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 : 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
Last Answer : Area of the parallelogram
Description : The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th
Last Answer : Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.
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°
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 : 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
Description : A cyclic parallelogram having unequal adjacent sides is necessarily a: -Maths 9th
Last Answer : answer:
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
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) ...
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°.
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°
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 : 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 : 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 : 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 : 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 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 : The diagonals of a quadrilateral are equal.Is it neccessary a parallelogram? -Maths 9th
Last Answer : Answer :- No,diagonals of a parallelogram bisect each other but may or may not be equal.
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
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 : 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]
Description : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -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
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 : How many diagonals can be drawn in a polygon of 15 sides? -Maths 9th
Description : 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
Description : If A = i + 2j + 3k and B = 3i - 2j + k, then the area of parallelogram formed from these vectors as the adjacent sides will be
Last Answer : If A = i + 2j + 3k and B = 3i - 2j + k, then the area of parallelogram formed from these ... units (C) 6√3 square units (D) 8√3 square units
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.