How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th
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How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Description : How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm. -Maths 9th

Last Answer : According to question ABCD is a square with diagonal 44 cm.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

A kite in the shape of a square with a diagonal 32 cm -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm -Maths 9th

Last Answer : As the diagonals of a square bisect each other at right angle ∴ AM = DM = 32/2 = 16 cm Area of shade I = Area of shade II = Area of △ABD = 1/2 x AD x BM = 1/2 x 32 x 16 = 256 cm2 For the area of shade III ... - a2) = 8/4 root under( √4(6)2 - 82)) = 2 root under( √144 - 64) = 2 √80 = 85 cm2

A kite in the shape of a square with a diagonal 32 cm -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm -Maths 9th

Last Answer : As the diagonals of a square are equal bisect each other at right angle ∴ AD = BC = 32 cm and AM = DM = 32/2 = 16 cm Area of shade I = Area of shade II = Area of △ABD = 1/2 x AD x BM = ... cm2 Area of sheet of shade III required for making 40 kites = 40 x 8 √5 = 320 √5 cm2 Social, loving, caring.

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

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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...

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Last Answer : (i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A. ∴ ∠DAC=∠BAC ---- ( 1 ) Now, AB∥DC and AC as traversal, ∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 ) AD∥BC and AAC as traversal, ∴ ∠DAC= ... ---- ( 2 ) From ( 1 ) and ( 2 ), ⇒ AB=BC=CD=DA Hence, ABCD is a rhombus.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) Diagonal BD bisects ∠B as well as ∠D. -Maths 9th

Description : ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that: (i) ABCD is a square (ii) Diagonal BD bisects ∠B as well as ∠D. -Maths 9th

Last Answer : Solution: (i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C) ⇒ AD = CD (Sides opposite to equal angles of a triangle are equal) also, CD = AB (Opposite sides of a rectangle) ,AB = BC = CD = AD Thus ... interior angles) ⇒ ∠CBD = ∠ABD Thus, BD bisects ∠B Now, ∠CBD = ∠ADB ⇒ ∠CDB = ∠ADB Thus, BD bisects ∠D

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm

Description : ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm

Last Answer : (a) 5cm

ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Description : ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Last Answer : (b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if a' is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{ ... of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th

Description : The curved surface of a cylinder is developed into a square whose diagonal is 2√2 cm. The area of the base of the cylinder (in cm^2) is -Maths 9th

Last Answer : answer:

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

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.

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

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.

A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. The crease will divide BC in the ratio -Maths 9th

Description : A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. The crease will divide BC in the ratio -Maths 9th

Last Answer : answer:

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

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.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

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

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.

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Description : ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ -Maths 9th

Last Answer : Q Solution: (i) In ΔAPB and ΔCQD, ∠ABP = ∠CDQ (Alternate interior angles) ∠APB = ∠CQD (= 90o as AP and CQ are perpendiculars) AB = CD (ABCD is a parallelogram) , ΔAPB ≅ ΔCQD [AAS congruency] (ii) As ΔAPB ≅ ΔCQD. , AP = CQ [CPCT]

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Last Answer : . Solution: (i) In ΔADC and ΔCBA, AD = CB (Opposite sides of a parallelogram) DC = BA (Opposite sides of a parallelogram) AC = CA (Common Side) , ΔADC ≅ ΔCBA [SSS congruency] Thus, ∠ACD = ∠CAB by ... are equal) Also, AB = BC = CD = DA (Opposite sides of a parallelogram) Thus, ABCD is a rhombus.

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 .

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

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

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.

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 .

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

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

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.

In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Description : In Fig.8.38, AM and CN are perpendiculars to the diagonal BD of a paralelogram ABCD.Prove that AM = CN. -Maths 9th

Last Answer : Solution :-

The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Description : The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Last Answer : answer:

(–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Description : (–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Last Answer : 3\(x\) - 2y + 5 = 0 ⇒ -2y = -3\(x\) - 5 ⇒ y = \(rac{3}{2}\)\(x\) + \(rac{5}{2}\)On comparing with y = m\(x\) + c, we see that slope of given line = \(rac{3}{2}\)As the required line is perpendicular to the given line, ... - 4)⇒ 3(y - 5) = - 2\(x\) + 8 ⇒ 3y - 15 = -2\(x\) + 8 ⇒ 3y + 2\(x\) - 23 = 0

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

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

In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Last Answer : answer:

O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).

O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Description : O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th

Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).

The diagonal of a square is 10cm. Find its area. -Maths 9th

Description : The diagonal of a square is 10cm. Find its area. -Maths 9th

Last Answer : Solution :-

The diagonal of a square A is (a + b). The diagonal of a square whose area is twice the area of square A is. -Maths 9th

Description : The diagonal of a square A is (a + b). The diagonal of a square whose area is twice the area of square A is. -Maths 9th

Last Answer : (c) (a+b)√2.Area of new square = 2 (Area of the square A)= \(2 imesrac{ ext{(diagonal)}^2}{2}=2 imesrac{(a+b)^2}{2}= (a+b)^2\)∴ Each side of new square = (a + b) ⇒ Diagonal of new square = side √2 = (a+b)√2.

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th

Last Answer : Cost of painting =

The area of a rhombus 10 cm2 .If one if its diagonal is 4 cm,then find the other diagonal. -Maths 9th

Description : The area of a rhombus 10 cm2 .If one if its diagonal is 4 cm,then find the other diagonal. -Maths 9th

Last Answer : hope its clear

If the side of a rhombus is 10 cm and the diagonal is 16 cm,...... -Maths 9th

Description : If the side of a rhombus is 10 cm and the diagonal is 16 cm,...... -Maths 9th

Last Answer : True. AC = 16 cm BD = ? and AB = 10 cm As the diagonals of a rhombus bisect each other at 90° ∴ OA = 1/2AC = 1/2 x 16 = 8cm OB = 1/2 BD ∴ OA2 + OB2 = AB2 82 + OB2 = 102 ⇒ OB2 = 100 - 64 OB2 = 36 ... ∴ BD = 2 x OB = 2 x 6 = 12 cm Area of rhombus = 1/2 AC x BD = 1/2 x 16 x 12 = 96cm 2

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

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

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°.

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

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

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

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

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

Last Answer : (c) We know that, area of parallelogram is the product of its any side and the corresponding altitude (or height). Here, when AB is base, then height is DL. Area of parallelogram = AB x DL and when AD is ... = DC x DL and when BC is base, then height is not given. Hence, option (c) is correct.

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

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

Last Answer : (c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB (i) We know that, the perpendicular distance between two ... AB+BE + EM+ AM [∴ CD = AB = EM] Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO). -Maths 9th

Description : In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO). -Maths 9th

Last Answer : Given, ABCD and AEFD are two parallelograms. To prove ar (APEA) = ar (AQFD) Proof In quadrilateral PQDA, AP || DQ [since, in parallelogram ABCD, AB || CD ] and PQ || AD [since, in parallelogram AEFD ... APFD) = ar (parallelogram AEFD) - ar (quadrilateral APFD) ⇒ ar (AQFD) = ar (APEA) Hence proved.