AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th

AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th
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According to question 4q 2 = p 2+ 3r 2.
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AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th

Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th

Last Answer : According to question 4q 2 = p 2+ 3r 2.

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

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

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]

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]

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 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 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.

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 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 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, 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, 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, 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, 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.

The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th

Description : The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th

Last Answer : Let PQ be a chord of a circle with centre O and radius 13cm such that PQ = 24cm. From O, draw OM perpendicular PQ and join OP. As, the perpendicular from the centre of a circle to a chord bisects the chord. ∴ PM ... Hence, the distance of the chord from the centre is 5cm.

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

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.

AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

Description : AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

Last Answer : AB and CD are equal and parallel chords of a circle with centre O. Then AC passes through the centre O. [Agree `//` Disagree]

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Description : Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Last Answer : (c) ParallelogramAB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)CD =\(\sqrt{(4-1)^2+(3-2) ... = \(2\sqrt{13}\)AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.

Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

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:

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 = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

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 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 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

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

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 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 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

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

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Description : The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of other chord from the centre? -Maths 9th

Last Answer : There are two parallel chords of length 8 cm and 6 cm. The distance between center and shortest chord (6cm chord) is 4cm. So, the perpendicular distance from the center to the shortest chord is 4cm. The ... 32+42 =5. Since the radius is 5. The distance from center to largest chord is 52−42 =3.

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

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.

In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Description : In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Last Answer : AB is the longest chord because it is passing through the Centre hence it is a diameter ThereforeAB>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.]

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]

In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Description : In the given figure, O is the centre of the circle, then compare the chords. -Maths 9th

Last Answer : AB is the longest chord because it is passing through the Centre hence it is a diameter ThereforeAB>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.]

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]

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.