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

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

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( √45(45 – 41)(45 – 40)(45 - 9))  = root under( √45 x 4 x 5 x 36 ) = 180 m2 = 1800000 cm2  Number of rose beds =   Total area /  Area needed for one rose bed = 1800000/900  = 2000
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Related questions

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

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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

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 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

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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

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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

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In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

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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

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Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Description : Find the median of the following observations: 46,64,87,41,58,77,35,90,55,92,33. If 92 is replaced by 99 and 41 by 43. find the new median -Maths 9th

Last Answer : Here's ur answer Hope it has helped u..

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

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

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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

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

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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

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

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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

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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.

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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

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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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

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

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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

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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

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A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

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A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

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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

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.

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Description : A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front. -Maths 9th

Last Answer : Let ABCD is a rectangular plot having a measurement of 40 m long and 15 m front. ∴ Length of inner-rectangle, EF = 40 - 3 - 3 = 34 m and breadth of inner-rectangle, FG =15 - 2 - 2 = ... [∴ area of a rectangle = length x breadth] Hence, the largest area where the house can be constructed in 374 m2.

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

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 : Solution of this question

A river 3 m deep and 40 m wide -Maths 9th

Description : A river 3 m deep and 40 m wide -Maths 9th

Last Answer : Length of water canal in one minute = 2 x 1000 m/60 = 100/3 m Volume of water flowing into the sea in one minute = l x b x h = (100 x 40 x 3)/3 = 4000 m 3 = 4000 x 1000 L = 4000000 L ( Since,1 m3 = 1000 L)

The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

Description : The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. -Maths 9th

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A person runs opposite to that of chandigarh train at a speed of 10km/hr. If the relative speed between train and the boy running in opposite direction is 25km/hr. What is the length of the train, if it takes 10 seconds to cross a boy, when is at rest? A) 36.6 m B) 40.6 m C) 43.6m D) 41.6m

Description :  A person runs opposite to that of chandigarh train at a speed of 10km/hr. If the relative speed between train and the boy running in opposite direction is 25km/hr. What is the length of the train, if it takes 10 seconds to cross a boy, when is at rest? A) 36.6 m B) 40.6 m C) 43.6m D) 41.6m

Last Answer : ANSWER : D  Explanation: Speed of person = 10km/hr  As the train and the running person are moving in opposite direction,therir speed values are added to find the relative speed.  Relative speed = speed of train + speed of ... s  Distance = speed * time = 25 / 6 * 10 = 250/6 m/s = 41.6 m

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

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 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

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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

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