What angles do the diagonals of a square cross each other?

1 Answer

Answer :

90 decgrees

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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 : 4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O. To show that, AC = BD AO = OC and ∠AOB = 90° Proof, In ΔABC and ΔBAD, AB = BA (Common) ∠ABC = ∠BAD = ... = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles

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

Last Answer : Solution :-

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

Last Answer : We have a quadrilateral ABCD such that angleO is the mid-point of AC and BD. Also AC ⊥ BD. Now, in ΔAOD and ΔAOB, we have AO = AO [Common] OD = OB [ ... i.e. The rhombus ABCD is having one angle equal to 90°. Thus, ABCD is a square.

Description : Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Proof: (i) In a ΔABC and ΔBAD, AB=AB ( common line) BC=AD ( opppsite sides of a square) ∠ABC=∠BAD ( = 90° ) ΔABC≅ΔBAD( By SAS property) AC=BD ( by CPCT). (ii) In a ΔOAD and ΔOCB, ... ) ( by CPCT ∠AOB+∠AOD=180° (linear pair) 2∠AOB=180° ∠AOB=∠AOD=90° ∴AC and BD bisect each other at right angles.

Description : Do diagonals of a trapezium cross at right angles?

Last Answer : No

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 .

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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.

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.

Description : ABCD is a square. Another square EFGH with the same area is placed on the square ABCD such that the point of intersection of diagonals of square -Maths 9th

Last Answer : (a) 32 (2 - √2)As is seen in the given figure, the sides of one square are parallel to the diagonals of another square. Also, square ABCD and EFGH have same area.⇒ Sides of square ABCD and square EFGH are 4 cm each. Let ... four Δs outside ABCD= 16 +16 (3 -2√2)= 16 + (4 - 2√2) = 32 (2 - √2) cm2.

Description : In a flow net (A) Flow lines and equipotential lines cross each other at right angles (B) Fields are rectangles whose length is twice the breadth (C) Smaller the dimensions of the field, ... gradient and velocity of flow through it (D) For homogeneous soil, the curves are smooth and circular

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Description : What is the amount of radiant energy received each second over each square meter that is at right angles to the sun’s rays at the top of the atmosphere?  a. 1400 J  b. 6000 J  c. 10000 J  d. 800 J

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Last Answer : Feel Free to Answer

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 : 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 : The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Last Answer : According to question parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°.

Description : Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer. -Maths 9th

Last Answer : No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other.

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.

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 : 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 : The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Last Answer : According to question parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°.

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.

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Last Answer : No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other.

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Last Answer : answer:

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Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : How many interior angles are in a square-based pyramid?

Last Answer : 20 angles

Description : Is ever rhombus with four right angles a square?

Last Answer : A square is technically a rhombus - but a rhombus is never a square. The definition of a rhombus is that it has two pairsof parallel sides. A square is defined as having all foursides equal,

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Description : Did you know that the diagonals of the faces of a cube can be joined to form the framework of a tetrahedron?

Last Answer : I just drew it. Amasing!

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°

Description : In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y. -Maths 9th

Last Answer : Since diagonals of a rhombus bisect each other at right angle . ∴ In △AOB , we have ∠OAB + ∠x + 90° = 180° ∠x = 180° - 90° - 35° [∵ ∠ OAB = 35°] = 55° Also, ∠DAO = ∠BAO = 35° ∴ ∠y + ∠DAO + ∠BAO + ∠x ... 180° ⇒ ∠y = 180° - 125° = 55° Hence the values of x and y are x = 55°, y = 55°.

Description : If the diagonals of a parallelogram are equal, then show that it is a rectangle. -Maths 9th

Last Answer : Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle . Proof : In △ABC and △ABD AB = AB [common] AC = BD [given] BC = AD [opp . sides of a | | gm] ⇒ △ABC ≅ △BAD [ ... ∵ ∠ABC = ∠BAD] ⇒ 2∠ABC = 180° ⇒ ∠ABC = 1 /2 180° = 90° Hence, parallelogram ABCD 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 : ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Last Answer : Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median ... AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 8 = 16 cm2

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)

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

Description : A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is -Maths 9th

Last Answer : The acute angle between the diagonals is given below.

Description : Diagonals of a rectangle are equal and perpendicular. Is this statement true ? -Maths 9th

Last Answer : No, diagonals of a rectangle are equal but need not be perpendicular.

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.

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 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.

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]