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

1 Answer

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 triangular tile = under root( √s(s -a)(s -b)(s - c))  = under root( √36(36 – 28)(36 - 9)(36 – 35)) =  under root( √36 x 8 x 27 x 1) = 36 √6 cm2  So, area of 16 triangular tiles = 16 x 36 √6 cm2   = 576 √6 cm2 = 576 x 2.45 = 1411.2 cm2 Hence, cost of polishing the tiles at the rate of ₹ 1/2 per cm2  =  ₹ 1/2 x 1411.2 = ₹ 705.60

Related questions

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

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

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

Description : An umbrella is made by stitching 10 triangular -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 ... √6 = 1000 √6 cm2 Cloth of each design required for 20 umbrellas = 20 x 1000 √6 = 20,000 √6 cm2 Helpful, caring, loving.

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.

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.

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

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

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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

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.

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

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

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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

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

Last Answer : hope its clear

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.

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.

Description : Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

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Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th

Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.

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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

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Description : What size rectangular floor can be completely covered by using 3x3 ft tiles or 5x5 ft tiles you cant cut tiles or combine them?

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Description : What size rectangular floor can be completely covered by using 3x3 ft tiles or 5x5 ft tiles you cant cut tiles or combine them?

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Description : Vinyl Floor Tiles?

Last Answer : form_title= Vinyl Floor Tiles form_header= Easy to clean and easy to install- Vinyl. Do you need to remove old flooring?*= () Yes () No What color do you want the tile?*= _ [50] What are the dimensions of the area?*= _ [50]

Description : Common colors of ceramic floor tiles?

Last Answer : Common colors of ceramic floor tiles include white and beige. The best color to match white walls and brown curtains would probably be yellow.

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Description : How to Lay Floor Tiles

Last Answer : How to Lay Floor Tiles Laying down floor tiles on your own ensures that you put them down exactly as you want them. Doing it correctly is also crucial, to make sure that they have a long life to be useful ... or rounded areas. Work using small nips at a time. Leave the tiles to set for 24 hours.

Description : Where can I get the cheapest floral design wallpaper online?

Last Answer : I performed a simple Google search for "free wallpapers" and found dozens of sites that will give you desktop wallpaper for free. Some have better images than others, and some are certainly better organized than ... you are looking for. They have all kinds and more than you would find in the store.

Description : Can you get a degree in floral design?

Last Answer : You can definitely earn certification in floral design, but I don't think colleges offer a bachelor in this field. Community colleges have associate degrees in horticulture, which include flower design classes.

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

Description : A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

Last Answer : According to question find the the total area of the design and the remaining area of the tiles.

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.

Description : 7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Last Answer : answer:

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.

Description : An R.C.C. lintel is spanning an opening of 2 m span in a brick wall. The height of the roof is 2.9 m above the floor level and that of the opening is 2.1 m above the floor level. The lintel is to be ... (B) UDL of wall (C) UDL of wall + load from the roof (D) Triangular load + load from the roof

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Description : What is the answer Josh has a job to tile a room 16 ft x 18 ft with 1 ft. square tiles. How many tiles does he need?

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Description : “The familiarity with political institutions of the colonial rule helped to develop an agreement over the institutional design.” Justify the statement. -Maths 9th

Last Answer : answer:

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Description : The number of planks of dimensions (4 m x 50cm x 20cm) that can be stored in a pit which is 16 m long, 12 m wide and 40 m deep is -Maths 9th

Last Answer : Length of the plank=4m=400cm Breadth=50cm Height=20cm Volume of the plank=L*B*H =400*50*20 =400000cm^3 Length of the pit=16m=1600cm Breadth=12m=1200cm Height=4m=400cm Volume of the pit= L ... *1200*400 =768000000cm^3 Number of planks that can be fitted= 768000000/400000 =1920 planks is the answer.

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3