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 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 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 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 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 : 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 uniform ring of resistance `R` there are two points `A` and `B` such that `/_ ACB = theta`, where `C` is the centre of the ring. The equivalent r
Last Answer : In a uniform ring of resistance `R` there are two points `A` and `B` such that `/_ ACB = theta`, where `C` ... ))` D. `(R )/(4pi^(2))(2pi-theta)theta`
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 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 : 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 : 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 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 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 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 : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th
Last Answer : NEED ANSWER
Last Answer : This answer was deleted by our moderators...
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 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 : 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 : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of radius `a`. Net magnetic field at centre
Last Answer : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of ... (ni)/(a)mu_(0)tan.(pi)/(n)` D.
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 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.
Description : If points A (4, 3) and B (x, 5) are on the circle with centre O (2, 3), find the value of x. -Maths 9th
Last Answer : Since A and B lie on the circle having centre O. Therefore, OA = OB root(4−2)2+(3−3)2=(root(x−2)2+(5−3)22=root(x−2)2+4(x−2)2+4=4 (x−2)2=0 x−2=0 x=2 hope it helps thank you
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 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 : 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 : A current l flows through a closed loop as shown in figure .The magnetic field at the centre O is
Last Answer : A current l flows through a closed loop as shown in figure .The magnetic field at the centre O is A. ... )l)/(2piR)(theta+sintheta)` D. None of these
Description : In the figure shown there are two semicircles of radii `r_(1)` and `r_(2)` in which a current `i` is flowing. The magnetic induction at the centre `O`
Last Answer : In the figure shown there are two semicircles of radii `r_(1)` and `r_(2)` in which a current `i` is flowing. The ... 4)[(r_(1)-r_(2))/(r_(1)r_(2))]`
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 : 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 `Delta ABC, /_ A= /_C= 50^(@)`. The longest side of `Delta ABC `is `__________________`.
Last Answer : In `Delta ABC, /_ A= /_C= 50^(@)`. The longest side of `Delta ABC `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 `Delta ABC`, if `/_ A lt /_B lt 45^(@)`, then ABC is a`//` an `"__________________"` triangle.
Last Answer : In `Delta ABC`, if `/_ A lt /_B lt 45^(@)`, then ABC is a`//` an `"__________________"` triangle.
Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th
Last Answer : According to question ∠BAC = ∠BDC.
Description : Why use the VCB at High Transmission System? Why can't use ACB?
Last Answer : Actually the thing is vacuum has high arc quenching property compare to air because in VCB, the die electric strengths equal to 8 times of air. That y always vacuum used as in HT breaker and air used as in LT.