A cyclic parallelogram having unequal adjacent sides is necessarily a: -Maths 9th

A cyclic parallelogram having unequal adjacent sides is necessarily a: -Maths 9th
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Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

Description : Prove that the figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. -Maths 9th

Last Answer : Solution :-

The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Description : The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Last Answer : For △BCD: Let a = 17 cm, b = 12 cm, c = 25 cm So its semi-perimeter, s = (a + b + c)/2 = (17 + 12 + 25)/2 = 27 cm ∴ Area of △BCD = root under(√(s -a)(s - b)(s - c)) = ... △BCD = 2 x 90 = 180 cm2 Also, area of parallelogram ABCD = DC x AE ∴ 180 = 12 x AE ⇒ AE = 180/12 = 15 cm

The adjacent sides of a parallelogram are 2a and a. If the angle between them is 60°, then one of the diagonals of the parallelogram is -Maths 9th

Description : The adjacent sides of a parallelogram are 2a and a. If the angle between them is 60°, then one of the diagonals of the parallelogram is -Maths 9th

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If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Description : If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Last Answer : Let the two adjacent angles be x° and 2x° . In a parallelogram, sum of the adjacent angles are 180°. ∴ x + 2x = 180° ⇒ 3x = 180° ⇒ x = 60° Thus , the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

If an angle of a parallelogram is two - third of its adjacent angle , then find the smallest angle of the parallelogram . -Maths 9th

Description : If an angle of a parallelogram is two - third of its adjacent angle , then find the smallest angle of the parallelogram . -Maths 9th

Last Answer : In a parallelogram ABCD, Let ∠A be x and ∠B be 2x / 3 ∴ ∠A + ∠B = 180° ⇒ x + 2x / 3 = 180° ⇒ 5x / 3 = 180° ⇒ x° = 180° × 3 / 5 = 108° ∠A = 108° , ∠B = 2 / 3 × 108° = 72°

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Description : If one of a parallelogram is twice of its adjacent angle , find the angles of the parallelogram . -Maths 9th

Last Answer : Let the two adjacent angles be x° and 2x° . In a parallelogram, sum of the adjacent angles are 180°. ∴ x + 2x = 180° ⇒ 3x = 180° ⇒ x = 60° Thus , the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

If an angle of a parallelogram is two - third of its adjacent angle , then find the smallest angle of the parallelogram . -Maths 9th

Description : If an angle of a parallelogram is two - third of its adjacent angle , then find the smallest angle of the parallelogram . -Maths 9th

Last Answer : In a parallelogram ABCD, Let ∠A be x and ∠B be 2x / 3 ∴ ∠A + ∠B = 180° ⇒ x + 2x / 3 = 180° ⇒ 5x / 3 = 180° ⇒ x° = 180° × 3 / 5 = 108° ∠A = 108° , ∠B = 2 / 3 × 108° = 72°

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

Description : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

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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

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

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Description : If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Last Answer : Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, ∠OAD = ∠OBC and ∠ODA = ∠OCB [since, same segments subtends equal angle to the circle] AB = BC [ ... is DOC on both sides, we get ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC ⇒ ΔADC ≅ ΔBCD AC = BD [by CPCT]

If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Description : If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Last Answer : Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

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

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

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Description : If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. -Maths 9th

Last Answer : Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, ∠OAD = ∠OBC and ∠ODA = ∠OCB [since, same segments subtends equal angle to the circle] AB = BC [ ... is DOC on both sides, we get ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC ⇒ ΔADC ≅ ΔBCD AC = BD [by CPCT]

If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Description : If non-parallel sides of a trapezium are equal, prove that it is cyclic. -Maths 9th

Last Answer : Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : We know that, each angle of a rectangle is right angle (i.e., 90°) and its opposite sides are equal and parallel. To construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm, use the ... AD and CD. Thus, ABCD is the required rectangle with adjacent sides of length 5 cm and 3.5 cm.

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Description : If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Last Answer : Solution :-

A rhombus has sides of length 1 and area 1/2 Find the angle between the two adjacent sides of the rhombus -Maths 9th

Description : A rhombus has sides of length 1 and area 1/2 Find the angle between the two adjacent sides of the rhombus -Maths 9th

Last Answer : answer:

The adjacent sides of a rectangle are 16 cm and 8 cm. Find the area of the rectangle. -Maths 9th

Description : The adjacent sides of a rectangle are 16 cm and 8 cm. Find the area of the rectangle. -Maths 9th

Last Answer : area of rectangle is l×b 16×8 =128cm sq .area of rectangle is 128cm sq

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

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

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.

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 .

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

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 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

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 mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

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

Last Answer : According to question prove that the area of the 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 .

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

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 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

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 mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

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

Last Answer : According to question prove that the area of the parallelogram

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : Area of the parallelogram

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 : 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

Last Answer : Solution :-

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 :-

In parallelogram PQRS, PQ = 10cm. The altitudes corresponding to the sides PQ and SP are respectively 6cm and 8cm. Find SP. -Maths 9th

Description : In parallelogram PQRS, PQ = 10cm. The altitudes corresponding to the sides PQ and SP are respectively 6cm and 8cm. Find SP. -Maths 9th

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If two parallelogram PQAD and PQBC arw on the opposite sides of PQ prove that ABCD is a paralellogram -Maths 9th

Description : If two parallelogram PQAD and PQBC arw on the opposite sides of PQ prove that ABCD is a paralellogram -Maths 9th

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X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

Description : X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -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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