Quadrilateral ABCD is inscribed in a circle with AD as diameter. If $A D = 4, A B = B C = 1$ , then $A C =$

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$\frac{\sqrt{15}}{2}$

$\frac{\sqrt{5}}{2}$

$\frac{7}{2}$

None of these

answer is A.

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Detailed Solution

$A D = 4, A B = B C = 1$ (given)

$∠ A C D = 90 °$ [ $∵$ angle in a semi circle]

Let $∠ A D C = α$ and $A C = x$

$∠ A B C = 180 ° - α$

From $x^{2} = 1^{2} + 1^{2} - 2 (1) (1) \cos (180 ° - α)$

$x^{2} = 2 + 2 \cos α$

$x^{2} = 2 + 2. \frac{C D}{4}$

$x^{2} = 2 + \frac{1}{2} \sqrt{A D^{2} - A C^{2}}$

$x^{2} = 2 + \frac{1}{2} \sqrt{16 - x^{2}}$

$x^{2} - 2 = \frac{1}{2} \sqrt{16 - x^{2}}$

Squaring on both sides

$x^{4} + 4 - 4 x^{2} = \frac{1}{4} (16 - x^{2})$

$4 x^{4} + 16 - 16 x^{2} = 16 - x^{2}$

$4 x^{4} - 15 x^{2} = 0$

$x^{2} (4 x^{2} - 15) = 0$

$4 x^{2} - 15 = 0$

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