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 : 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 : 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 : 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 : 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 : 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 : 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 : 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 : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th
Last Answer : clear .
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 : 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 : 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 : 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 : 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 : 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 : 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 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 : 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 Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th
Last Answer : Solution :-
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 = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)
Description : In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th
Last Answer : answer:
Description : In a parallelogram ABCD, AE is perpendicular to DC and CF is perpendicular to AD. If AB = 10 cm, AE = 6 cm and CF = 8 cm, then find AD. -Maths 9th
Last Answer : Given, Parallelogram ABCD pAE = 8cm AB = 16cm CF = 10cm In a parallelogram, we know that opposite sides are equal. Therefore, CD = AB = 16cm To find the value of AD, the base is multiplied with height. Area of parallelogram = b x h 16 x 8 = AD x 10 128 = 10AD AD = 12.8cm
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 : 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 : 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 : 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)
Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th
Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)
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 .
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 : 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 : 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 : 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 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 : ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. -Maths 9th
Last Answer : According to question prove that ar (DCYX) = 7/9 ar (XYBA).