The angles of a quadrilateral ABCD are `x^(@),(x+1)^(@),(x+2)^(@)` and `(x+3)^(@)`, taken in the same order. Then the quadrilateral ABCD is necessaril

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The angles of a quadrilateral ABCD are `x^(@),(x+1)^(@),(x+2)^(@)` and `(x+3 ... taken in the same order. Then the quadrilateral ABCD is necessarily a

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Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

Description : If angles A, B,C and D of the quadrilateral ABCD, taken in order are in the ratio 3 :7:6:4, then ABCD is a -Maths 9th

Last Answer : (c) Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4. Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°. 3x + 7x + 6x + 4x = 360° => 20x = 360° => x=360°/20° = 18°

Description : ABCD is a quadrilateral in which `/_A= 60^(@), /_ B= 70^(@), /_ C =110^(@)` and `/_ D =120^(@)`. The number of pairs of parallel lines is `"__________

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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

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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

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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 : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

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Description : Three Angles of a quadrilateral ABCD are equal.Is it a parallelogram? -Maths 9th

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Description : The side of a quadrilateral ABCD are 6cm,12cm,8cm,12cm,4cm (taken in oder) respectively and the angle between the 1st two side is a right angle. Find area of tiresome by herons fourmula -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 ... 1+6-√)cm2 =24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2 241+6cm2 .

Description : The side of a quadrilateral ABCD are 6cm,12cm,8cm,12cm,4cm (taken in oder) respectively and the angle between the 1st two side is a right angle. Find area of tiresome by herons fourmula -Maths 9th

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Description : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

Description : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

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 : 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

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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

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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 the following figure ABCD is a cyclic quadrilateral, the sum of degree measures of ∠A and ∠D is: (SOURCE: Fig. 3.7, Exercise 3.7, Chapter 3, PAIR of LINEAR EQUATIONS in TWO VARIABLES, NCERT, Class X) (a) 120° (b) 180° (c) 230°(d) 250°

Last Answer : (c) 230°

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

Description : 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

Last Answer : Solution: Let ABCD be a quadrilateral whose diagonals bisect each other at right angles. Given that, OA = OC OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90° To show that, if the ... a parallelogram. , ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle. Hence Proved.

Description : If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Last Answer : Quadrilateral will be Rhombus .

Description : If the diagonals of a quadrilateral bisect each other at right angles , then name the quadrilateral . -Maths 9th

Last Answer : Quadrilateral will be Rhombus .

Description : Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

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Description : If A, B, C, D are the successive angles of a cyclic quadrilateral, then what is cos A + cos B + cos C + cos D equal to: -Maths 9th

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Description : Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

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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.)

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 : The diagonals of a quadrilateral ABCD are perpendicular to each other. -Maths 9th

Last Answer : Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P,Q,R and S are mid points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS. To ... 90° Thus, PQRS is a parallelogram whose one angle is 90°. ∴ PQRS is a rectangle.

Description : BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

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)

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 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 : ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. -Maths 9th

Last Answer : According to question p rove that ∠CBD +∠CDB = 1/2 ∠BAD.

Description : The diagonals of a quadrilateral ABCD are perpendicular to each other. -Maths 9th

Last Answer : Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P,Q,R and S are mid points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS. To ... 90° Thus, PQRS is a parallelogram whose one angle is 90°. ∴ PQRS is a rectangle.

Description : BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

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)

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 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,

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Last Answer : According to question p rove that ∠CBD +∠CDB = 1/2 ∠BAD.

Description : Show that in a quadrilateral ABCD,AB + BC + CD + DA > AC + BD. -Maths 9th

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

Last Answer : Solution :- Parallelogram.

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 cyclic quadrilateral. Show that the incentres of the triangles ABC, BCD, CDA and DAB form a rectangle. -Maths 9th

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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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