Question

15. In the given figure, a circle with centre O,
is inscribed in a quadrilateral ABCD such
that it touches the side BC, AB, AD and
CD at points P, Q, R and S respectively.
If AB = 29 cm, AD = 23 cm, angle B = 90° and
DS = 5 cm then find the radius of the
circle.
[CBSE 2008,'13]


$spam = report \\ 20thx = follow \:$
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Answer 1

Aɴsᴡᴇʀ:-

The radius of the circle is 11 cm

Exᴘʟᴀɴᴀᴛɪᴏɴ:-

Given:-

• AB = 29 cm
• AD = 23 cm
• DS = 5 cm
• Angle B = 90°

To find:-

Find the radius of the circle

Solution:-

DS = DR [tangents from D]

So,

$\pink\implies$AR = AD - DR

$\pink\implies$AR = 23 - 5 cm

$\pink\implies$AR = 18 cm

Again,

$\purple\longrightarrow$AR = AQ = 18 cm [tangents from A]

So,

$\red\implies$QB = AB - AQ

$\red\implies$QB= 29 - 18 cm

$\red\implies$QB = 11 cm

$\pink\longrightarrow$Now, In quadrilateral OPBQ,

Angle B = 90°

$\green\longrightarrow$Also, Angle OPB = Angle OQB = 90°

Tangent is perpendicular to radius at point of contact

$\red\longrightarrow$ So, angle POQ = 90°,that is OPBQ is a rectangle.

Further since , PA = PB

.°.OPBQ is a square.

$\orange\longrightarrow$Therefore, Radius OP = BQ = 11 cm

The radius of the circle is 11 cm

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