BD = 30\ cm $
$FC = DC = 7\ cm$
Let $AE = AF = x$
$AB = BE + AE$
$= (30 + x)$
$AC = AF + FC$
$= (7 + x)$
$BC = BD + DC$
$= 30 + 7$
$= 37\ cm$
In right angled triangle $ABC$,
By Pythagoras theorem,
$BC^2 = AB^2 + AC^2$
$(37)^2 = (30 + x)^2 + (7 + x)^2$
$1369 = 900 + 60x + x^2 + 49 + 14x + x^2$
$2x^2 + 74x + 949 - 1369 = 0$
$2x^2+ 74x - 420 = 0$
$x^2 + 37x - 210 = 0$
$x^2 + 42x - 5x - 210 = 0$
$x (x + 42) - 5 (x + 42) = 0$
$(x - 5) (x + 42) = 0$
$(x - 5) = 0$ or $(x + 42) = 0$
$x = 5$ or $x = - 42$
$x = 5$ (Since x cannot be negative)
This implies,
$AF = 5\ cm$.
The length of $AF$ is $5\ cm$.