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The give figure shows a circle with centre O and radius 4 cm circumscribed by $\triangle ABC$. BC touches the circle at D such that BD = 6 cm, DC= 10 cm. Find the length of AE.
The give figure shows a circle with centre O and radius 4 cm circumscribed by $\triangle ABC$. BC touches the circle at D such that BD = 6 cm, DC= 10 cm. Find the length of AE.
This Question has 1 answers.
$ \text{Given:} $
A circle with center O and radius 4 cm is inscribed in $ \triangle ABC. $
BC is tangent to the circle at D, $\text{ with } BD = 6 \text{ cm and } DC = 10 \text{ cm}. $
Let the tangents drawn from A, B, and C to the incircle be AE, AF, BD, BE, CF, and CD.
Since tangents drawn from an external point to a circle are equal:
$ \text{Let } AE = AF = x, $ $\quad BD = BE = 6 \text{ cm}, $ $\quad DC = CF = 10 \text{ cm}. $
$ \text{The perimeter of } \triangle ABC \text{ is:} $
AB + BC + CA = (AE + EB) + (BD + DC) + (CF + FA)
$ = (x + 6) + (6 + 10) + (10 + x) $
$ = x + 6 + 16 + 10 + x $
$ = 2x + 32. $
Using the formula for the semi-perimeter:
$ s = \frac{\text{Perimeter}}{2} = \frac{2x + 32}{2} = x + 16. $
$ \text{Since } AE = s - (BD + DC), $
$ AE = (x + 16) - (6 + 10)$
A circle with center O and radius 4 cm is inscribed in $ \triangle ABC. $
BC is tangent to the circle at D, $\text{ with } BD = 6 \text{ cm and } DC = 10 \text{ cm}. $
Let the tangents drawn from A, B, and C to the incircle be AE, AF, BD, BE, CF, and CD.
Since tangents drawn from an external point to a circle are equal:
$ \text{Let } AE = AF = x, $ $\quad BD = BE = 6 \text{ cm}, $ $\quad DC = CF = 10 \text{ cm}. $
$ \text{The perimeter of } \triangle ABC \text{ is:} $
AB + BC + CA = (AE + EB) + (BD + DC) + (CF + FA)
$ = (x + 6) + (6 + 10) + (10 + x) $
$ = x + 6 + 16 + 10 + x $
$ = 2x + 32. $
Using the formula for the semi-perimeter:
$ s = \frac{\text{Perimeter}}{2} = \frac{2x + 32}{2} = x + 16. $
$ \text{Since } AE = s - (BD + DC), $
$ AE = (x + 16) - (6 + 10)$
$ = x + 16 - 16 = x. $
$ \text{Thus, } AE = 8 \text{ cm}. $
$ \text{Thus, } AE = 8 \text{ cm}. $
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