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(B) Given that $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$.We know that the lengths of tangents drawn from an ex

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$20$

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(B) Given that $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$.We know that the lengths of tangents drawn from an external point to a circle are equal,so $PA = PB = 10 \, cm$.The perimeter of $	riangle PCD = PC + CD + PD$.Since $CD$ is a tangent at point $E$,it can be written as $CD = CE + ED$.Also,$CE = CA$ and $ED = DB$ (tangents from points $C$ and $D$ respectively).Substituting these into the perimeter formula:Perimeter of $	riangle PCD = PC + (CE + ED) + PD = PC + CA + DB + PD$.Since $PC + CA = PA$ and $PD + DB = PB$,we get:Perimeter of $	riangle PCD = PA + PB = 10 + 10 = 20 \, cm$.

From an external point $P$,two tangents,$PA$ and $PB$ are drawn to a circle with centre $O$. At one point $E$ on the circle,a tangent is drawn which intersects $PA$ and $PB$ at $C$ and $D$,respectively. If $PA = 10 \, cm$,find the perimeter of the triangle $PCD$ (in $cm$).

Similar Questions

$\stackrel{\leftrightarrow}{PA}$ and $\stackrel{\leftrightarrow}{PB}$ are tangents to the circle $\odot(O, r)$ at points $A$ and $B$ respectively. If $m\angle OPB = 35^\circ$,then $m\angle AOB = \ldots$ (in $^\circ$)

Write 'True' or 'False' and give reasons for your answer.
If the angle between two tangents drawn from a point $P$ to a circle of radius $a$ and center $O$ is $90^{\circ}$,then $OP = a\sqrt{2}$.

If the angle between two radii of a circle is $130^{\circ}$,the angle between the tangents at the ends of the radii is: (in $^{\circ}$)

$\overline{ PA }$ and $\overline{ PB }$ are the tangents to $\odot( O , r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 65^{\circ}$,then $m \angle AOB = \ldots \ldots \ldots . .$ (in $^{\circ}$)

Let $s$ denote the semi-perimeter of a triangle $ABC$ in which $BC = a, CA = b, AB = c$. If a circle touches the sides $BC, CA, AB$ at $D, E, F$ respectively,prove that $BD = s - b$.

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