If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th
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Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof Since, CD||EF arc EC = arc PD … (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF …(ii)
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Related questions

If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Description : If two chords AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle. -Maths 9th

Last Answer : Given In a circle AYDZBWCX, two chords AB and CD intersect at right angles. To prove arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle. Construction Draw a diameter EF parallel to CD having centre M. Proof ... (i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF (ii)

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Description : If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Last Answer : Explanation of this question

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Description : If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

Last Answer : Explanation of this question

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th

Last Answer : Draw OM perpendicular AB and ON perpendicular AC Join OA. In right △OAM, OA2 = OM2 + AM2 ⇒ r2 = p2 + (1/2AB)2 (Since,OM perpendicular AB, ∴ OM bisects AB ) ⇒ 1/4AB2 = r2 - p2 or AB2 = 4r2 - 4p2 ... ) and (ii), we have 4r2 - 4p2 = 16r2 - 16q2 or r2 - p2 = 4r2 - 4q2 or 4q2 = 3r2 + p2

In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

Description : In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

Last Answer : Solution :- Let OP be perpendicular from O on line l. Since the perpendicular from the centre of a circle to a chord,bisects the chord.Therefore, AP = DP ...(i) BP = CP ...(ii) Subtracting (ii) from (i), we get AP - BP = DP - CP ⇒ AB = CD

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Description : In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Last Answer : Given In a figure, two chords AB and CD intersecting each other at point E.

In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Description : In figure, AB and CD are two chords of a circle intersecting each other at point E. -Maths 9th

Last Answer : Given In a figure, two chords AB and CD intersecting each other at point E.

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

ABCD is a parallelogram.The circle through A,B and C intersect CD (produce if necessary) at E.Prove that AE = AD. -Maths 9th

Description : ABCD is a parallelogram.The circle through A,B and C intersect CD (produce if necessary) at E.Prove that AE = AD. -Maths 9th

Last Answer : Solution :- ∠ABC + ∠AEC = 1800 (Opposite angles of cyclic quadrilateral) .. . (i) ∠ADE + ∠ADC = 1800 (Linear pair) But ∠ADC = ∠ABC (Opposite angles of ||gm) ∴ ∠ADE + ∠ABC = 1800 .. (ii) ... ∠ABC + ∠AEC = ∠ADE + ∠ABC ⇒ ∠AEC = ∠ADE ⇒ AD = AE (sides opposite to equal angles)

A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Description : A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Last Answer : answer:

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Description : 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

Prove that, the bisector of any two consecutive angles of parallelogram intersect at right angle. -Maths 9th

Description : Prove that, the bisector of any two consecutive angles of parallelogram intersect at right angle. -Maths 9th

Last Answer : Construct a Parallelogram ABCD where AB is Parallel to CD and BC is Parallel to AD. Make Angle Bisectors of Angle A and Angle B and let them join at O and let Angle OAB be X and ... Proved that the bisector of any two consecutive angles of parallelogram intersect at right angle I HOPE IT HELPS

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Description : If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Last Answer : Given : In a circle C(O,r) , ∠AOB = ∠COD To Prove : Chord AB = Chord CD . Proof : In △AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] ∠AOB = ∠COD [given] ⇒ △AOB ≅ △COD [by SAS congruence axiom] ⇒ Chord AB = Chord CD [c.p.c.t]

If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Description : If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Last Answer : Given : In a circle C(O,r) , ∠AOB = ∠COD To Prove : Chord AB = Chord CD . Proof : In △AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] ∠AOB = ∠COD [given] ⇒ △AOB ≅ △COD [by SAS congruence axiom] ⇒ Chord AB = Chord CD [c.p.c.t]

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Description : 1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD

AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th

Description : AC and BD are chords of a circle that bisect each other. Prove that AC and BD are diameters and ABCD is a rectangle. -Maths 9th

Last Answer : Solution :- Let AC and BD bisect each other at point 0. Then, OA = OC and OB = OD In triangles AOB and COD, we have OA = OC OB = OD and ∠ AOB = ∠ COD (Vertically opposite angles) ∴ △ AOB ... ∠ADC Also, ∠BAD = 90° = ∠BCD Also, AB = CD and BC = DA (Proved above) Hence, ABCD is a rectangle.

If two chords intersect inside a circle the angles formed are called inscribed angles.?

Description : If two chords intersect inside a circle the angles formed are called inscribed angles.?

Last Answer : yes

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Description : ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Last Answer : (b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if a' is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{ ... of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

If two intersecting chords of a circle make equal angles ...... -Maths 9th

Description : If two intersecting chords of a circle make equal angles ...... -Maths 9th

Last Answer : Given: AB and CD are two chords of a circle with centre O, intersecting at point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ. To prove: AB = CD Construction: Draw OL perpendicular AB and ... the circle. Since, chords of a circle which are equidistant from the centre are equal. ∴ AB = CD

Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Description : Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Last Answer : Given : In a circle C(O,r), chord AB = chord CD. To Prove : ∠AOB = ∠COD. Proof : In△AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] Chord AB = Chord CD [given] ⇒ △AOB ≅ △COD [by SSS congruence axiom] ⇒ ∠AOB = ∠COD. [c.p.c.t.]

Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Description : Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Last Answer : Given : In a circle C(O,r), chord AB = chord CD. To Prove : ∠AOB = ∠COD. Proof : In△AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] Chord AB = Chord CD [given] ⇒ △AOB ≅ △COD [by SSS congruence axiom] ⇒ ∠AOB = ∠COD. [c.p.c.t.]

AB and AC are two equal chords of a circle. -Maths 9th

Description : AB and AC are two equal chords of a circle. -Maths 9th

Last Answer : Given AS and AC are two equal chords whose centre is O. To prove Centre O lies on the bisector of ∠BAC. Construction Join SC, draw bisector AD of ∠BAC. Proof In ΔSAM and ΔCAM, AS = AC [given] ∠BAM = ∠CAM [given]