In an isosceles triangle $A B C, A B = A C$ . A circle passing through $B$ and touching $A C$ at its middle point D, intersects $A B$ at $M$ . If $A M = 3 c m$ , then $A B = . . . .$

6 c m

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9 c m

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12 c m

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15 c m

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N o n e o f t h e s e

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Solution

The correct option is C $12 c m$
$\begin{matrix} O D ⊥ A C (sin c e A C i s tan g e n t t o c i r c l e) w h e n t a n g e n t a n d a s e c a n t a r e d r a w n f r o m o n e sin g l e p o i n t t o a c i r c l e, t h e n, A D^{2} = A M \cdot A B A D = \frac{A C}{2} = \frac{A B}{2} (g i v e n) {(\frac{A B}{2})}^{2} = A M \cdot A B \frac{A B^{2}}{2} = A M \cdot A B A B = 4 A M = 4 \times 3 = 12 c m \end{matrix}$

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In an isosceles triangle ABC, AB=AC. A circle passing through B and touching AC at its middle point D, intersects AB at M. If AM=3 cm, then AB=....

The correct option is C 12 cmOD⊥AC(sinceACistangenttocircle)whentangentandasecantaredrawnfromonesinglepointtoacircle,then,AD2=AM⋅ABAD=AC2=AB2(given)(AB2)2=AM⋅ABAB22=AM⋅ABAB=4AM=4×3=12cm

In an isosceles triangle $A B C, A B = A C$ . A circle passing through $B$ and touching $A C$ at its middle point D, intersects $A B$ at $M$ . If $A M = 3 c m$ , then $A B = . . . .$