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# in the given figure, a chord ab of the circle with centre o and radius 10 cm, that subtends a right angle at the centre of the circle. find the area of the minor segment aqbp. hence find the area of major segment albqa

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### Mohammed

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## In figure there is a chord AB of a circle with centre O and radius 10 cm , that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP . Hence find the area of major segment ALBQA .

Click here👆to get an answer to your question ✍️ In figure there is a chord AB of a circle with centre O and radius 10 cm , that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP . Hence find the area of major segment ALBQA . Question

## In figure there is a chord AB of a circle with centre O and radius 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Hence find the area of major segment ALBQA. Hard Open in App Solution Verified by Toppr ∠ALB= 2 1 ​ ∠AOB= 2 90 ​ =45 0

LQ is perpendicular bisection on AB. Hence by isoceles triangles property.

LA=LB OB=10cm, & OA=10cm AB= 10 2 +10 2 ​ =10 2 ​ cm QB= 2 AB ​ =5 2 ​ cm OQ= OB 2 −QB 2 ​ = 100−50 ​ = 50 ​ =5 2 ​ cm LQ=LO+OQ=(10+5 2 ​ )cm Area of ALBQA= 2 1 ​ ×AB×LQ= 2 1 ​ ×10 2 ​ ×(10+5 2 ​ ) =50(1+ 2 ​ )cm 2 Area of AQBPA= 4 πr 2 ​ −Ar.ofAOB = 4 π×10 2 ​ − 2 1 ​ ×10 2 ⇒28.54cm 2 32 32

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## In the given figure, a chord AB of the circle with centre O and radius 10 cm, that subtends a right angle at the

In the given figure, a chord AB of the circle with centre O and radius 10 cm, that subtends a right angle ... of major segment A ∠ LBQA. (Use π =3.14) ## In the given figure, a chord AB of the circle with centre O and radius 10 cm, that subtends a right angle at the

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## In Fig. 5 is a Chord Ab of a Circle, with Centre O and Radius 10 Cm, that Subtends a Right Angle at the Centre of the Circle. Find the Area of the Minor Segment Aqbp.

In Fig. 5 is a Chord Ab of a Circle, with Centre O and Radius 10 Cm, that Subtends a Right Angle at the Centre of the Circle. Find the Area of the Minor Segment Aqbp. In fig. 5 is a chord AB of a circle, with centre O and radius 10 cm, that subtends a right angle at the centre of the circle. Find the area of the minor segment AQBP. Hence find the area of major segment ALBQA. (use π = 3.14) ### SOLUTION Given: Radius of the circle, r = 10 cm

Area of the circle  =

π(10)2=3.14×100=314cm2

Area of the sector OAPB

=90o360o×π(10)2 =14×314[From(1)] =78.5cm2 Area of ∆BOA =12×BO×OA =12×10×10 =50cm2

Area of the minor segment AQBP = Area of the sector OAPB − Area of the triangle BOA

= (78.5 − 50) cm2 = 28.5 cm2

Area of the major segment ALBQ = Area of the circle − Area of the minor segment AQBP

= (314 − 28.5) cm2 = 285.5 cm2

Concept: Tangent to a Circle

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2015-2016 (March) Foreign Set 1

Q 14 Q 13 Q 15

### APPEARS IN

2015-2016 (March) Foreign Set 1 (with solutions)

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Tangent to a Circle video tutorial 00:10:01 Tangent to a Circle video tutorial 00:04:35 Tangent to a Circle video tutorial 00:11:38 Tangent to a Circle video tutorial 00:05:07

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Mohammed 4 day ago

Guys, does anyone know the answer?