Question

In Q. No. 1, if PB = 10 cm, what is the perimeter of Δ PCD?

Answer 1

The perimeter of Δ PCD is 20 cm.

Step-by-step explanation:

Given,

PB = 10 cm

We have to find the perimeter of the Δ PCD.

Solution,

Since PA and PB are two tangents drawn to the circle from point P.

So according to tangent theorem which states that;

"Two tangents drawn from one external point are equal in length".

So we can say that;

$PA=PB=10\ cm$

Again,

CA = CQ  ⇒ 1

DQ = DB  ⇒ 2

according to tangent theorem.

Now we know that the perimeter of triangle is equal to sum of all sides of the triangle.

$Perimeter =PC+CD +PD$

Now we can write CD as;

CD = CQ +QD

And from 1 and 2 we can say that;

CD = CA + DB ⇒ 3

So we can say that;

$Perimeter =PC+CA+DB +PD$

And we can also say that;

PC+CA = PA   and

PD+DB = PB

$\therefore Perimeter = PA+PB=10+10=20\ cm$

Hence The perimeter of Δ PCD is 20 cm.

Answer 2

Answer:

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Step-by-step explanation:

$\bold{The \: perimeter \: of \: Δ PCD \: is \: 20 cm.}</p><p></p><p></p><p> \\ Given,</p><p></p><p> \\ PB = 10 cm</p><p></p><p> \\ We \: have \: to \: find \: the \: perimeter \\ \: of \: the \: Δ PCD.</p><p></p><p> \\ Solution,</p><p></p><p> \\ Since \: PA \: and \: PB \: are \: two \: \\ tangents \: drawn \: to \: the \: circle \\ from \: point \: P.</p><p></p><p> \\ So \: according \: to \: \: tangent \: the \: \\ orem \: which \: states \: that;</p><p></p><p> \\ "Two \: tangents \: drawn \: from \: one \: \\ external \: point \: are \: equal \: in \: length".</p><p></p><p> \\ So \: we \: can \: say \: that;</p><p></p><p> \\ PA=PB=10\ cm \\ PA=PB=10 cm</p><p></p><p> \\ Again,</p><p></p><p> \\ CA = CQ ⇒ 1</p><p></p><p> \\ DQ = DB ⇒ 2</p><p></p><p> \\ according \: to \: tangent \: theorem.</p><p></p><p> \\ Now \: we \: know \: that \: the \: perimeter \: \\ of \: triangle \: is \: equal \: to \: sum \: of \: \\ all \: sides \: of \: the \: triangle.</p><p></p><p> \\ Perimeter =PC+CD +PD \\ Perimeter=PC+CD+PD</p><p></p><p> \\ Now \: we \: can \: write \: CD \: as;</p><p></p><p> \\ CD = CQ +QD</p><p></p><p> \\ And \: from \: 1 \: and \: 2 \: we \: can \: say \: that \\ CD = CA + DB ⇒ 3</p><p></p><p> \\ So \: we \: can \: say \: that;</p><p></p><p> \\ Perimeter =PC+CA+DB +Pd \\ Perimeter=PC+CA+DB+PD</p><p></p><p> \\ And \: we \: can \: also \: say \: that; \\ </p><p></p><p>PC+CA = PA \\ and</p><p></p><p> \\ PD+DB = PB</p><p></p><p> \\ \begin{gathered}\therefore Perimeter = PA+PB=10+10=20\ cm\\\end{gathered} </p><p> \\ ∴Perimeter=PA+PB=10+10=20 cm</p><p> </p><p> </p><p></p><p> \\ Hence \: The \: perimeter \: of \: Δ PCD \: is \: \bold{20 cm.}$

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