If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter -Maths 9th

If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter -Maths 9th
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The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

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, 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

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th

Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.

If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th

Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Description : If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th

Last Answer : Solution :-

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

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.

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

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.

Prove that if the opposire anles of a parallelogram are equal then it is a rectangle -Maths 9th

Description : Prove that if the opposire anles of a parallelogram are equal then it is a rectangle -Maths 9th

Last Answer : This answer was deleted by our moderators...

Calculate the perimeter of a rectangle whose area is 25x2 – 35x + 12. -Maths 9th

Description : Calculate the perimeter of a rectangle whose area is 25x2 – 35x + 12. -Maths 9th

Last Answer : Given, Area of rectangle = 25x2 - 35x + 12 We know, area of rectangle = length breadth So, by factoring 25x2 - 35x + 12, the length and breadth can be obtained. 25x2 - 35x + 12 = 25x2 - 15x - 20x + 12 => 25x2 - 35x + ... )] = 2(5x - 3 + 5x - 4) = 2(10x - 7) = 20x - 14 So, the perimeter = 20x - 14

Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 cm²

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

A triangle and a parallelogram have the same base and the same area. -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. -Maths 9th

Last Answer : Let a = 26 cm, b = 28 cm, c = 30 cm ∴ s = (a + b + c)/2 = (26 + 28 + 30)/2 = 84/2 = 42 ∴ Area of triangle = root under (√s(s - a )(s - b)(s - c) = root ... cm2 Now, Area of parallelogram = Area of triangle ⇒ Base x height = 336 ⇒ 28 x height = 336 ⇒ height = 336/28 = 12 cm

The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Description : Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Last Answer : Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC ... and ∠PSR = 90° Thus, PQRS is a quadrilateral whose each angle is 90°. Hence, PQRS is a rectangle.

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Description : Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Last Answer : Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC ... and ∠PSR = 90° Thus, PQRS is a quadrilateral whose each angle is 90°. Hence, PQRS is a rectangle.

Prove that the bisector of the angles of a parallelogram enclose a rectangle. -Maths 9th

Description : Prove that the bisector of the angles of a parallelogram enclose a rectangle. -Maths 9th

Last Answer : Solution :-

Prove that the angle bisectors of a parallelogram form a rectangle. -Maths 9th

Description : Prove that the angle bisectors of a parallelogram form a rectangle. -Maths 9th

Last Answer : answer:

The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

Last Answer : Given, base of an isosceles triangle =24 cm Area of isosceles triangle =192 sq.cm Area = 21​×b×h ∴192=224.h​ ∴h=16 cm Side=h2+122​=256+144​=20 cm Perimeter of triangle =2a+b =2(20)+24=64 cm

The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Description : The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Last Answer : Let the sides of triangle be 3x,4x,5x Perimeter =3x + 4x + 5x=144 cm 12x=144 ∴x=12 Then sides of triangle are 3x=3 12=36 cm, 4x=4 12=48 cm, 5x=5 12=60 cm. Now, Semi perimeter, s=2 Sum of sides of ... , Area of triangle =s (s−a)(s−b)(s−c) = 72(72−36)(72−48)(72−60) = 72 36 24 12 = 864 cm2

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Description : The base and the corresponding altitude of a parallelogram are 10 cm and 7 cm, respectively. Find its area. -Maths 9th

Last Answer : Area of parallelogram = Base × Corresponding altitude = 10 × 7 = 70 cm2.

Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Description : Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Last Answer : Suppose AL and PM are the altitudes corresponding to equal bases AB and PQ of ||gm ABCD and PQRS respectively . Since the ||gm are between the same parallels PB and SC. ∴ AL = PM Now, ar(||gm ABCD) = AB AL ar(|| ... PM But, AB = PQ [given] AL = PM [proved] ∴ ar(||gm ABCD) = ar(||gm PQRS)

Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Description : Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Last Answer : Suppose AL and PM are the altitudes corresponding to equal bases AB and PQ of ||gm ABCD and PQRS respectively . Since the ||gm are between the same parallels PB and SC. ∴ AL = PM Now, ar(||gm ABCD) = AB AL ar(|| ... PM But, AB = PQ [given] AL = PM [proved] ∴ ar(||gm ABCD) = ar(||gm PQRS)

Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Description : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Last Answer : We know that diagnol of a parallelogram bisect it in two triangles of equal area area of triangle =1÷2×b×h so. 1÷2×b×h+1÷2×b×h=b×h Hence,proved

Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Description : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Last Answer : We know that diagnol of a parallelogram bisect it in two triangles of equal area area of triangle =1÷2×b×h so. 1÷2×b×h+1÷2×b×h=b×h Hence,proved

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Last Answer : Weknowthatthediagonalofaparallelogram(∥gm)dividesitintotwocongruenttriangles.SoAreaof∥gmABCD=2 Areaof△BCD.AccordingtoHeron′sformulathearea(A)oftrianglewithsidesa,b&cisgivenasA=2[s(s−a)(s−b) ... 90=180Areaof∥gm=base heightHeightofaltitudefromvertexAonsideCDoftheof∥gm=baseCDareaof∥gmABCD =12180 =15cm

Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th

Last Answer : =3 x 3 x 5 x 2 cm2 Area of parallelogram ABCD = 2 x 90 = 180 cm2 (ii) Let altitude of a parallelogram be h. Also, area of parallelogram ABCD =Base x Altitude ⇒ 180 = DC x h [from Eq. (ii)] ... h ∴ h = 180/12= 15 cm Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th

Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C)​ [Heron's Formula] S=2120+140+160​=210 m A=210(210−120)(210−140)(210−160)​=210015​ sq. m

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x​ 2 ⇒h= 4 4y 2 −x 2 ​ ​ cm ∴Area= 2 1​ ×base×h = 2 x​ 4 4y 2 −x 2 ​ ​ cm 2

A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Description : A rectangle inscribed in a triangle has its base coinciding with the base b of the triangle. If the altitude of the triangle is h, and the -Maths 9th

Last Answer : answer:

The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Description : The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th

Last Answer : (b) 9 : 4√3.Let each side of the square = a cm. Then, Area of square = a2 cm2 Also, let r be the radius of the circle. Then, πr2 = a2 Let each side of the equilateral triangle = b cm. Then 3b = 4a ⇒ ... ratio between area of circle and area of equilateral Δ is a2 : \(rac{4\sqrt3a^2}{9}\) = 9 : 4√3.

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Description : The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Last Answer : Let the smallest side of the triangle be x cm long So, second side =(x+4)cm and third side =(2x−6)cm Given, perimeter =50cm Therefore, x+(x+4)+(2x−6)=50 ⇒4x=52 ⇒x=13cm So, first side of the triangle is ... =25cm Therefore, area of Δ=s(s−a)(s−b)(s−c) =25(25−13)(25−17)(25−20) =25 12 8 5 = 2030 cm2

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Description : The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th

Last Answer : According to question find the area of triangle.

If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Description : If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Last Answer : Let ABCD be the quadrilateral from which a circular sheet is cut off touching each side of the quadrilateral. Also, given AB + BC + CD + DA = 2p ...(i) Circumference of the circle = 2πr ⇒ ... = pr.∴ Required remaining area = Area of quadrilateral - Area of circle = pr - πr2 = r(p - πr).

A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Last Answer : answer:

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)

P is the mid-point of the side CD of a parallelogram ABCD. -Maths 9th

Description : P is the mid-point of the side CD of a parallelogram ABCD. -Maths 9th

Last Answer : According to question prove that DA = AR and CQ = QR.