When a and b are positive reals, prove that (a^4+b^4)(a^5+b^5) is -Maths 9th

When a and b are positive reals, prove that (a^4+b^4)(a^5+b^5) is -Maths 9th
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If a^2 + b^2 + c^2 = 1, x^2 + y^2 + z^2 = 1, where a, b, c, x, y, z are positive reals then ax + by + cz is -Maths 9th

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If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

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If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

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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

Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

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Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).

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Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

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Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

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Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

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Last Answer : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ΔABD, S and P are the mid- ... SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other.

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

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

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Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

An equation to prove -Maths 9th

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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

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.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

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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

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.

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If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

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Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

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

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.

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

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ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

If a and b are two rational numbers, prove that a + b, a - b, ab are rational numbers. -Maths 9th

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Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

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Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

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ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

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)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

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Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). -Maths 9th

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If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

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Prove that through a given point, we can draw only one perpendicular to a given line. -Maths 9th

Description : Prove that through a given point, we can draw only one perpendicular to a given line. -Maths 9th

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Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. -Maths 9th

Description : Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. -Maths 9th

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Prove that a triangle must have at least two acute angles. -Maths 9th

Description : Prove that a triangle must have at least two acute angles. -Maths 9th

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state and prove angle sum property of triangle -Maths 9th

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P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Description : Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Last Answer : Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC ... and ∠PSR = 90° Thus, PQRS is a quadrilateral whose each angle is 90°. Hence, PQRS is a rectangle.

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

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Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

Description : If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). -Maths 9th

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In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO). -Maths 9th

Description : In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO). -Maths 9th

Last Answer : Given, ABCD and AEFD are two parallelograms. To prove ar (APEA) = ar (AQFD) Proof In quadrilateral PQDA, AP || DQ [since, in parallelogram ABCD, AB || CD ] and PQ || AD [since, in parallelogram AEFD ... APFD) = ar (parallelogram AEFD) - ar (quadrilateral APFD) ⇒ ar (AQFD) = ar (APEA) Hence proved.

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

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

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Description : A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Last Answer : According to question prove that the perpendicular bisectors of AB, BC and CA are concurrent.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : Given : E and F are mid points of 2 chords AB and CD respectively. Line EF passes through centre. To prove : AB||CD ∠ OFC = ∠ OEA = 90° as line drawn through the centre to bisect the ... EF as traversal for lines AB and CD as alternate interior angles on same side are equal. Therefore, AB || CD