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
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.
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°
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
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]
Description : In the above figure, O is the centre of the circle AB,AD and CD are the chords . If `/_ ADC=130^(@)` then fid `/_ ACB`.
Last Answer : In the above figure, O is the centre of the circle AB,AD and CD are the chords . If `/_ ADC=130^(@)` then fid `/_ ACB`.
Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th
Last Answer : answer:
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
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.]
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.
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
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.
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
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.
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
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.
Description : In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. -Maths 9th
Last Answer : In △OBC, OB = BC ⇒ ∠BOC = ∠BCO = y ...[angles opp. to equal sides are equal] ∠OBA is the exterior angle of △BOC So, ∠ABO = 2y ...[ext. angle is equal to the sum of int. opp. angles] Similarly, ∠AOD is the exterior angle of △AOC ∴ x = 2y + y = 3y
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.
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]
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.
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)
Description : MN and PS are two equal chords of a circle drawn on either side of centre O of the circle . Both the chords are produced to meet at point A. If the ra
Last Answer : MN and PS are two equal chords of a circle drawn on either side of centre O of the circle . Both the chords ... 12 `cm and OA `=17` cm , then find NA.
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]
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
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.
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.
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.
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
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.
Description : In Fig. 10.17, O is the centre of a circle, find the value of x. -Maths 9th
Last Answer : Solution :- In △APB, ∠APB = 90° (Angle in the semi- circle) ∠APE + ∠ABP + ∠BAP = 180° 90° + 40° + ∠BAP = 180° ⇒ ∠BAP = 180° - 130° ⇒ ∠BAP = 50° Now, ∠BQP = ∠BAP (Angles in the same segment) ∴ x = 50°
Description : If O is the centre of the circle and chord AB = OA and the area of triangle AOB -Maths 9th
Last Answer : (b) 16π cm2.AB = OA ⇒ AB = OA = OB (radii of circle are equal) ⇒ ΔAOB is equilateral. ∴ If ‘r’ is the radius of the circle,then area of ΔAOB = \(rac{\sqrt3}{4}\)side2⇒ \(rac{\sqrt3}{4}\)(r)2 = 4√3 (given)⇒ r2 = 16 ⇒ r = 4∴ Area of circle = πr2 = 16π cm2.
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
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.