Express vectors AC in terms of vectors AB and CB shown in the following figure.

Express vectors AC in terms of vectors AB and CB shown in the following figure.
by

1 Answer

Answer :

Express vector \(\overset\longrightarrow{AC}\) in terms of vectors \(\overset\longrightarrow{AB} ... {CB}\) shown in the following figure.
by
selected by

Related questions

`AB` is a wire of uniform resistance. The galvanometer `G` shows no deflection when the length `AC = 20 cm` and `CB = 80 cm`. The resistance `R` is eq

Description : `AB` is a wire of uniform resistance. The galvanometer `G` shows no deflection when the length `AC = 20 cm` and `CB = 80 cm`. The resistance `R` is eq

Last Answer : `AB` is a wire of uniform resistance. The galvanometer `G` shows no deflection when the length `AC = 20 cm ... ` B. `8Omega` C. `20Omega` D. `40Omega`

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

The order of the chemical reaction as shown in the bellow figure, whose rate equation is given as -rA = KCA 2 . CB is (A) 0 (B) 1 (C) 2 (D) 3

Description : The order of the chemical reaction as shown in the bellow figure, whose rate equation is given as -rA = KCA 2 . CB is (A) 0 (B) 1 (C) 2 (D) 3

Last Answer : (D) 3

What is the resultant of vectors shown in the figure below?

Description : What is the resultant of vectors shown in the figure below?

Last Answer : What is the resultant of vectors shown in the figure below?

The bearing of AB is 190° and that of CB is 260° 30'. The included angle ABC, is (A) 80° 30' (B) 99° 30' (C) 70° 30' (D) None of these

Description : The bearing of AB is 190° and that of CB is 260° 30'. The included angle ABC, is (A) 80° 30' (B) 99° 30' (C) 70° 30' (D) None of these

Last Answer : (C) 70° 30'

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.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

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.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

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, 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

Last Answer : answer:

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 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

Last Answer : answer:

In the figure, find teh perimeter of the shaded region where. ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively. -Maths 10th

Description : In the figure, find teh perimeter of the shaded region where. ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively. -Maths 10th

Last Answer : Given: Diameter of semicircle AEB= 2.8 cm RADIUS OF SEMICIRCLE(AEB)= 2.8/2= 1.4 cm Diameter of semicircle BFC=1.4 cm RADIUS OF SEMICIRCLE(BFC)= 1.4/2= 0.7 cm Diameter of semicircle ADC= 2.8+1.4 = ... ; .6 = 13.2 cm Hence, the perimeter of the shaded region is 13.2 cm HOPE THIS WILL HELP YOU...

In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units and AC=2 units What is the relationship among AB,BC and C

Description : In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units and AC=2 units What is the relationship among AB,BC and C

Last Answer : In the following figure, (not to scale) ABC is triangle, where AB=1 unit, BC=`sqrt(3)` units ... units What is the relationship among AB,BC and CA?

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 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.

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 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.

Figure abc is rectangle. segment AB is 6cm long , segment BC is 8 cm long , and segment AC is 10 cm long. what is the area of triangle abc?

Description : Figure abc is rectangle. segment AB is 6cm long , segment BC is 8 cm long , and segment AC is 10 cm long. what is the area of triangle abc?

Last Answer : 28

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

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

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

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

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

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

In the part of circuit shown in figure, match the following two columns for `V_(AB)`

Description : In the part of circuit shown in figure, match the following two columns for `V_(AB)`

Last Answer : In the part of circuit shown in figure, match the following two columns for `V_(AB)`

In the figure `V_(ab)` versus time graph along an inductor is shown. Match the following

Description : In the figure `V_(ab)` versus time graph along an inductor is shown. Match the following

Last Answer : In the figure `V_(ab)` versus time graph along an inductor is shown. Match the following

In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab) "at" t=5s`

Description : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab) "at" t=5s`

Last Answer : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. ... 50 V B. 35.5 V C. 46.5 V D. 40.2 V

In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab)=(V_(a)-V_(b)) at t=3s`

Description : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. Find `V_(ab)=(V_(a)-V_(b)) at t=3s`

Last Answer : In the circuit shown in figure q varies with time t as `q=(t^(2)=16)`. Here q is in coulomb and t in second. ... `18.5 V` C. `-25.5 V` D. `22.5 V`

Derive an expression for cross product of two vectors and express it in determinant form.

Description : Derive an expression for cross product of two vectors and express it in determinant form.

Last Answer : Derive an expression for cross product of two vectors and express it in determinant form.

An uncharged capacity ER C=`100muF` with a resistor is connected with AC source as shown in the figure If R= is `50Omega` and switch S is closed at t=

Description : An uncharged capacity ER C=`100muF` with a resistor is connected with AC source as shown in the figure If R= is `50Omega` and switch S is closed at t=

Last Answer : An uncharged capacity ER C=`100muF` with a resistor is connected with AC source as shown in the figure If R ... `(v_(A)-V_9B)` is k volt. Calculate K

The potentiak difference across a 2H inductor as a function of time is shown in the figure. At time t=0, current is zero. Current versus time graph ac

Description : The potentiak difference across a 2H inductor as a function of time is shown in the figure. At time t=0, current is zero. Current versus time graph ac

Last Answer : The potentiak difference across a 2H inductor as a function of time is shown in the figure. At time t=0, ... across the inductor will be A. B. C. D.

The device in figure 'B' shown in the illustration can be used to control an AC load yet maintain its high gain by ______________. EL-0091 A. connecting the load to the output side of the bridge B. reversing the current flow through the control winding C. connecting the load to the input side of the bridge D. reversing the current flow through the bias winding

Description : The device in figure 'B' shown in the illustration can be used to control an AC load yet maintain its high gain by ______________. EL-0091 A. connecting the load to the output side of the ... connecting the load to the input side of the bridge D. reversing the current flow through the bias winding

Last Answer : Answer: C

What type of current would flow through the load in figure 'B' shown in the illustration? EL-0091 A. AC only B. DC only C. AC or DC depending on control winding polarity D. AC or DC depending on instantaneous polarity of the source

Description : What type of current would flow through the load in figure 'B' shown in the illustration? EL-0091 A. AC only B. DC only C. AC or DC depending on control winding polarity D. AC or DC depending on instantaneous polarity of the source

Last Answer : Answer: B

The purpose of the bias winding in figure 'B' shown in the illustration is for ____________. EL-0091 A. changing the direction of current in the control winding B. setting the operating point of the device C. allowing the use of either AC or DC in the load circuit D. changing the direction of current through the load

Description : The purpose of the bias winding in figure 'B' shown in the illustration is for ____________. EL-0091 A. changing the direction of current in the control winding B. setting the operating point of the ... use of either AC or DC in the load circuit D. changing the direction of current through the load

Last Answer : Answer: B

For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is

Description : For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is

Last Answer : For the circuit shown in figure, the Boolean expression for the output y in terms of inputs P, Q, R and S is P+Q+R+S

Define and explain the following terms:  i. Zero vector (Null vector)  ii. Resultant vector  iii. Negative vectors iv. Equal vectors 

Description : Define and explain the following terms:  i. Zero vector (Null vector)  ii. Resultant vector  iii. Negative vectors iv. Equal vectors 

Last Answer : Define and explain the following terms: i. Zero vector (Null vector) ii. Resultant vector ... vectors iv. Equal vectors v. Position vector

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

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

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.

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

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.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

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

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.

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

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.

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

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........

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

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

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

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

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

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.

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,

AB and AC are two equal chords of a circle. -Maths 9th

Description : AB and AC are two equal chords of a circle. -Maths 9th

Last Answer : Given AS and AC are two equal chords whose centre is O. To prove Centre O lies on the bisector of ∠BAC. Construction Join SC, draw bisector AD of ∠BAC. Proof In ΔSAM and ΔCAM, AS = AC [given] ∠BAM = ∠CAM [given]

AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th

Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. -Maths 9th

Last Answer : According to question 4q 2 = p 2+ 3r 2.

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

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.

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

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.