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 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, `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 : 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 : 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 : 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 : 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 : 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 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 `Delta ABC`, if `/_A=80^(@)` and `AB=AC`, then `/_ B = ___________________`.
Last Answer : In `Delta ABC`, if `/_A=80^(@)` and `AB=AC`, then `/_ B = ___________________`.
Description : In the above figure (not to scale) , `AB||DE` and `EC||GF`. If `/_ EGF=100^(@)` and `/_ ECF=40^(@),` find the following . (i) `/_ABC` (ii) `/_GFC` (ii
Last Answer : In the above figure (not to scale) , `AB||DE` and `EC||GF`. If `/_ EGF=100^(@)` and `/_ ECF=40^ ... following . (i) `/_ABC` (ii) `/_GFC` (iii) `/_GDF`
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 : The Karnaugh map for a Boolean function is given as The simplified Boolean equation for the above Karnaugh Map is (A) AB + CD + AB’ + AD (B) AB + AC + AD + BCD (C) AB + AD + BC + ACD (D) AB + AC + BC + BCD
Last Answer : (B) AB + AC + AD + BCD
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 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 : 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 : 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
Last Answer : answer:
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 coordinates of point A , where AB is the diameter of a circle whose centre is (1,3) and B is (5,4),are: (a) (-3,2) (b) (4,-9) (c) (2,-3) (d) (1/2,-2)
Last Answer : (a) (-3,2)
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 : In the following figure, if `/_ AOB =60^(@)` then `/_ ACB=30^(@)`. [True `//` False `//` Cannot say ]
Last Answer : In the following figure, if `/_ AOB =60^(@)` then `/_ ACB=30^(@)`. [True `//` False `//` Cannot say ]
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 : ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm
Last Answer : (a) 5cm
Description : In given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then (a) DE || BC (b) DF || AC (c) EF || AB (d) none of
Last Answer : (c) EF || AB
Description : ABCD is a quadrilateral in which `/_A= 60^(@), /_ B= 70^(@), /_ C =110^(@)` and `/_ D =120^(@)`. The number of pairs of parallel lines is `"__________
Last Answer : ABCD is a quadrilateral in which `/_A= 60^(@), /_ B= 70^(@), /_ C =110^(@)` and ... )`. The number of pairs of parallel lines is `"________________"`.
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 : 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 : In the figure below ( not to scale) `bar(CD)||bar(RS) /_EMG=90^(@),/_GMD= gamma^(@),/_CME= x^(@) ` and `gamma^(@)=(x^(@))/(2)`. `/_ FNS : /_FNR `is
Last Answer : In the figure below ( not to scale) `bar(CD)||bar(RS) /_EMG=90^(@),/_GMD= gamma^(@),/_CME= x^(@) ` and `gamma^(@)=(x^(@))/(2)`. `/_ FNS : /_FNR `is
Description : In the figure below ( not to scale), ABC is a straight line. If `/_ FBE =60^(@), /_ CBG =120^(@), /_ ABG = x^(@), /_ ABF= gamm^(@)` and `/_ CBE = z^(@
Last Answer : In the figure below ( not to scale), ABC is a straight line. If `/_ FBE =60^(@), /_ CBG =120^(@), /_ ABG ... )`, then `( x^(@)+z^(@)) : gamma^(@)` is
Description : In the above figure ( not to scale ) the sides BA,BC and CA of` Delta ABC` are produced to D,F, and E respectively such that `/_ ACF= 120^(@)` and `/_
Last Answer : In the above figure ( not to scale ) the sides BA,BC and CA of` Delta ABC` are produced to D,F, and E ... /_ BAE= 150^(@)`. Then `/_ ABC = ________`.
Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th
Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th
Description : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@)` and `/_ ABO=20^(@)`. If `/_ OCB=(1)/(2) /_ A
Last Answer : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@) ... OCB=(1)/(2) /_ ACO,` then find `/_ BOC.`
Description : In the adjoining figure, if ∠BAC = 90° and AD ⊥ BC, then (а) BD.CD = BC2 (b) AB.AC = BC2 (c) BD.CD = AD2 (d) AB.AC = AD2
Last Answer : (c) BD.CD = AD2
Description : In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. If the value of π is taken 3, what is the perimeter of the shaded region? (a) 58 cm (b) 116 cm (c) 29 cm (d) 156 cm
Last Answer : (a) 58 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 given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th
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 : 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 : 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.