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

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

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 =

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

Perimeter of the rhombus is 100 m and its diagonal is 40m. Find the area of rhombus. -Maths 9th

Description : Perimeter of the rhombus is 100 m and its diagonal is 40m. Find the area of rhombus. -Maths 9th

Last Answer : Perimeter of rhombus =4 side ⇒ 100=4 side ⇒ side= 4 100 ⇒ side=25 We know diagonals of a rhombus divides the rhombus in two equilateral triangle. Now, we are going to find area of 1 equilateral triangle. Semi perimeter = ... ) = 45 5 20 20 = 90000 =300m 2 ⇒ Area of rhombus =2 300m 2 =600m 2

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Description : A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus. -Maths 9th

Last Answer : According to question parallelogram bisects one of its angles.

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.

(–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

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.

PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Description : PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th

Last Answer : Solution :- As diagonal of the parallelogram divides it into two triangles of equal area. Since, area (△SRQ ) = 1/2 area(PQRS) area (△SRQ ) = 1/2 x 180 ... = 90 cm2 (Given) This is not possible unless area (△SRQ ) = area (△ASR ) So, the given statement is false.

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:

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.

A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Description : A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Last Answer : we know that , all sides of a rhombus are equal and the diagonals of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4cm and 6cm ... ) From Eqs. (i) and (ii), AB = BC = CD = DA Hence , ABCD is a rhombus.

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Description : A rhombus whose diagonals are 4 cm and 6 cm in lengths. -Maths 9th

Last Answer : We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps. 1.Draw ... B and D. 6.Now, join AB, BC, CD, and DA . Thus, ABCD is the required rhombus.

A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th

Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c​=210+10+12​=16cm ar△ABC=16(16−10)(16−10)(16−12)​=16×6×6×4​=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

The area of a trapezium is 475 cm2 and the height is 19 cm. -Maths 9th

Description : The area of a trapezium is 475 cm2 and the height is 19 cm. -Maths 9th

Last Answer : Let the length of one parallel side of trapezium is a. So length of other parallel side = a + 4 Given the area of trapezium = 475 & the height = 19 Now area of a trapezium = 21 height ... =23∴Length of smalle rparallel side of trapezium =23 & Length of other parallel side of trapezium = 23+4=27

The area of a trapezium is 475 cm2 and the height is 19 cm. -Maths 9th

Description : The area of a trapezium is 475 cm2 and the height is 19 cm. -Maths 9th

Last Answer : According to question find the lengths of its two parallel sides

According to I.S. : 456 specifications, the safe diagonal tensile stress for M 150 grade concrete, is (A) 5 kg/cm2 (B) 10 kg/cm2 (C) 15 kg/cm2 (D) 20 kg/cm

Description : According to I.S. : 456 specifications, the safe diagonal tensile stress for M 150 grade concrete, is (A) 5 kg/cm2 (B) 10 kg/cm2 (C) 15 kg/cm2 (D) 20 kg/cm

Last Answer : Answer: Option A

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. -Maths 9th

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.

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

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 : NEED ANSWER

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 -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.

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 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.

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 : A rhombus has sides of length 1 and area 1/2 Find the angle between the two adjacent sides of the rhombus -Maths 9th

Last Answer : answer:

A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Description : A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Last Answer : hope its clear

If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Description : If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Here, we are joining A and C. In ΔABC P is the mid point of AB Q is the mid point of BC PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and ... RS=PS=RQ[All sides are equal] ∴ PQRS is a parallelogram with all sides equal ∴ So PQRS is a rhombus.

If ABCD is a rhombus, then -Maths 9th

Description : If ABCD is a rhombus, then -Maths 9th

Last Answer : answer:

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

Description : ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

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

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.

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Last Answer : answer:

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Last Answer : Total surface area of a cube = 726 cm2 6 × (side)2 = 726 (side)2 = 121 side = 11 cm Hence, the length of the edge of cube is 11 cm.

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Description : The total surface area of a cube is 726 cm2. Find the length of its edge . -Maths 9th

Last Answer : Total surface area of a cube = 726 cm2 6 × (side)2 = 726 (side)2 = 121 side = 11 cm Hence, the length of the edge of cube is 11 cm.

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is -Maths 9th

Description : An isosceles right triangle has area 8 cm2. The length of its hypotenuse is -Maths 9th

Last Answer : (a) Given, area of an isosceles right triangle = 8 cm2 Area of an isosceles triangle = 1/2 (Base x Height) ⇒ 8 = 1/2 (Base x Base) [∴ base = height, as triangle is an ... √32 cm [taking positive square root because length is always positive] Hence, the length of its hypotenuse is √32 cm.

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is -Maths 9th

Description : An isosceles right triangle has area 8 cm2. The length of its hypotenuse is -Maths 9th

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

An isosceles right triangle has area 8 cm2 . Find the length of its hypotenuse. -Maths 9th

Description : An isosceles right triangle has area 8 cm2 . Find the length of its hypotenuse. -Maths 9th

Last Answer : Area = 1/2a2 ⇒ 1/2a2 = 8 ⇒ a2 = 16 cm ⇒ a = 4 cm Hypotenuse = √2a = √2.4 = 4√2 cm.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

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°.

The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th

Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.