In the figure given below O is the centre of the circle. If QR

In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:

(i) m∠BDC

(ii) m∠BEC

(iii) m∠BAC


Consider the followiong figure:

(i) Given that BD is a diameter of the circle.

The angle in a semi circle is a right angle.

So, we can write, 

Also, given that DBC = 58°

In BDC, by angle sum property, we have

DBC + BCD + BDC = 180°

 58° + 90° + BDC = 180°

 148° + BDC = 180°

 BDC = 180° - 148°

 BDC = 32°

(ii) BECD is cyclic quadrilateral.

we know that, opposite angles are supplementary

 BEC + BCD = 180°

 BEC + 32° = 180°

 BEC = 148°

(iii) Angles in the same segment are equal.

 BAC = BDC = 32° 


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In the figure given below 'O' is the centre of the circle. If QR = OP and ∠ORP = 20°. Find the value of 'x' giving reasons.

                   


Now,

OP = QR       .........GIVEN

So,  OP = OT = OQ = QR

In  RQP

RQ = QO

So,   QRO = QOR = 20°So,  by sum of angles in  RQPRQO = 140°

Now

RQO + OQP = 180°       .........linear pairOQP = 40°In POQOQ = PO       .......radiiSo,   QPO = OQP = 40°So,  by sum of angles in  OQPPOQ = 100°

Now,

POT + POQ + QOR = 180°       .........angles in straight linex = 60°


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The given figure represents a kite with a circular and a semicircular motifs stuck on it. The radius of a circle is  2.5 cm and the semicircle is  2 cm. If diagonals  AC  and  BD  are of lengths  12 cm  and  8 cm  respectively, find the area of the:

 (i) Shaded part. Give your answer correct to the nearest whole number.

(ii) Unshaded part.

                                         


(i) Area of the shaded part  =  Area of the circle + area of the semicircle

                                        = π  2.5 2 + π  2 22= π  6.25 + 2 = 227  8.25  26 cm2

 

 ( ii ) Area of kite = product of the diagonals2                          = AC x BD2 = 12 x 82                          = 48 cm2

Area of the unshaded part = Area of the kite  - Area of the shaded pa                                
                                       = 48 - 26

                                       = 22 cm2


In the figure given, O is the centre of the circle. DAE = 70o, Find giving suitable reasons the measure of: 

(i) BCD

(ii) BOD

(iii) OBD

                                             


( i )   DAE = 70°                ..............( given )BAD + DAE = 180°    ..............( linear pair ) BAD + 70° = 180° BAD  = 110°

Since ABCD is a cyclic quadrilateral, sum of the measures of the opposite angles are supplementary.

So,  BCD  +  BAD  = 180° BCD  + 110° = 180° BCD  = 70°( ii )  BOD   =  2 BCD            ............( Inscribed angle theorem ) BOD   = 2 ( 70° ) = 140°( iii )  In   OBD,OB = OD                                        .............( radii of some circle ) OBD = ODB

By Angle Sum property,

OBD   + ODB + BOD     = 180°  2 OBD  +  BOD = 180°  2 OBD  +  140° = 180°  2 OBD  = 40° OBD  = 20°


In the figure given below,  AD  is a diameter.  O  is the centre of the circle.
AD  is parallel to  BC  and  CBD = 30°. Find:

(i)   OBD 

(ii)  AOB

(iii)  BED

                            


( i )  AD is parallel to  BC, i. e.  OD is parallel to  BC  and  BD  is transversal.

  ODB = CBD = 32°        .............( Alternate angles )In OBD,OD = OB                                      .............( Radii of the same circle )  ODB = OBD = 32°

 

( ii ) AD  is parallel to  BC, i.e.  AO is parallel to  BC  and  OB  is transversal.

 

 AOB = OBC                     .................( Alternate angles ) OBC =  OBD + DBC OBC = 32° + 32° OBC = 64° AOB = 64°

 

( iii )  In OAB,OA = OB                  ...............( Radii of the same circle ) OAB = OBA = X  ( say ) OAB + OBA + AOB = 180° x + x + 64° = 180° 2 x  = 180° - 64° 2 x  = 116° x  = 58° OAB = 58°i.e.,   DAB = 58° DAB = BED =  58°               ..............( Angles inscribed in the same arc are equal. )


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