The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

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According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

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Description : The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only, if -Maths 9th

Last Answer : According to question mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only.

Description : Name the quadrilateral formed by joining the mid - points of the sides of any quadrilateral ABCD. -Maths 9th

Last Answer : Solution :- Parallelogram.

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

Description : The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if -Maths 9th

Last Answer : According to question the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle,

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 :-

Description : ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th

Last Answer : According to the given statement, the figure will be a shown alongside; using mid-point theorem: In △ABC,PQ∥AC and PQ=21 AC .......(1) In △ADC,SR∥AC and SR=21 AC .... ... are perpendicular to each other) ∴PQ⊥QR(angle between two lines = angle between their parallels) Hence PQRS is a rectangle.

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

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,

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,

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

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

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

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Now, In ΔACD, R and S are the mid points of CD and DA respectively. , ... , PQRS is parallelogram. PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

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

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.

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 : 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.

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.

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.

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Description : Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

Description : Show that the quadrilateral formed by joining the consecutive sides of a square is also a square. -Maths 9th

Last Answer : According to question quadrilateral formed by joining the consecutive sides of a square is also a square.

Description : Show that the quadrilateral formed by joining the consecutive sides of a square is also a square. -Maths 9th

Last Answer : According to question quadrilateral formed by joining the consecutive sides of a square is also a square.

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.

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.

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

Description : A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

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Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

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Description : Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

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Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

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.

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

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.

Description : Prove that the line segment joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and equal to half of their difference. -Maths 9th

Last Answer : In a parallelogram ABCD, the bisector of ∠ A also bisects BC at X.Prove that AD = 2AB.

Description : Proof to show that the quadrilateral formed by joining the midpoints of a square is a square. -Maths 9th

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Description : Proof to show that the quadrilateral formed by joining the midpoints of a square is a square. -Maths 9th

Last Answer : This answer was deleted by our moderators...

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 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

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

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

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

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

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

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

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.