Math, asked by Anonymous, 4 days ago


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A quadrilateral ABCD such that ∠A = 90°. A circle with center O touches the sides AB, BC, CD, DA at E, F, G and H respectively. If CD = 30cm, AD = 15cm and CF = 20cm, find the radius of the circle​

Answers

Answered by sreeragsreelakam2008
2

Answer:

Since tangents to a circle is perpendicular to the radius through the point.

∴∠ORD=∠OSD=90

o

It is given that ∠D=90

o

. Also, OR=OS. Therefore, ORDS is a square.

Since tangents from an exterior point to a circle are equal in length.

∴BP=BQ CQ=CR and, DR=DS.

Now,

BP=BQ

⇒ BQ=27 [∵BP=27 cm (Given)]

⇒BC−CQ=27

⇒38−CQ=27

∵ BC=38cm ⇒CQ=11 cm

⇒CR=11 [∵CR=CQ]

⇒CD−DR=11

⇒25−DR=11 [∵CD=25cm]

⇒DR=14 cm

But, ORDS is a square. ⇒ OR=DR=14 cm.

⇒ r=14 cm

Answered by aditiupadhyayj
2

Answer:

answer is as follows:-

Step-by-step explanation:

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Question

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ABCD is a quadrilateral such that ∠D=90

o

. A circle C(O,r) touch the sides AB, BC, CD and DA at P,Q,R and S respectively. If BC=38cm, CD=25cm and BP=27cm, find r.

Medium

Solution

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Since tangents to a circle is perpendicular to the radius through the point.

∴∠ORD=∠OSD=90

o

It is given that ∠D=90

o

. Also, OR=OS. Therefore, ORDS is a square.

Since tangents from an exterior point to a circle are equal in length.

∴BP=BQ CQ=CR and, DR=DS.

Now,

BP=BQ

⇒ BQ=27 [∵BP=27 cm (Given)]

⇒BC−CQ=27

⇒38−CQ=27

∵ BC=38cm ⇒CQ=11 cm

⇒CR=11 [∵CR=CQ]

⇒CD−DR=11

⇒25−DR=11 [∵CD=25cm]

⇒DR=14 cm

But, ORDS is a square. ⇒ OR=DR=14 cm.

⇒ r=14 cm.

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