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

A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th
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According to question find the the total area of the design and the remaining area of the tiles.
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A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. -Maths 9th

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.

A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 48 cm x 36 cm. From each of its corners, a square of 8 cm is cut-off and an open box is made of the remaining sheet. Find the volume of the box. -Maths 9th

Last Answer : When squares of 8 cm is cutt-off from rectangulare sheet then, Length of box (l) = (98 - 8 - 8) = 32 cm Breadth of box (b) = (36 - 8 - 8) = 20 cm Height of box (h) = 8cm ∴ Volume of box = lbh = 32 x 20 x 8 = 5120 cm3

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

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

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

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 : According to question find the volume of this liquid.

The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th

Description : The volume of a certain rectangular solid is 8 cm^3. Its total surface area is 32 cm^2 and its three dimensions are in geometric progression. -Maths 9th

Last Answer : (b) 32 Let the edges of the solid be a, ar, ar2. Then, Volume = a x ar x ar2 = a3r3 = (ar)3. Given (ar)3 = 8 ⇒ ar = 2 Also, surface area = 2(a x ar + ar x ar2 + a × ar2) = 2(a2r + ... Given, 2ar (a + ar + ar2) = 32 ⇒ 4(a + ar + ar2) = 32 ; Sum of lengths of all edges = 32.

A rectangular box has dimensions x, y and z units, where x < y < z. If one dimension is only increased by one unit, -Maths 9th

Description : A rectangular box has dimensions x, y and z units, where x < y < z. If one dimension is only increased by one unit, -Maths 9th

Last Answer : answer:

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

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

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

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Description : A metallic sheet is of rectangular shape with dimensions 28m × 36m. From each of its corners, a square is cut off so as to make an open box. -Maths 9th

Last Answer : R.E.F image Volume of box =l×b×h From the diagram l=48−2(8) ∵ Two square formed side =32m b=36−2(8) =20m Also h=8m from question ∴ Volume =32×20×8 =5120m3

If two rectangular sheets each of dimensions 2x and 2y form the curved surfaces of two different cylinders, then the ratio -Maths 9th

Description : If two rectangular sheets each of dimensions 2x and 2y form the curved surfaces of two different cylinders, then the ratio -Maths 9th

Last Answer : answer:

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

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm x 25 cm. -Maths 9th

Last Answer : According to question find the lengths QC and PD.

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 rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Description : A rectangular piece of paper is 22 cm long and 10 cm wide. -Maths 9th

Last Answer : Since rectangular piece of paper of rolled along its length. ∴ 2πr = 22 r = 22 × 7 / 2 × 22 = 3.5 cm Height of cyclinder (h) = 10 cm ∴ Volume of cyclinder = πr2h = 22 / 7 × 3.5 × 3.5 × 10 = 385 cm3.

A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Last Answer : answer:

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.

The dimensions of a rectangle ABCD are 51 cm × 25 cm. -Maths 9th

Description : The dimensions of a rectangle ABCD are 51 cm × 25 cm. -Maths 9th

Last Answer : Area of rectangle ABCD = AB x BC = 51 x 25 = 1275 cm2 Area of trapezium PBCQ = 5/6 x 1275 = 6375/6 cm2 Let QC = 9x cm and PB = 8x cm ∴ Area of trapezium PBCQ = 1/2(QC + PB) x BC ⇒ 6375/6 = 1/2(9x + 8x) x 25 ⇒ 17x ... 6375/6 x 2/17 x 25 ⇒ x = 5 ∴ QC = 9 x 5 cm = 45 cm and PB = 8 x 5 cm = 40 cm

Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Description : Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Last Answer : For surface I: It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm = (0.75 x 3.3) cm2 = 2.475 cm2 (approx.) For surface II: It is a rectangle with length 6.5 cm and breadth 1 cm. ∴ Area of ... surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2 = 19.275 cm2 = 19.3 cm2 (approx.)

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm... -Maths 9th

Last Answer : Given : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in the fig. These pieces are arranged ... length of coloured tape required = 30 cm (b) The values are : Happiness, beauty, Knowledge.

Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that AB||HC || GD || FE. Also BC=CD=DE, and GF=6 cm. He wants to decorate the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Description : Sohan wants to show gratitude towards his teacher by giving her a card made by him. He has three pieces of trapezium pasted one above the other as shown in fig. These pieces are arranged in a way that ... the card by putting up a colored tape on the non-parallel sides of the trapezium.. -Maths 9th

Last Answer : Let us consider the following lay out of the greeting card. Trapeziums are arranged in such a way that AB || HC || GD || FE. Also BC=CD=DE and GF=6 cm and DE = 4cm. If three parallel lines make equal ... HG+GF+BC+CD+DE = 6+6+6+4+4+4=30 cm. (b) The values are: Happiness, beauty, Knowledge.

(i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. Find the dimensions of this box if we want to use least material fo

Description : (i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. Find the dimensions of this box if we want to use least material fo

Last Answer : (i) The base of an open rectangular box is square and its volume is ` 256 cm^(3)`. ... dimensions of this window from which maximum light can admit.

If iron has the same dimensions of cm times cm times cm if the mass of this rectangular shaped object is g?

Description : If iron has the same dimensions of cm times cm times cm if the mass of this rectangular shaped object is g?

Last Answer : Need answer

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

The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Description : The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th

Last Answer : Let a = 8cm, b = 15cm, c = 17cm s = (a + b + c)/2 = (8 + 15/ + 17)/2 = 40/2 = 20cm ∴ Area = root under √s(s - a)(s - b)(s - c) = root under √20(20 - 8)(20 - 15)(20 - 17) = root under √20 x 12 x 5 x 3 = 60 cm2

The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Description : The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th

Last Answer : For △BCD: Let a = 17 cm, b = 12 cm, c = 25 cm So its semi-perimeter, s = (a + b + c)/2 = (17 + 12 + 25)/2 = 27 cm ∴ Area of △BCD = root under(√(s -a)(s - b)(s - c)) = ... △BCD = 2 x 90 = 180 cm2 Also, area of parallelogram ABCD = DC x AE ∴ 180 = 12 x AE ⇒ AE = 180/12 = 15 cm

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.

In the figure shown here, QS = SR, QU = SU, PW = WS and ST || RV. What is the value of -Maths 9th

Description : In the figure shown here, QS = SR, QU = SU, PW = WS and ST || RV. What is the value of -Maths 9th

Last Answer : answer:

A child accidentally drops a square ceramic tile. It breaks into smaller pieces as shown in Figure P. To avoid a scolding, she plans to compose a mural of broken pieces to gift to her mother. Which figure out of the given options could she make?

Description : A child accidentally drops a square ceramic tile. It breaks into smaller pieces as shown in Figure P. To avoid a scolding, she plans to compose a mural of broken pieces to gift to her mother. Which figure out of the given options could she make?

Last Answer : C 

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

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

what- possible side lengths for a rectangular frame are 6 inches, 10 inches, and 12 inches.If all measurements are whole numbers, which dimensions can be used to frame a picture that has an area between 45 and 50 square inches?

Description : what- possible side lengths for a rectangular frame are 6 inches, 10 inches, and 12 inches.If all measurements are whole numbers, which dimensions can be used to frame a picture that has an area between 45 and 50 square inches?

Last Answer : 6 in. by 8 in., 10 in. by 5 in., 12 in. by 4 in.

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

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.

In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

The length of a rectangular piece of fabric is twice its width . identify the geometrical representation of this situation when represented as an equation in two variables. Width and length be represented as x and y respectively . -Maths 9th

Description : The length of a rectangular piece of fabric is twice its width . identify the geometrical representation of this situation when represented as an equation in two variables. Width and length be represented as x and y respectively . -Maths 9th

Last Answer : answer:

Find the length of the longest pole that can be put in a room of dimensions 10 m x 10 m x 5 m. -Maths 9th

Description : Find the length of the longest pole that can be put in a room of dimensions 10 m x 10 m x 5 m. -Maths 9th

Last Answer : Here, we have a cuboid with dimensions l = length = 10 m, b = breadth = 10 m and h = height = 5 m Now, length of longest pole = diagonal of cuboid ∴ Required length = √(l2 + b2 + h2) = √(100 + 100 + 25) = √225 = 15 cm