159
In the given figure, $\Delta ABC$ circumscribes a circle of radius of $4 \text{ cm}$. If $AD = 7 \text{ cm}$, $BD = 8 \text{ cm}$, and area $(\Delta ABC) = 84 \text{ cm}^2$, find the lengths of BC and AC.
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Solution:
As length of tangents from an external point to a circle are equal so,
$AE = AD = 7 \text{ cm}$ (1/2 Mark)
$BF = BD = 8 \text{ cm}$ (1/2 Mark)
Let $CF = CE = x \text{ cm}$
Then, $\text{ar}(\Delta BOC) + \text{ar}(\Delta AOB) + \text{ar}(\Delta AOC) = \text{ar}(\Delta ABC)$
$\frac{1}{2} (8+x)4 + \frac{1}{2} (15)4 + \frac{1}{2} (7 + x)4 = 84$ (2 Marks)
$\implies 60 + 4x = 84$ (1/2 Mark)
$4x = 24$
$x = 6$ (1/2 Mark)
$:: BC = 8 + 6 = 14 \text{ cm}$ (1/2 Mark)
and $AC = 7+ 6 = 13 \text{ cm}$ (1/2 Mark)
As length of tangents from an external point to a circle are equal so,
$AE = AD = 7 \text{ cm}$ (1/2 Mark)
$BF = BD = 8 \text{ cm}$ (1/2 Mark)
Let $CF = CE = x \text{ cm}$
Then, $\text{ar}(\Delta BOC) + \text{ar}(\Delta AOB) + \text{ar}(\Delta AOC) = \text{ar}(\Delta ABC)$
$\frac{1}{2} (8+x)4 + \frac{1}{2} (15)4 + \frac{1}{2} (7 + x)4 = 84$ (2 Marks)
$\implies 60 + 4x = 84$ (1/2 Mark)
$4x = 24$
$x = 6$ (1/2 Mark)
$:: BC = 8 + 6 = 14 \text{ cm}$ (1/2 Mark)
and $AC = 7+ 6 = 13 \text{ cm}$ (1/2 Mark)