In Fig., a tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre. (use π = 22/7)

In the given figure,

For conical portion,

radius = 3/2 = 1.5m and length is 2.8m

∴ S

∴ S

= π x 1.5 x 2.1

= 6.3π m

Area of canvas used for making tent = S

=4.2π +6.3π

= 10.5π

=10.5 x (22/7)

=33m

Thus, total cost of the canvas at the rate of Rs. 500 per m

Rs. 16500 is total cost.

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In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.

Side of square = 28 cm and radius of each circle = 28/2 cm

Area of the shaded region

= Area of the square + Area of the two circles − Area of the two quadrants

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From each end of a solid metal cylinder, metal was scooped out in the hemispherical form of the same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. ( π = 22/7)

Height of the solid metal cylinder, h = 10 cm

Radius of the solid metal cylinder, r = 4.2 cm

therefore,

Radius of each hemisphere = radius of the solid metal cylinder, r = 4.2 cm

Now,

Volume of the rest of the cylinder = Volume of cylinder - 2 x volume of each hemisphere

Thickness of the cylindrical wire = 1.4 cm

Therefore,

Radius of the cylindrical wire, R = 1.4/2 = 0.7 cm

Let the length of the wire be H cm.

It is given that the rest of the cylinder is melted and converted into a cylindrical wire.

therefore,

Volume of the cylindrical wire = Volume of the rest of the cylinder

⇒ π x 0.7 x 0.7 x H = π x (4.2)^{2} x (4.4)

Hence, the length of the wire is 158.4 cm

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The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

Let the height and radius of the given cone be H and R respectively.

The cone is divided into two parts by drawing a plane through the mid point of its

axis and parallel to the base.

Upper part is a smaller cone and the bottom part is the frustum of the cone.

⇒ OC = CA = h/2

Let the radius of smaller cone be r cm.

In ΔOCD and ΔOAB,

∠OCD = ∠OAB = 90°

∠COD = ∠AOB (common)

∴ ΔOCD ∼ ΔOAB (AA similarity)

⇒ OA/OC = AB/CD = OB/OD

⇒ h / h/2 = R/r

⇒ R = 2r

the radius and height of the cone OCD are r cm and h/2 cm

therefore the volume of the cone OCD = 1/3 x π x r^{2} x h/2 = 1/6 πr^{2}h

Volume of the cone OAD = 1/3 x π x R^{2} x h = 1/3 x π x 4r^{2} x h

The volume of the frustum = Volume of the cone OAD - Volume of the cone OCD

= (1/3 x π x 4r^{2} x h) – (1/3 x π x r^{2} x h/2)

= 7/6 πr^{2}h

Ratio of the volume of the two parts = Volume of the cone OCD : volume of the frustum

= 1/6 πr^{2}h : 7/6 πr^{2}h

= 1 : 7

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In Figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

Area of the square lawn PQRS = 42 m x 42 m

Let OP = OS = xm

So, x^{2} + x^{2} = (42)^{2}

⇒ 2x^{2} = 42 x 42

⇒ x^{2} = 21 x 42

Now,

area of sector POS=

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In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of the hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park. Write your views on recycling of water.

Diameter of cylinder (d) = 2 m

Radius of cylinder (r) = 1 m

Height of cylinder (H) = 5 m

Volume of cylindrical tank, Vc = πr^{2}H = π×(1)2×5=5π m

Length of the park (l) = 25 m

Breadth of park (b) = 20 m

the height of standing water in the park = h

Volume of water in the park = lbh = 25×20×h

Now water from the tank is used to irrigate the park. So,

Volume of cylindrical tank = Volume of water in the park

⇒5π=25×20×h

⇒5π/25×20=h

⇒h=π/100 m

⇒h=0.0314 m

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.

It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

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