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 Meritnation’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.

Question 1:

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

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 cm2

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 × ) cm2

= (60 + 8 × 1.73) cm2

= 73.84 cm2

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?

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) m2

= 1.68 m2

Area of the rectangular base = Length × Breadth

= (1.4 × 0.7) m2

= 0.98 m2

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

= (1.68 + 0.98) m2

= 2.66 m2

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

= 2 × 2.66 m2

= 5.32 m2

Cost of painting per sq. metre = Rs.75

Cost of painting 5.32 m2 area = Rs.75 × 5.32 = Rs.399

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?

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 cm2

= 510 cm2

Base area of the rectangular prism = Length × Breadth

= 5 × 12 cm2

= 60 cm2

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

= (510 + 2 × 60) cm2

= (510 + 120) cm2

= 630 cm2

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?

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 cm3

= 6.25 × 1.73 × 5 cm3

= 54.06 cm3 (approx.)

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?

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 cm2

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

Volume of the tank = Base area × Height

= 2500 × 80 cm3

= 200000 cm3

We know that 1 L = 1000 cm3

200000 cm3 = Thus, the water tank can contain 200 litres of water.

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?

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 m3 = 1000 L

10.38 m3 = 1000 × 10.38 L = 10380 L

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

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?

Side of the base of the square prism = 16 cm

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

= (16 × 16) cm2

= 256 cm2

Height of the water level = 10 cm

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

= 256 × 10 cm3

= 2560 cm3

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

Volume of the cube = Side × Side × Side

= 8 × 8 × 8 cm3

= 512 cm3

New volume of water in the vessel = (2560 + 512) cm3

= 3072 cm3

Base area × New height = 3072

256 cm2 × New height = 3072 cm3

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

Thus, the water level will rise by 2 cm.

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?

Base area of the square prism = 144 cm2

(Side)2 = 144 cm2

Side = Volume of the square prism = 3600 cm3

Base area × Height = 3600 cm3

144 cm2 × Height = 3600 cm3

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) cm2

= 1200 cm2

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

= (2 × 144 + 1200) cm2

= 1488 cm2

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?

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 = 2πr

= (2 × 3.14 × 1.25) m

= 7.85 m

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

= (7.85 × 2.5) m2

= 19.625 m2

Cost of cementing per square metre = Rs.350

Cost of cementing 19.625 m2 area = Rs.350 × 19.625 = Rs.6868.75

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? 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 = 2πr

= (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) m2

= 3.014 m2

Thus, 3.014 m2 area of the ground is levelled.

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.

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 m2

= (42 × 4.5) m2

= 189 m2

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

= × 2πr × h

= (3.14 × 1.5 × 18) m2

= 84.78 m2

Area of two semicircles at the ends of the cylinder = 2 × × π × r2

= (3.14 × 1.5 × 1.5) m2

= 7.065 m2

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) m2

= 280.845 m2

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

Cost of making the tent of area 280.845 m2 = Rs.2000 × 280.845 = Rs.561690

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?

Base radius (r1) of the solid metal cylinder = 15 cm

Height (h1) of the solid metal cylinder = 32 cm

Area of base of the solid metal cylinder = πr12

= π(15)2 cm2

= 225π cm2

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

= (225π × 32) cm3

= 7200π cm3

Base radius (r2) of the new cylinder = 20 cm

Let the height of the new cylinder be h2 cm.

Base area of the new cylinder = πr22

= π(20)2 cm2

= 400π cm2

Volume of the new cylinder = (400π × h2) cm3

= 400πh2 cm3

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

7200π = 400πh2

h2 = Therefore, the height of the new cylinder is 18 cm.

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?

Base radius (r1) of the solid metal cylinder = 18 cm

Height (h1) of the solid metal cylinder = 40 cm

Base area of the solid metal cylinder = πr12

= π(18)2 cm2

= 324π cm2

Volume of the solid metal cylinder = Base area × Height

= (324π × 40) cm3

= 12960π cm3

Base radius (r2) of the new cylinder = 2 cm

Height (h2) of the new cylinder = 5 cm

Base area of the new cylinder = πr22

= π(2)2 cm2

= 4π cm2

Volume of the new cylinder = (4π × 5) cm3

= 20π cm3

Number of cylinders that can be made =  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?

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

Let the height of the cylinders be h units.

Base area of the first cylinder = πr2

= π(3x)2 sq. units

= 9πx2 sq. units

Volume of the first cylinder = Base area × Height

= 9πx2h cubic units

Base area of the second cylinder = π(4x)2 = 16πx2 sq. units

Volume of the second cylinder = 16πx2h cubic units

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

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?

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

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

Base area of the first cylinder = πr2

= π(2x)2 sq. units

= 4πx2 sq. units

Volume of the first cylinder = Base area × Height

= 4πx2 × 5y cubic units

= 20πx2y cubic units

Base area of the second cylinder = π(3x)2 sq. units = 9πx2 sq. units

Volume of the second cylinder = 9πx2 × 4y cubic units = 36πx2y 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 cm3

Volume of the second cylinder = × Volume of the second cylinder 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?

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

2πr = 10 cm Base area = πr2 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

2πr = 6 cm Base area = πr2 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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