Description : A chord of a circle of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. -Maths 10th
Last Answer : Area of the minor segment = { pi × 90 /360 - sin 45 × cos 45 } × r × r ={ 3.14 ×1/4 - 1÷√2 ×1÷√2 } × 20 × 20 = { 3.14 ... Area of major segment = area of circle - area of minor segment = 1256 - 114 = 1142 HOPE IT HELPS YOU
Description : The distances of two chords AB and CD from the centre of a circle are 6 cm and 8 cm respectively. Then, which chord is longer?
Last Answer : The distances of two chords AB and CD from the centre of a circle are 6 cm and 8 cm respectively. Then, which chord is longer?
Description : The length of normal chord of parabola `y^(2)=4x`, which subtends an angle of `90^(@)` at the vertex is :
Last Answer : The length of normal chord of parabola `y^(2)=4x`, which subtends an angle of `90^(@)` at the vertex is : A. 1 B. `1/2` C. `4/3` D. `1/4`
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 : 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 figure below, CD is a chord of the semi circle with centre O. OA is the radius of the circle. If `CD=10` cm, `AB=2` cm and `bar(OA)_|_bar(CD)`
Last Answer : In the figure below, CD is a chord of the semi circle with centre O. OA is the radius of the ... |_bar(CD)` the length of OB is `"_____________"`
Description : In a circle of radius 14 cm, an arc subtends an angle of 45 O at the centre, then the area of the sector is (a) 71 cm 2(b) 76 cm 2 (c) 77 cm 2 (d) 154 cm 2
Last Answer : (c) 77 cm 2
Description : In the given figure, AB is the diameter and `/_ADC = 2 /_BDC`. If `/_ BCD =70^(@)`, then find the angle made by AC at the centre of the circle.
Last Answer : In the given figure, AB is the diameter and `/_ADC = 2 /_BDC`. If `/_ BCD =70^(@)`, then find the angle made by AC at the centre of the circle.
Description : An arc of a circle of raduis `R` subtends an angle `(pi)/2` at the centre. It carriers a current `i`. The magnetic field at the centre will be
Last Answer : An arc of a circle of raduis `R` subtends an angle `(pi)/2` at the centre. It carriers a current `i`. The ... (0)i)/(4R)` D. `(2mu_(0)i)/(5R)`
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 : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th
Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm
Description : In the figure above (not to scale), AB is the diameter of the circle with centre O. If `/_ ACO=30^(@),` then find `/_ BOC`.
Last Answer : In the figure above (not to scale), AB is the diameter of the circle with centre O. If `/_ ACO=30^(@),` then find `/_ BOC`.
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 : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th
Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.
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 : 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.
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.
Description : In the figure above (not to scale), `AB=AC` and `/_BAO=25^(@)`. Find `/_BOC,` if O is the centre of the circle.
Last Answer : In the figure above (not to scale), `AB=AC` and `/_BAO=25^(@)`. Find `/_BOC,` if O is the centre of the circle.
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 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
Description : The radius of a circle is 10cm. The length of a chord is 12 cm. Then the distance of the chord from the centre is `"__________________"`.
Last Answer : The radius of a circle is 10cm. The length of a chord is 12 cm. Then the distance of the chord from the centre is `"__________________"`.
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 : 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 : 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:
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
Description : In the figure below, `bar(MN)` is the diameter of the circle with centre O. `bar(NP)` bisects the `/_ANM`. If `/_ NMA =33^(@)`, then find `/_ ANP`.
Last Answer : In the figure below, `bar(MN)` is the diameter of the circle with centre O. `bar(NP)` bisects the `/_ANM`. If `/_ NMA =33^(@)`, then find `/_ ANP`.
Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th
Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm
Description : What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th
Last Answer : Solution :- Perpendicular line from the centre bisect the chord.
Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th
Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.
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 : 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 : PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to chord PS at M. If `bar(PS) =30 cm` and `bar(OM) =
Last Answer : PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to ... =8 cm`, then find the radius of the circle.
Description : A portion of an embankment having a uniform up-gradient 1 in 500 is circular with radius 1000 m of the centre line. It subtends 180° at the centre. If the height of the bank is 1 m at the lower end, and side slopes 2 ... earth work involved. (A) 26,000 m3 (B) 26,500 m3 (C) 27,000 m3 (D) 27,500 m
Last Answer : (D) 27,500 m3
Description : The length of ——ab (the minor arc) is 40 cm. What is the circumference of circle c inside angle is 60 degrees?
Last Answer : s = 2 π r (θ/360°), C = πr^240cm = 2πr(1/6)(40*6)/2π = rr = 38.1972cmC = π*(38.1972)^2C = 4,583.66 cm
Description : A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. -Maths 9th
Last Answer : Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e., AB = BO …(i) Join OA, AC and BC. Since, OA = OB= Radius of circle OA = AS = BO
Description : A circle has radius √2 cm. It is divided into two segments by a chord of length 2cm.Prove that the angle subtended by the chord at a point in major segment is 45 degree . -Maths 9th
Last Answer : Given radius =2 cm Therefore AO=2 cm Let OD be the perpendicular from O on AB And AB =2cm Therefore AD=1cm (perpendicular from the centre bisects the chord) Now in triangle AOD, AO=2 cm ... by a chord at the centre is double of the angle made by the chord at any poin on the circumference)
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 : 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 : In the figure (not to scale ), O is the centre of the circle and `/_OBA=30^(@)`. Find `/_ ACB`
Last Answer : In the figure (not to scale ), O is the centre of the circle and `/_OBA=30^(@)`. Find `/_ ACB`