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

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
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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 : 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

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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.

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 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 : 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

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E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

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ABCD is a trapezium where AB and CD are non-parallel sides. If the vertices A, B, C and D are concyclic, then -Maths 9th

Description : ABCD is a trapezium where AB and CD are non-parallel sides. If the vertices A, B, C and D are concyclic, then -Maths 9th

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The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Description : The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

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ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. -Maths 9th

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ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm . E and F are the mid - points of the non - parallel sides . -Maths 9th

Last Answer : Clearly, EF = AB + DC / 2 = a + b / 2 Let h be the height , then ar(Trap. ABFE) : ar(Trap. EFCD) ⇒ 1/2 [a + (a+b / 2)] × h : 1/2 [b + (a+b / 2)] × h ⇒ 2a + a + b / 2 : 2b + a + b / 2 ⇒ 3a + b : 3b + a

ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. -Maths 9th

Description : ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. -Maths 9th

Last Answer : Solution of this question

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)

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)

The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

Description : The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

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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 a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th

Description : In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th

Last Answer : answer:

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

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.

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

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

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,

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Description : P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Last Answer : Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively. To show PRQS is a parallelogram. Proof Since, ABCD is a parallelogram. AB||CD ⇒ AP || QC

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

Last Answer : Join AQ and PC . Since ABCD is a parallelogram . ⇒ AB | | DC ⇒ AP | | QC ∵ AP and QC are parts of AB and DC respectively] Also, AP = CQ [given] Thus, APCQ is a parallelogram . We know that diagonals of a parallelogram bisect each other . Hence AC and PQ bisect each other .

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,

P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Description : P and Q are the mid-points of the opposite sides AB and CD of a parallelogram ABCD. -Maths 9th

Last Answer : Given In a parallelogram ABCD, P and Q are the mid-points of AS and CD, respectively. To show PRQS is a parallelogram. Proof Since, ABCD is a parallelogram. AB||CD ⇒ AP || QC

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 : 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

Last Answer : Solution :-

If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Description : If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Here, we are joining A and C. In ΔABC P is the mid point of AB Q is the mid point of BC PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and ... RS=PS=RQ[All sides are equal] ∴ PQRS is a parallelogram with all sides equal ∴ So PQRS is a rhombus.

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

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. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -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.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.

The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Description : The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. -Maths 9th

Description : Diagonals AC and BD of a parallelogram ABCD intersect each other at O. -Maths 9th

Last Answer : According to parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Description : The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

Diagonals AC and BD of a parallelogram ABCD intersect each other at O. -Maths 9th

Description : Diagonals AC and BD of a parallelogram ABCD intersect each other at O. -Maths 9th

Last Answer : According to parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Last Answer : NEED ANSWER

Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Last Answer : Its given ABCD is a IIgram and AC and BD are its diagonals intersecting at point O. . Given : angle BOC = 900 angle BDC = 500 To find : angle OAB Answer : i) ...

ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Description : ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Last Answer : answer:

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Description : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Last Answer : Draw AM ⟂ BD and CL ⟂ BD. Now, ar(△APB) × ar(△CPD) = {1/2 PB × AM} × {1/2 DP × CL} = {1/2 PB × CL} × {1/2 DP × AM} ar(△BPC) × ar(△APD) Hence, ar(△APB) × ar(△CPD) = ar(△APD) × ar(△BPC)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Description : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Last Answer : Draw AM ⟂ BD and CL ⟂ BD. Now, ar(△APB) × ar(△CPD) = {1/2 PB × AM} × {1/2 DP × CL} = {1/2 PB × CL} × {1/2 DP × AM} ar(△BPC) × ar(△APD) Hence, ar(△APB) × ar(△CPD) = ar(△APD) × ar(△BPC)

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

Description : 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

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.

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

Description : 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

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.