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

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
by

1 Answer

Answer :

Cost of painting  =
by

Related questions

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 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 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]

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 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

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

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

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.

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.

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.

(–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.

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

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

Description : A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

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

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 :-

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.

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.

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.

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.

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:

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:

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

A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Description : A sphere is cut into two equal halves and both the halves are painted from all the sides. The radius of the sphere is r unit and the -Maths 9th

Last Answer : answer:

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Description : The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th

Last Answer : Let length, breadth, and height of the rectangular hall be l, b, and h respectively. Area of four walls = 2lh+2bh = 2(l+b)h Perimeter of the floor of hall = 2(l+b) = 250 m Area of four walls = 2( ... of paining the walls is Rs. 15000. 15000 = 2500h Or h = 6 Therefore, the height of the hall is 6 m.

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.

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.

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:

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 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 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.

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 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

If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Description : If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th

Last Answer : Solution :-

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Description : The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

Last Answer : (c) 67.5 mGiven 2s = a + b + c ⇒ 240 = 78 + 50 + Third side ⇒ Third side = 240 m - 128 m = 112 m.∴ Area of Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)= \(\sqrt{120(120-78)(120-50)(120-112)}\)= \(\sqrt{120 imes42 imes70 ... (rac{1}{2}\)x b x h ∴ \(rac{1}{2}\) x 50 x h = 1680 ⇒ h = \(rac{1680}{25}\) = 67.5 m.

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)​where, s=2a+b+c​s=250+65+65​s=90 Area of triangle = 90(40)(25)(25)​Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th

Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)​where, s=2a+b+c​s=250+65+65​s=90 Area of triangle = 90(40)(25)(25)​Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500

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.

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: