The area of the parallelogram ABCD is 90 cm2. -Maths 9th

The area of the parallelogram ABCD is 90 cm2. -Maths 9th
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Given, area of parallelogram, ABCD = 90 cm2 1.We know that, parallelograms on the same base and between the same parallel are equal in areas. Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF. So, ar (ΔBEF) = ar (ABCD) = 90 cm2 2.We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is equal to half of the area of the parallelogram. Here, ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD. So, ar (ΔABD) = 1/2 ar (ABCD) = 1/2 x 90 = 45 cm2 [∴ ar (ABCD) = 90 cm2] 3.Here, ABEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF. ar (ΔBEF) = 1/2 ar (ABEF) = 1/2 x 90 = 45 cm2 [∴ ar (ABEF) = 90 cm2, from part (i)]
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The area of the parallelogram ABCD is 90 cm2. -Maths 9th

Description : The area of the parallelogram ABCD is 90 cm2. -Maths 9th

Last Answer : Given, area of parallelogram, ABCD = 90 cm2 1.We know that, parallelograms on the same base and between the same parallel are equal in areas. Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB ... (ABEF) = 1/2 x 90 = 45 cm2 [∴ ar (ABEF) = 90 cm2, from part (i)]

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Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

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Description : In the figure, the area of parallelogram ABCD is -Maths 9th

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Description : In figure, if parallelogram ABCD and rectangle ABEM are of equal area, then -Maths 9th

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ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

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P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

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In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

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ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

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

ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

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ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

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ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

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The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

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

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

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Diagonals AC and BD of a parallelogram ABCD intersect each other at O. -Maths 9th

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Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

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Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

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In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Description : In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

The diagonals AC and BD of parallelogram ABCD intersect at the point O. -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°

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD . -Maths 9th

Last Answer : In gm ABCD , AP and CQ are perpendicular from the vertices A and C on diagonal BD. Show that : (i) AAPB ≅ ACQD (ii) AP = CQ .

ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Description : ABCD is a parallelogram and line segments AX, CY bisect the angles A and C, respectively. -Maths 9th

Last Answer : Since opposite angles are equal in a parallelogram . Therefore , in parallelogram ABCD , we have ∠A = ∠C ⇒ 1 / 2 ∠A = 1 / 2 ∠C ⇒ ∠1 = ∠2 ---- i) [∵ AX and CY are bisectors of ∠A and ∠C ... intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal . ∴ AX | | CY .

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 .

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

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...

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

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

ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

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

ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Last Answer : In △ADF and △ECF , we have ∠ADF = ∠ECF [alt.int.∠s] AD = EC [ ∵ AD = BC and BC = EC] ∠DFA = ∠CFE [vert. opp. ∠s] ∴ By AAS congruence rule , △ADF ≅ △ECF ⇒ DF = CF [c.p.c.t.] ⇒ ar(△ADF) = ar(△ECF) ... 3 = 6 cm2 [∵ar(△DFB) = 3 cm2] Thus, ar(||gm ABCD) = 2 ar(△BDC) = 2 6 = 12 cm2

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)