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 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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 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 : 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 : 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 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 : 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 : 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 : 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 : If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by (A) a = S.R/L (B) a = L.S/R (C) a = L.R/S (D) None of these
Last Answer : (A) a = S.R/L
Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th
Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ
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 : In a circle , chord AB subtends an angle of `60^(@)` at the centre and chord CD subtends `120^(@)`, at it. Then which chord is longer ?
Last Answer : In a circle , chord AB subtends an angle of `60^(@)` at the centre and chord CD subtends `120^(@)`, at it. Then which chord is longer ?
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 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 a purely cohesive soil, the critical centre lies at the intersection of (A) Perpendicular bisector of slope and the locus of the centre (B) Perpendicular drawn at 1/3rd slope from toe and the ... C) Perpendicular drawn at 2/3rd slope from toe and the locus of the centre (D) Directional angles
Last Answer : Answer: Option D
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 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`
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 : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the circle solve the following :
Last Answer : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the ... D. `x^(2)+y^(2)+6x+1=0`
Last Answer : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the ... 0` D. `x^(2)+y^(2)-8y=0`
Description : Prove that the length of the tangents drawn from an external point to a circle are equal, hence show that the centre lies on the bisector of the angle between the two tangents? -SST 10th
Last Answer : Given: PT and TQ are two tangent drawn from an external point T to the circle C (O, r). To prove: 1. PT = TQ 2. ∠OTP = ∠OTQ Construction: Join OT. Proof: We know that, a tangent to ... point to a circle are equal. ∠OTP = ∠OTQ, ∴ Centre lies on the bisector of the angle between the two tangents.
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 : 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 : Find the back bearing of the following lines having fore bearing as given below: (i) PQ = N 55° 0' E (ii) EF == 280° 0' (iii) CD = S 58° 30' W (iv) OM= 180° 0'
Last Answer : i) F.B OF PQ = N 55°0 ’ E B.B Of PQ = S 55°0 ’W ii) F.B. Of EF = 280°0 ’ B.B OF EF = F.B -180°= 280°0 ’ -180° = 100°0 ’ iii) F.B OF CD = S 58°30’W B.B OF CD = N 58°30’ E iv) F.B OF GH = 180° B.B OF GH = F.B -180°=180°-180°= 0°
Description : While setting a plane table at a station it was found that the error in centering was 30 cm away from the ray of length 40 m drawn from the station. If the scale of the plan is 1 cm = 2 cm, the displacement of the end of ... true position will be (A) 0.02 cm (B) 0.15 cm (C) 0.2 cm (D) 0.1 cm
Last Answer : (B) 0.15 cm
Description : If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. -Maths 9th
Last Answer : Given: Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect ... 0° Thus, AP = BP and ∠APO = ∠BPO = 9 0° Hence, OO' is the perpendicular bisector of AB.
Description : Top chord of a truss is continuous over joints l apart. Effective lengths of the member in the plane perpendicular to the truss is (a) 0.7 l (b) 0.85 l (c ) l (d) 1.5 l
Last Answer : (c ) l
Description : A bed of spherical particles (specific gravity 2.5) of uniform size 1500 μm is 0.5 m in diameter and 0.5 m high. In packed bed state, the porosity may be taken as 0.4. Ergun's equation for the above fluid-particle ... fluidisation velocity, VOM is (A) 12 mm/s (B) 16 mm/s (C) 24 mm/s (D) 28 mm/s
Last Answer : (B) 16 mm/s
Description : A bed of spherical particles (specific gravity 2.5) of uniform size 1500 μm is 0.5 m in diameter and 0.5 m high. In packed bed state, the porosity may be taken as 0.4. Ergun's equation for the above fluid-particle system ... What is the porosity of the fluidised bed? (A) 0.2 (B) 0.5 (C) 0.7 (D) 0.8
Last Answer : (C) 0.7
Description : A bar graph is drawn to the scale 1 cm = 4x units .The length of the bar represeting a quantity 1000 units is 1.25 cm . Find x
Last Answer : A bar graph is drawn to the scale 1 cm = 4x units .The length of the bar represeting a quantity 1000 units is 1.25 cm . Find x
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 .