Description : Two circles with centre O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through B intersecting the circles at P and Q. Prove that PQ = 2 OO'. -Maths 9th
Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ 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 : 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 : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th
Last Answer : According to question prove that PB = PD.
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 : In Fig. 10.20, two circles intersects at two points A and B.AD and AC are diameters to the circles. Prove that B lies on the line A segment DC. -Maths 9th
Last Answer : Solution :- Jion AB ∠ABD = 90° (Angle in a semicircle) Similarly, ∠ABC = 90° So, ∠ABD + ∠ABC = 90° + 90° = 180° Therefore,DBC is a line i.e.,B lies on the segment DC.
Description : Two congruent circles intersect each other at point A and B.Through A any line segment PAQ is drawn so that P,Q lie on the two circles.Prove that BP = BQ. -Maths 9th
Last Answer : Solution :- Let, O and O' be the centres of two congruent circles. As, AB is the common chord of these circles. ∴ ACB = ADB As congruent arcs subtent equal angles at the centre. ∠AOB = ∠AO'B ⇒ 1/2∠AOB = 1/2∠AO'B ⇒ ∠BPA = ∠BQA ⇒ BP = BQ (Sides opposite to equal angles)
Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th
Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved
Description : In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th
Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th
Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th
Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.
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 : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th
Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF=ABFB=1 from (1) ⇒LF=LD so, DF=2DL
Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th
Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.
Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th
Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.
Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th
Last Answer : Solution :-
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 : Two circles with centres O and O’ intersect at two points A and B. -Maths 9th
Last Answer : Given, draw two circles having centres O and O’ intersect at points A and 8.
Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th
Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.
Description : ABCD is a parallelogram.The circle through A,B and C intersect CD (produce if necessary) at E.Prove that AE = AD. -Maths 9th
Last Answer : Solution :- ∠ABC + ∠AEC = 1800 (Opposite angles of cyclic quadrilateral) .. . (i) ∠ADE + ∠ADC = 1800 (Linear pair) But ∠ADC = ∠ABC (Opposite angles of ||gm) ∴ ∠ADE + ∠ABC = 1800 .. (ii) ... ∠ABC + ∠AEC = ∠ADE + ∠ABC ⇒ ∠AEC = ∠ADE ⇒ AD = AE (sides opposite to equal angles)
Description : In Fig. 10.33, if OA = 10cm, AB = 16 cm and OD perpendicular to AB. Find the value of CD. -Maths 9th
Last Answer : Solution :- As OD is perpendicular to AB ⇒ AC = AB (Perpendicular from the centre to the chord bisects the chord) ∴ AC = AB/2 = 8cm In right △OCA, OA2 = AC2 + OC2 (102) = 82 + OC2 OC2 = 100 - 64 OC2 = 36 ... = 6cm CD = OD - OC = 10 - 6 = 4cm [∴ OA = OD = 10cm (radii)]
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 : 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 : 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 Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th
Last Answer : hope its clearhope its clear
Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th
Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.
Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th
Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.
Description : In Fig. 6.17, AB| |CD. Find the value of x. -Maths 9th
Description : In Fig. 6.19, AB| |CD. Determine x. -Maths 9th
Description : In Fig. 6.20, find x if AB| |CD| |EF. -Maths 9th
Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th
Description : ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th
Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th
Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.
Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th
Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)
Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th
Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.
Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th
Last Answer : Solution of this question
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 : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th
Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.