At end A of a chord AB of a circle tangent is drawn. A point D is taken on tangent such that $∠ A B E = ∠ A B D$ where E is some point on the circumference If BE = $3$ and BD = $6$ Find AD

3

units

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6

units

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3 \sqrt{2}

units

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3 \sqrt{3}

units

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Solution

The correct option is C $3 \sqrt{2}$ units
According to the tangent – secant theorem length of tangent segment squared equals product of secant segment and its external segment

$⟹ A D^{2} = B D \times E D$

$⟹ A D = \sqrt{6 \times 3} = 3 \sqrt{2}$

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At end A of a chord AB of a circle tangent is drawn. A point D is taken on tangent such that ∠ABE=∠ABD where E is some point on the circumference If BE = 3 and BD = 6 Find AD

The correct option is C 3√2 unitsAccording to the tangent – secant theorem length of tangent segment squared equals product of secant segment and its external segment ⟹AD2=BD×ED ⟹AD=√6×3=3√2

At end A of a chord AB of a circle tangent is drawn. A point D is taken on tangent such that $∠ A B E = ∠ A B D$ where E is some point on the circumference If BE = $3$ and BD = $6$ Find AD

The correct option is C $3 \sqrt{2}$ units
According to the tangent – secant theorem length of tangent segment squared equals product of secant segment and its external segment

$⟹ A D^{2} = B D \times E D$

$⟹ A D = \sqrt{6 \times 3} = 3 \sqrt{2}$