There is a slide in a park,one of its side walls has been painted.... -Maths 9th

There is a slide in a park,one of its side walls has been painted.... -Maths 9th
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Let the dimensions of triangular shape wall be  a = 15 m,  b = 6 m, c = 11 m  ∴  s = (a + b + c)/2 = (15 + 6 + 11)/2 = 32/2 = 16 m Area painted = Area of triangle  = root under(√s(s - a)(s - b)(s - c) = root under (√16(16 - 15)(16 - 6)(16 - 11)) = root under (√16 x 1 x 10 x 5) = 20√2 m2
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There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Description : There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Last Answer : Let the sides of the wall be a = 15m, b = 11m, c = 6m Semi-perimeter, Thus, the required area painted in colour = 20√2 m2

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.

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

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

Last Answer : hope its clear

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

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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 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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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

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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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Two hemispherical domes are to be painted -Maths 9th

Description : Two hemispherical domes are to be painted -Maths 9th

Last Answer : Let radii of the bases of two domes be r and R. ∴ 2πr = 17.6 ⇒ 2 x 22/7 x r = 17.6 ⇒ r = 17.6 x 7/2 x 22 = 2.8 cm and 2 πR = 70.4 ⇒ 2 x 22/7 x R = 70.4 ⇒ R = 70.4 x 7/2 x 22 = ... .2 x 11.2 = 49.28 + 788.48 cm2 = 837.76 cm2 Cost of painting at the rate of ₹10 per cm2 = 837.76 x 10 = ₹8377.6

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 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

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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 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

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Old-fashioned straight-backed armchair, freshly painted walls: how to keep them apart?

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What should I use for drawing (and erasing) on painted walls?

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Can you use tri sodium phosphate to clean medium gloss painted walls?

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Last Answer : um just buy paint thinner

Is the ceiling or walls painted first?

Description : What starts?

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An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Description : Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Last Answer : Solution :- Let R, S and M represent the position of Reshma, Salma and Mandeep respectively. Clearly △RSM is an isosceles triangle as RS = SM = 6m Join OS which intersect RM at A. In △ROS and △MOS OR = OM ( ... . ∴ RM = 2RA RM = 2 x 4.8 = 9.6m Hence, distance between Reshma and Mandeep is 9.6m.

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

Last Answer : (d) 3 : 1Let ABC be the triangular flower bed of side lengths a, b and c respectively. Then Area of ΔABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(rac{a+b+c}{2}\)Now according to the given condition,ΔPQR ... of ΔABC - Area of ΔABC= 3 (Area of ΔABC)∴ Reqd. Ratio = Area of Path : Area of ΔABC = 3 : 1.

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 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

Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 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).

Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Description : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Last Answer : We know that diagnol of a parallelogram bisect it in two triangles of equal area area of triangle =1÷2×b×h so. 1÷2×b×h+1÷2×b×h=b×h Hence,proved

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.

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).

Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Description : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Last Answer : We know that diagnol of a parallelogram bisect it in two triangles of equal area area of triangle =1÷2×b×h so. 1÷2×b×h+1÷2×b×h=b×h Hence,proved

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.

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th

Last Answer : Area of the triangle =

The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th

Description : The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th

Last Answer : area of trapezium = 39 cm sq height of trpezium = 6 cm one || side = 5 cm let other be x area of trapezium = 1/2 × ( sum of || sides) × h 39 = 1/2 × ( 5+x) × 6 39×2 = (5+x) × 6 78/6 = 5+x 13-5 = x 8 = x x = 8cm

Construct an equilateral triangle, given its side and justify the construction. -Maths 9th

Description : Construct an equilateral triangle, given its side and justify the construction. -Maths 9th

Last Answer : Steps of Construction (i) Draw a ray AX with initial point A. (ii) Taking A as centre and radius equal to length of side of the triangle draw an arc intersecting the ray AX at B. (iii) Taking B as ... required triangle. Justification Arcs AB, AC and BC are of the same radii Since, AB = BC = CA

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th

Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.

One side of an equilateral triangle is 4 cm.Find its area. -Maths 9th

Description : One side of an equilateral triangle is 4 cm.Find its area. -Maths 9th

Last Answer : Area of equilateral triangle = √3/4a2 = √3/4 x 42 = 4√3 cm2.

One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Description : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th

Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.

The percentage increase in the area of a triangle if each of its side is doubled is -Maths 9th

Description : The percentage increase in the area of a triangle if each of its side is doubled is -Maths 9th

Last Answer : (c) 300%.Let a, b, c be the sides of the original triangle and s be its semi-perimeter.Then, s = \(rac{1}{2}\)(a+b+c)Let s1 be the semi-perimeter of the new triangle. Then,s1 = \(rac{1}{2}\) (2a + 2b ... increase in area = \(\bigg(rac{A_1-A}{A} imes100\bigg)\)%= \(rac{(4A-A)}{A} imes100\)% = 300%.

The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Description : The base of a right prism is an equilateral triangle with a side 6 cm and its height is 18 cm. Find its volume, -Maths 9th

Last Answer : Volume of a right prism = Area of base height. Since the base is an equilateral triangle of side 6 cm, Area of base = 3√434 x (side)2 = (3√4 62)(34 62)cm2 = 3√434 x 36 cm2 = 93-√93 cm2 ∴ Volume = (93-√93 x18) ... ) = (324 + 2 9√3 ) cm2 = (324 + 18√3 ) cm2 = (324 + 31.176) cm2 = 355.176 cm2.

An equilateral triangle with side a is revolved about one of its sides as axis. What is the volume of the solid of revolution thus obtained ? -Maths 9th

Description : An equilateral triangle with side a is revolved about one of its sides as axis. What is the volume of the solid of revolution thus obtained ? -Maths 9th

Last Answer : answer:

A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Description : A square has its side equal to the radius of the sphere. The square revolves round a side to generate a surface of total area S. -Maths 9th

Last Answer : answer:

A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Last Answer : answer:

A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

Description : A rigid square plate ABCD of unit side rotates in its own plane about the middle-point of CD until the new position of A coincides with -Maths 9th

Last Answer : answer:

Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th

Description : Draw an equiliateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is of the side of the first. -Maths 9th

Last Answer : hope this helps you