Mathematics Part II Solutions Solutions for Class 9 Math Chapter 5 Prisms are provided here with simple step-by-step explanations. These solutions for Prisms are extremely popular among Class 9 students for Math Prisms Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 199:

#### Question 1:

A solid triangular prism with equilateral base is 5 centimetres high and its base perimeter is 12 centimetres. Find its surface area.

#### Answer:

Height of the prism = 5 cm

Base perimeter = 12 cm

Lateral surface area of the prism = Base perimeter × Height of the prism

= 12 cm × 5 cm

= 60 cm^{2}

Base of the prism is an equilateral triangle with perimeter 12 cm.

⇒ 3 × length of side = 12 cm

⇒ Length of each side = 4 cm

Area of the equilateral triangle =

∴ Surface area of the prism = Lateral surface area of the prism + 2 × Area of the equilateral triangle

= (60 + 2 × ) cm^{2}

= (60 + 8 × 1.73) cm^{2}

= 73.84 cm^{2}

#### Page No 199:

#### Question 2:

A water tank has two faces congruent isosceles trapeziums and the other faces rectangles. Its dimensions are as shown below:

What would be the cost of painting its inside and out at the rate of 75 rupees per square metre?

#### Answer:

Height of the water tank = 0.4 m

Length of the rectangular base = 1.4 m

Breadth of the rectangular base = 0.7 m

Base perimeter = 2 (Length + Breadth)

= 2 (1.4 + 0.7) m

= 4.2 m

Lateral surface area of the water tank = Base perimeter × Height

= (4.2 × 0.4) m^{2}

= 1.68 m^{2}

Area of the rectangular base = Length × Breadth

= (1.4 × 0.7) m^{2}

= 0.98 m^{2}

Surface area of the water tank = Lateral surface area + Area of the rectangular base

= (1.68 + 0.98) m^{2}

= 2.66 m^{2}

Area to be painted = 2 × Surface area of the water tank

= 2 × 2.66 m^{2}

= 5.32 m^{2}

Cost of painting per sq. metre = Rs.75

∴ Cost of painting 5.32 m^{2} area = Rs.75 × 5.32 = Rs.399

#### Page No 199:

#### Question 3:

Two identical prisms with base a right angled triangle are joined together to form a rectangular prism, as in the picture below:

What is the surface area of this rectangular prism?

#### Answer:

Lengths of sides of the base of the triangular prism are 12 cm and 5 cm.

Height of the triangular prism = 15 cm

Two triangular prisms are joined together to form a rectangular prism.

∴ Length of the base of the rectangular prism = 12 cm

Breadth of the base of the rectangular prism = 5 cm

Height of the rectangular prism = 15 cm

Base perimeter of the rectangular prism = 2 (Length + Breadth)

= 2 (12 + 5) cm

= 2 × 17 cm

= 34 cm

Lateral surface area of the rectangular prism = Height × Base perimeter

= 15 × 34 cm^{2}

= 510 cm^{2}

Base area of the rectangular prism = Length × Breadth

= 5 × 12 cm^{2}

= 60 cm^{2}

Surface area of the rectangular prism = Lateral surface area + 2 × Base area

= (510 + 2 × 60) cm^{2}

= (510 + 120) cm^{2}^{ }

= 630 cm^{2}

#### Page No 203:

#### Question 1:

The base of a prism is an equilateral triangle of perimeter 15 centimetres and its height is 5 centimetres. What is its volume?

#### Answer:

Perimeter of the equilateral triangle at the base = 15 cm

∴ 3 × Length of each side = 15 cm

⇒ Length of each side = 5 cm

Height of the prism = 5 cm

Base area of the prism =

Volume of the prism = Base area × Height

= × 5 cm^{3}

= 6.25 × 1.73 × 5 cm^{3}

= 54.06 cm^{3}^{ }(approx.)

#### Page No 203:

#### Question 2:

A water tank in the shape of a prism has isosceles trapeziums as bases whose dimensions are shown below.

The length of the tank is 80 centimetres. How many litres of water can it contain?

#### Answer:

Parallel sides of the trapezium are 75 cm and 50 cm.

Height of the trapezium = 40 cm

Area of the trapezium = × Sum of parallel sides × Height

∴ Base area of the tank = 2500 cm^{2}

Length of the tank = Height of the tank = 80 cm

Volume of the tank = Base area × Height

= 2500 × 80 cm^{3}

= 200000 cm^{3}

We know that 1 L = 1000 cm^{3}

∴ 200000 cm^{3} =

Thus, the water tank can contain 200 litres of water.

#### Page No 203:

#### Question 3:

In the school ground, there is a pit in the shape of a regular hexagon, to collect rain-water. Each side of the pit is 2 metres long and the pit is 3 metres deep. It contains water 1 metre high. How many litres is it?

#### Answer:

Height of the pit = 3 m

Height of water in the pit = 1 m

Length of side of the regular hexagon = 2 m

Diagonals of a regular hexagon divide it into six equilateral triangles.

∴ Area of the regular hexagon =

Volume of water = Area of the regular hexagon × Height of water

We know that 1 m^{3}* *= 1000* *L

∴ 10.38 m^{3}* *= 1000 × 10.38 L = 10380 L

Thus, the pit contains 10380 litres of rain-water.

#### Page No 203:

#### Question 4:

A vessel is in the shape of a square prism, with each side of the base 16 centimetres long. It contains water 10 centimetres high. If a cube of side 8 centimetres is completely immersed in it, how high would the water level rise?

#### Answer:

Side of the base of the square prism = 16 cm

Area of the base of the square prism = Side × Side

= (16 × 16) cm^{2}

= 256 cm^{2}

Height of the water level = 10 cm

Quantity of water in the vessel = Base area × Height of water

= 256 × 10 cm^{3}

= 2560 cm^{3}

Length of side of cube immersed in the vessel = 8 cm

Volume of the cube = Side × Side × Side

= 8 × 8 × 8 cm^{3}

= 512 cm^{3}

∴ New volume of water in the vessel = (2560 + 512) cm^{3}

= 3072 cm^{3}

⇒ Base area × New height = 3072

⇒ 256 cm^{2}^{ }× New height = 3072 cm^{3}

⇒ New height =

∴ Increase in the water level = (12 − 10) cm = 2 cm

Thus, the water level will rise by 2 cm.

#### Page No 204:

#### Question 1:

The volume of a square prism is 3600 cubic centimetres and its base area is 144 square centimetres. What is its surface area?

#### Answer:

Base area of the square prism = 144 cm^{2}

⇒ (Side)^{2} = 144 cm^{2}

⇒ Side =

Volume of the square prism = 3600 cm^{3}

⇒ Base area × Height = 3600 cm^{3}

⇒ 144 cm^{2} × Height = 3600 cm^{3}

⇒ Height =

Perimeter of the base = 4 × side = 4 × 12 cm = 48 cm

Lateral surface area of the square prism = Perimeter of the base × Height

= (48 × 25) cm^{2}

= 1200 cm^{2}

Surface area of the square prism = 2 × Base area + Lateral surface area

= (2 × 144 + 1200) cm^{2}

= 1488 cm^{2}

#### Page No 206:

#### Question 1:

The inner diameter of a well is 2.5 metres and it is 8 metres deep. What would be the cost of cementing its inside from the top to a depth of metres at the rate of 350 rupees per square metre?

#### Answer:

Diametre of the well = 2.5 m

Radius (*r*) of the well = = 1.25 m

Depth (*h*) of the well = 8 m

Depth of the portion to be cemented =

Base perimeter = 2π*r*

= (2 × 3.14 × 1.25) m

= 7.85 m

Curved surface area of the cemented portion = Base perimeter × Height

= (7.85 × 2.5) m^{2}

= 19.625 m^{2}

Cost of cementing per square metre = Rs.350

∴ Cost of cementing 19.625 m^{2} area = Rs.350 × 19.625 = Rs.6868.75

#### Page No 206:

#### Question 2:

A road-roller is of diameter 80 centimetres and length 1.2 metres. What is the area of ground levelled, as it rolls once?

#### Answer:

Diametre of the road roller = 80 cm = 0.8 m (1 m = 100 cm)

Radius (*r*) of the road roller =

Length of the road roller = 1.2 m

Perimeter of the circular base = 2π*r*

= (2 × 3.14 × 0.4) m

= 2.512 m

If the road roller rolls once, it covers area equal to its curved surface area.

Curved surface area of the road roller = Perimeter of the base × Length of the road roller

= (2.72 × 1.2) m^{2}

= 3.014 m^{2}

Thus, 3.014 m^{2} area of the ground is levelled.

#### Page No 207:

#### Question 1:

The tent for an exhibition is as shown below, with a half-cylinder on top of a rectangular prism.

What is the cost of making it, at the rate of 2000 rupees per square metre.

#### Answer:

Length of the rectangular prism = 18 m

Breadth of the rectangular prism = 3 m

Diametre of the cylinder = Breadth of the rectangular prism = 3 m

∴ Radius (*r*) of the cylinder =

Height (*h*) of the cylinder = 18 m

Height of the rectangular prism = Total height − Radius of the cylinder

= (6 − 1.5) m

= 4.5 m

Lateral surface area of the rectangular prism = Base perimeter × Height

= 2 (Length + Breadth) × Height

= 2(18 + 3) × 4.5 m^{2}

= (42 × 4.5) m^{2}

= 189 m^{2}

Lateral surface of the half cylinder = × Base perimeter × Height

= × 2π*r* × *h*

= (3.14 × 1.5 × 18) m^{2}

= 84.78 m^{2}

Area of two semicircles at the ends of the cylinder = 2 ×× π × *r*^{2}

= (3.14 × 1.5 × 1.5) m^{2}

= 7.065 m^{2}

Total area of the tent = Lateral surface area of the rectangular prism + Lateral surface of the half cylinder + Area of two semicircles at the ends of the cylinder

= (189 + 84.78 + 7.065) m^{2}

= 280.845 m^{2}

Cost of making the tent per sq. metre^{ }= Rs.2000

∴ Cost of making the tent of area 280.845 m^{2}^{ }= Rs.2000 × 280.845 = Rs.561690

#### Page No 209:

#### Question 1:

A solid metal cylinder of base radius 15 centimetres and height 32 centimetres is melted down and recast into a cylinder of base radius 20 centimetres. What is the height of this cylinder?

#### Answer:

Base radius (*r*_{1}) of the solid metal cylinder = 15 cm

Height (*h*_{1}) of the solid metal cylinder = 32 cm

Area of base of the solid metal cylinder = π*r*_{1}^{2}

= π(15)^{2} cm^{2}

= 225π cm^{2}

Volume of the solid metal cylinder = Area of base × Height

= (225π × 32) cm^{3}

= 7200π cm^{3}

Base radius (*r*_{2}) of the new cylinder = 20 cm

Let the height of the new cylinder be *h*_{2} cm.

Base area of the new cylinder = π*r*_{2}^{2}

= π(20)^{2} cm^{2}

= 400π cm^{2}

Volume of the new cylinder = (400π × *h*_{2}) cm^{3}

= 400π*h*_{2} cm^{3}

Now, volume of the solid metal cylinder = Volume of the new cylinder

⇒ 7200π = 400π*h*_{2}

⇒ *h*_{2}_{ }=

Therefore, the height of the new cylinder is 18 cm.

#### Page No 209:

#### Question 2:

A solid metal cylinder is of base radius 18 centimetres and height 40 centimetres. By melting this down, how many cylinders of base radius 2 centimetres and height 5 centimetres can be made?

#### Answer:

Base radius (*r*_{1}) of the solid metal cylinder = 18 cm

Height (*h*_{1}) of the solid metal cylinder = 40 cm

Base area of the solid metal cylinder = π*r*_{1}^{2}

= π(18)^{2} cm^{2}

= 324π cm^{2}

Volume of the solid metal cylinder = Base area × Height

= (324π × 40) cm^{3}

= 12960π cm^{3}

Base radius (*r*_{2}) of the new cylinder = 2 cm

Height (*h*_{2}) of the new cylinder = 5 cm

Base area of the new cylinder = π*r*_{2}^{2}

= π(2)^{2} cm^{2}

= 4π cm^{2}

Volume of the new cylinder = (4π × 5) cm^{3}

= 20π cm^{3}

Number of cylinders that can be made =

#### Page No 209:

#### Question 3:

The base radii of two cylinders of the same height are in the ratio 3 : 4. What is the ratio of their volumes?

#### Answer:

Let the base radii of the first and second cylinders be 3*x** *and 4*x* units respectively.

Let the height of the cylinders be *h* units.

Base area of the first cylinder = π*r*^{2}

= π(3*x*)^{2} sq. units

= 9π*x*^{2} sq. units

Volume of the first cylinder = Base area × Height

= 9π*x*^{2}*h* cubic units

Base area of the second cylinder = π(4*x*)^{2} = 16π*x*^{2} sq. units

Volume of the second cylinder = 16π*x*^{2}*h* cubic units

Ratio of the volumes =

Therefore, the ratio of the volumes of the first cylinder to the second cylinder is 9:16.

#### Page No 210:

#### Question 1:

The base radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 4. What is the ratio of their volumes?

The volume of the first cylinder is 720 cubic centimetres. What is the volume of the second?

#### Answer:

Let the base radii of the first and second cylinders be 2*x* and 3*x* units respectively.

Let the heights of the first and second cylinders be 5*y** *and 4*y* units respectively.

Base area of the first cylinder = π*r*^{2}

= π(2*x*)^{2} sq. units

= 4π*x*^{2}^{ }sq. units

Volume of the first cylinder = Base area × Height

= 4π*x*^{2} × 5*y* cubic units

= 20π*x*^{2}*y* cubic units

Base area of the second cylinder = π(3*x*)^{2} sq. units = 9π*x*^{2} sq. units

Volume of the second cylinder = 9π*x*^{2} × 4*y* cubic units = 36π*x*^{2}*y* cubic units

Ratio of the volumes =

Therefore, the ratio of the volumes of the first cylinder to the second cylinder is 5:9.

Volume of the first cylinder = 720 cm^{3}

∴ Volume of the second cylinder = × Volume of the second cylinder

#### Page No 210:

#### Question 2:

Two rectangular sheets, each of sides 10 centimetres and 6 centimetres are rolled to form cylinders one along the longer edge and the other along the shorter edge, to make two cylindrical cans. Which has the larger capacity?

#### Answer:

Length of the rectangular sheets = 10 cm

Breadth of the rectangular sheets = 6 cm

If we roll the sheet along the longer edge, the length will become the perimeter of the base of the cylinder and the breadth will become the height of the cylinder.

Perimeter of the base = 10 cm

⇒ 2π*r* = 10 cm

Base area = π*r*^{2}

Volume of the first cylinder = Base area × Height

If we roll the sheet along the shorter edge, the breadth will become the perimeter of the base of the cylinder and the length will become the height of the cylinder.

Perimeter of the base = 6 cm

⇒ 2π*r* = 6 cm

Base area = π*r*^{2}

Volume of the second cylinder = Base area × Height

The cylinder whose volume is greater will have larger capacity.

As , the second cylinder has larger capacity ._{ }

Thus, the cylinder which is rolled along its longer edge has larger capacity.

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