Question

From a point p which is at a distance of 13cm from the centre O of a circle of radius of 5 cm ,the pair of tangents PQ and PR to the circle are drawn .The. the area of quadrilateral PQOR is​

Answer 1

✬ Area = 60 cm² ✬

Step-by-step explanation:

Given:

• A point P is at a distance of 13 cm from centre O.
• Radius of circle is 5 cm.
• Two tangents i.e PQ & PR are drawn from P.

To Find:

• Area of quadrilateral PQOR ?

Solution: Let P be a point at a distance of 13 cm from O.

We know that the tangent and radius are perpendicular at point of contact and tangents drawn from a external points are also equal to each other.

Here we have

• PO = 13 cm
• PR = PQ
• OR = OQ = 5 cm

In right angled triangle PQR , by Pythagoras theorem

• OQ = Base
• OP = Hypotenuse
• PQ = Perpendicular

★ H² = B² + P² ★

$\implies{\rm }$ OP² = OQ² + PQ²

$\implies{\rm }$ 13² = 5² + PQ²

$\implies{\rm }$ 169 = 25 + PQ²

$\implies{\rm }$ √169 – 25 = PQ

$\implies{\rm }$ √144 = PQ

$\implies{\rm }$ 12 = PQ

Similarly PR will be also 12 cm.

Now area of quadrilateral PQOR will be

• ar(∆POQ + ∆POR)

★ Ar of triangle = 1/2 × Base × Height ★

• Base = OP = 5 cm
• Height = PQ = 12 cm

• { since, there are two triangles with same height and base }

$\implies{\rm }$ 2 × 1/2 × 5 × 12

$\implies{\rm }$ 5 × 12

$\implies{\rm }$ 60 cm²

Hence, area of quadrilateral PQOR will be 60 cm².

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• As shown in the figure,

1. The distance between point P to the center of the circle O is '13 cm' .
2. Radius of the circle is '5 cm' .
3. A pair of tangents PQ and PR to the circle are drawn .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• The area of quadrilateral PQOR .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

☞︎︎︎ See the attachment figure .

☯︎ It is clearly seen that, tangent PQ and PR are perpendicular to OQ & OR respectively .

☯︎ Hence both ∆POQ & ∆POR are right angled triangle .

$\bf{\implies\:PQ\:=\:\sqrt{(OP)^2\:-\:(OQ)^2}\:}$

$\rm{\implies\:PQ\:=\:\sqrt{(13)^2\:-\:(5)^2}\:}$

$\rm{\implies\:PQ\:=\:\sqrt{169\:-\:25}\:}$

$\rm{\implies\:PQ\:=\:\sqrt{144}\:}$

$\bf\blue{\implies\:PQ\:=\:12\:cm\:}$

$\huge\red\checkmark$ $\mathcal{\purple{Area\:of\:\triangle{POQ}\:=\:\dfrac{1}{2}\:\times{PQ}\times{OQ}\:}}$

$\rm{\implies\:Area\:of\:\triangle{POQ}\:=\:\dfrac{1}{2}\:\times{12}\times{5}\:}$

$\rm{\implies\:Area\:of\:\triangle{POQ}\:=\:6\times{5}\:}$

$\bf\green{\implies\:Area\:of\:\triangle{POQ}\:=\:30\:cm^2\:}$

Similarly,

$\huge\red\checkmark$ $\mathcal{\purple{Area\:of\:\triangle{POR}\:=\:\dfrac{1}{2}\:\times{PR}\times{OR}\:}}$

$\rm{\implies\:Area\:of\:\triangle{POR}\:=\:\dfrac{1}{2}\times{12}\times{5}\:}$

$\rm{\implies\:Area\:of\:\triangle{POR}\:=\:6\times{5}\:}$

$\bf\green{\implies\:Area\:of\:\triangle{POR}\:=\:30\:cm^2\:}$

__________________________

∴ Area of quadrilateral PQOR = Area of (∆POQ + ∆POR)

⇛ Area of quadrilateral PQOR = 30 cm² + 30 cm²

⇛ Area of quadrilateral PQOR = 60 cm²

___________________________

$\huge\red\therefore$ The area of quadrilateral PQOR is "60cm²" .