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The given figure shows a circle with centre O and radius $4 \text{ cm}$ circumscribed by $\triangle ABC$. BC touches the circle at D such that $BD = 6 \text{ cm}, DC = 10 \text{ cm}$. Find the length of AE.
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Let $AE = x \Rightarrow AF = x$ and $CE = 10 \text{ cm}, BF = 6 \text{ cm}$ (1 mark)
$s = \frac{16 + 10 + x + 6 + x}{2} = 16 + x$ ($\frac{1}{2}$ mark)
$\therefore \text{Area of } \triangle ABC = \sqrt{(16 + x)(x)(6)(10)}$ (1 mark)
Also, area of $\triangle ABC = \frac{1}{2} [16 \times 4 + (10 + x)4 + (6 + x)4]$ (1 mark)
Equating (i) & (ii), we get $x = \frac{64}{11}$ or $5.8$ ($1\frac{1}{2}$ marks)
Length of $AE = \frac{64}{11} \text{ cm or } 5.8 \text{ cm}$
$s = \frac{16 + 10 + x + 6 + x}{2} = 16 + x$ ($\frac{1}{2}$ mark)
$\therefore \text{Area of } \triangle ABC = \sqrt{(16 + x)(x)(6)(10)}$ (1 mark)
Also, area of $\triangle ABC = \frac{1}{2} [16 \times 4 + (10 + x)4 + (6 + x)4]$ (1 mark)
Equating (i) & (ii), we get $x = \frac{64}{11}$ or $5.8$ ($1\frac{1}{2}$ marks)
Length of $AE = \frac{64}{11} \text{ cm or } 5.8 \text{ cm}$