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

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
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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
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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

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Last Answer : Length (l) of room = 5m Breadth (b) of room = 4m Height (h) of room = 3m It can be observed that four walls and the ceiling of the room are to be white washed. Total area to be white washed = Area of walls + ... m2 area = Rs.7.50 (Given) Cost of white washing 74 m2 area = Rs. (74 7.50) = Rs. 555

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.

The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.

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.

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

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.

The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Description : The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

Last Answer : Let a = 41m, b = 40 m, c = 9 m. s = (a + b + c)/2 = (41 + 40 +9)/2 = 90/2 ⇒ s = 45 m Area of the triangular field = root under( √s(s - a)(s - b)(s -c)) = root under( ... x 5 x 36 ) = 180 m2 = 1800000 cm2 Number of rose beds = Total area / Area needed for one rose bed = 1800000/900 = 2000

The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Description : The inner diameter of a circular well is 3.5m. It is 10m deep. Find (i) its inner curved surface area, (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2. -Maths 9th

Last Answer : Inner radius of circular well, r = 3.5/2m = 1.75m Depth of circular well, say h = 10m (i) Inner curved surface area = 2πrh = (2 (22/7 ) 1.75 10) = 110 Therefore, the inner curved surface ... area = Rs (110 40) = Rs.4400 Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th

Last Answer : Let the sides of the triangular will be a = 122m, b = 12cm, c = 22m Semi-perimeter, s = a+b+c2 (122+120+224)m = 2642 m = 132m The area of the triangular side wall Rent for 1 year (i.e. 12 months) per m2 = ... = Rs. 5000 x 312 = Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.

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]

A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

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...

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Description : What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th

Last Answer : Lateral surface area of a prism = Perimeter of base Height ⇒ 840 = (5 + 5 + 8) Height ⇒ Height = 8401884018 = 46 cm. = Semi perimeter of the triangular base = 182182 = 9 cm ∴ Area of triangle = 9(9- ... 4 1 = 12 cm2 ∴ Required volume of prism = Area of base Height = (12 46) cm3 = 552cm3

A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)​where S=2a+b+c​∴Areaof△PQR= S=23+4+5​=6 S=6(6−3)(6−4)(6−5)​=3×2×3×2×1​=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.

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

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Description : A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. -Maths 9th

Last Answer : Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm Formula for Surface area of hemispherical bowl = 2πr2 = 2 (22/7) (5.25)2 = 173.25 Surface area of hemispherical ... of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

what- the sides of the triangular top of the Transamerica building in san francisco are about 65.2 meters long?

Description : what- the sides of the triangular top of the Transamerica building in san francisco are about 65.2 meters long?

Last Answer : 16

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Last Answer : Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : NEED ANSWER

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th

Last Answer : Area of the parallelogram

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Description : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th

Last Answer : In trapezium ABCD, we draw a perpendicular line CE to the line AB.

A horse is tied to a peg hammered at one of the corner of a rectangular grass field of 40 m by 24 m by a rope 14 m long. Over how much area of the field can the horse graze? (1) 154 m2 (2) 308 m2 (3) 240 m2 (4) 480 m2

Description : A horse is tied to a peg hammered at one of the corner of a rectangular grass field of 40 m by 24 m by a rope 14 m long. Over how much area of the field can the horse graze? (1) 154 m2 (2) 308 m2 (3) 240 m2 (4) 480 m2

Last Answer : (1) 154 m2

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : NEED ANSWER

A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Description : A cylindrical roller 2.5 m in length, 1.75 m in radius when rolled on a road was found to cover the area of 5500 m2. -Maths 9th

Last Answer : Solution of the question

If cosec |-sin |=l and sec |- cos |=m, prove that l2m2(l2+m2+3)=1 -Maths 9th

Description : If cosec |-sin |=l and sec |- cos |=m, prove that l2m2(l2+m2+3)=1 -Maths 9th

Last Answer : cosec(A) - sin(A) = l ⇒ 1/sin(A) - sin(A) = l ⇒ l² = 1/sin²(A) + sin²(A) - 2 --------- sec(A) - cos(A) = m ⇒ 1/cos(A) - cos(A) = m ⇒ m² = 1/cos²(A) + cos²(A) - 2 ---------- l²m² = [1/sin²(A) + ... A)) = = 1/(sin²(A)cos²(A)) ------------- ⇒ l²m² (l² + m² + 3) = sin²(A)cos²(A) / [sin²(A)cos²(A)] = 1

In what ratio must a person mix three kinds of dall costing him Rs 12.25,Rs 12.50 and Rs 12.75 per Kg so that the mixture may be worth Rs 12.65 per Kg? a) 11:7:1 b) 22:3:6 c) 14:3:4 d) 2:7:4 e) 3:12:32

Description : In what ratio must a person mix three kinds of dall costing him Rs 12.25,Rs 12.50 and Rs 12.75 per Kg so that the mixture may be worth Rs 12.65 per Kg? a) 11:7:1 b) 22:3:6 c) 14:3:4 d) 2:7:4 e) 3:12:32

Last Answer : E Step1: Mix dall of first and third kind to get a mixture worth Rs 12.25 per Kg. C.P of 1 Kg dall of 1st kind 1225p C.P of 1 Kg dall of 3rd kind 1275p So they must be mixed in the ... So the Mixed ratio = 3:8 Thus the quantities of dall 1st : 2nd : 3rd  = 3:12:32

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Last Answer : Length of a room (l) = 5m Breadth (b) = 4 m and height (h) = 3 m ∴ Area of 4 walls = 2(l + b) x h = 2(5 + 4) x 3 = 6 x 9 = 54 m² and area of ceiling = l x b = 5 x ... ∴ Total area = 54 + 20 = 74 m2 Rate of white washing = 7.50 per m² ∴ Total cost = ₹74 x 7.50 = ₹555

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Description : A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. Meter 1).Rs.680 2).Rs.860 3).Rs.608 4).None.

Last Answer : 1).Rs.680 Exp: overall area of the grassy plot is 110 x 65 =7150 sq. meters.The area of the grassy plot after applying the gravel path is 105 x 60=6300 sq. meters.Area of the gravel path is 7150 – 6300 = 850 sq.mtrs.Hence x 0.80=Rs.680 .

A trader sells 145 m of cloth for Rs 12325 at the profit of Rs 10 per metre of cloth. What is the cost price of 1 m of cloth? 1. Rs 65 2. Rs 75 3. Rs 95 4. Rs 85

Description : A trader sells 145 m of cloth for Rs 12325 at the profit of Rs 10 per metre of cloth. What is the cost price of 1 m of cloth? 1. Rs 65 2. Rs 75 3. Rs 95 4. Rs 85

Last Answer : 2. Rs 75

The cost of preparing a cricket ground of rectangular shape at 70 paisa per square meter is Rs.508.20. Find the perimeter of the field if its sides are in the ratio 3:2 a) 121m b) 96m c) 55m d) 110m

Description : The cost of preparing a cricket ground of rectangular shape at 70 paisa per square meter is Rs.508.20. Find the perimeter of the field if its sides are in the ratio 3:2 a) 121m b) 96m c) 55m d) 110m

Last Answer : d) 110m

In a rectangular field of dimension 60 m x 50 m, -Maths 9th

Description : In a rectangular field of dimension 60 m x 50 m, -Maths 9th

Last Answer : Area of rectangular field = length x breadth = 60 x 50 = 3,000 m2 Now, a = 50 m, b = 45 m, c = 35 m s = (a + b + c)/2 = (50 + 45 + 35)/2 = 130/2 = 65 m By Heron's formula ... m2 ( approximately ) Hence, the remaining area = Area of rectangle - Area of triangle = 3,000 - 764.85 = 2,235.15 m2

A wall have length 24m and breadth 16m. What will be total cost of painting it from both sides if rate of painting is Rs 8 per square metre? 1) Rs 5696 2) Rs 5856 3) Rs 6032 4) Rs 6144 5) Rs 6272

Description : A wall have length 24m and breadth 16m. What will be total cost of painting it from both sides if rate of painting is Rs 8 per square metre? 1) Rs 5696 2) Rs 5856 3) Rs 6032 4) Rs 6144 5) Rs 6272

Last Answer : 4) Rs 6144

The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

Description : The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

Last Answer : s=2a+b+c​=26+8+10​=12 By Heron's formula, Area of the triangle =s(s−a)(s−b)(s−c)​=12(12−6)(12−8)(12−10)​=12(6)(4)(2)​=24cm2 Cost of painting =9×24 paise =216 paise = Rs. 2.16.

The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Last Answer : Rent paid by company is

The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

Description : The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

Last Answer : s=2a+b+c​=26+8+10​=12 By Heron's formula, Area of the triangle =s(s−a)(s−b)(s−c)​=12(12−6)(12−8)(12−10)​=12(6)(4)(2)​=24cm2 Cost of painting =9×24 paise =216 paise = Rs. 2.16.

The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Last Answer : Rent paid by company is

An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Last Answer : The sides of triangular pieces are 20 cm, 50 cm and 50 cm. Let a = 20 cm, b = 50 cm, c = 50 cm ∴ Semi - perimeter, s = (a + b + c)/2 = (20 + 50 + 50)/2 s = 60 cm ∴ Area of ... 40 x 10 x 10) = 200√6 cm2 Cloth of each design required = Area of 5 triangular pieces = 5 x 200√6 = 1000√6 cm2

A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Last Answer : Measures of the sides of the triangular tile are 28 cm, 9 cm and 35 cm. Let a = 28 cm, b = 9 cm, c = 35 cm Semi-perimeter, s = (a + b + c)/2 = (28 + 9 + 35)/2 = 36 cm ∴ Area of one ... 1411.2 cm2 Hence, cost of polishing the tiles at the rate of ₹ 1/2 per cm2 = ₹ 1/2 x 1411.2 = ₹ 705.60

The triangular side walls of a flyover is used for -Maths 9th

Description : The triangular side walls of a flyover is used for -Maths 9th

Last Answer : hope its clear

Triangular pieces of cardboards were cutout by some people -Maths 9th

Description : Triangular pieces of cardboards were cutout by some people -Maths 9th

Last Answer : The two cutouts may not be congruent. For example all equilateral triangles have equal angles but may have different sides. Environmental concern, cooperative, caring, social.