Question

Question :


A is a point at a distance of 13 cm from the centre O of circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.​

Answer 1

Answer:

The perimeter of the ΔABC  24 cm.

Step-by-step explanation:

Given :

OA = 13cm

Radius = OP = 5cm

Since AP is a tangent to the circle with center O and OP is its radius, OP ⊥ AP

Now, In ΔOPA ,

$\sf\\ \\\implies \angle OPA = 90\textdegree\\ \\\implies AP^{2} = OA^{2} - OP^{2}\quad\bigg[\textbf{Using Pythagoras Theorem.}\bigg]\\ \\ \implies 13^{2} - 5^{2}\\ \\ \implies 169 - 25$

So,

$\sf\\ \\\implies AP^{2} = 144\\ \\ \implies AP = \sqrt{144}\\ \\ \implies AP = 12$

Now,

$\sf\\ \\\implies AP = \dfrac{1}{2} \times Perimeter\;of \;\triangle ABC\\ \\ \implies 12 = \dfrac{1}{2} \times Perimeter\;of \;\triangle ABC\\ \\ \implies Perimeter\;of \;\triangle ABC\\ \\ \implies24cm$

Answer 2

Answer:

Perimeter of ∆ABC = 24 cm.

Step-by-step explanation:

$\huge\bold{Given\::}$

OA = 13 cm

OP = radius = 5 cm

As AP is a tengant to the circle,

In, ∆AOP

• <APO = 90°

Then,

${AP}^{2}={AO}^{2}-{OP}^{2}$

${AP}^{2}={13}^{2}-{5}^{2}$

${AP}^{2}=169-25$

${AP}^{2}=144$

$AP =\sqrt{144}$

★$AP =12\:cm$

In ∆ABC,

AP = 1/2 × Perimeter of ∆ABC

12 = 1/2 × Perimeter if ∆ABC

★Perimeter of ∆ABC =24 cm

$\huge\bold{Thank\:You}$

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