8 Math Mensuration

Finish Line & Beyond

Mensuration 1. Area of (i) a trapezium = half of the sum of the lengths of parallel sides × perpendicular distance between them.

A

E

B

D

C

F

The area of rectangle ABCD and areas of triangles AEB and DCF will give the area of the trapezium. AS you know Area of Rectangle ABCD = AD × AB Area of ΔAEB and ΔDCF = =

1 1 × EB × AB + × CF × AB 2 2

1 × AB( EB + CF ) 2

1 × AB( EB + CF ) + AD × AB 2 1 EB + CF + 2AD 1 = AB( ( EB + CF ) + AD) = AB × = × AB( AD + AD + EB + CF ) 2 2 2 1 = × AB × ( EF + AD) 2

Area of the Trapezium=

This makes it clear how the area of a trapezium is equal to half of the sum of length multiplied by the perpendicular distance between them (ii) a rhombus = half the product of its diagonals.

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Finish Line & Beyond A

B

O

D

C In the given rhombus if we calculate the areas of all four triangles then their sum total will give the area of the rhombus. As you know that diagonals bisect each other in a rhombus so AO=OC and BO = OD so all triangles will be congruent. Hence their areas will be equal. So area of the rhombus = 4 × Area of one triangle

1 × AO × BO 2 = 2 × AO × BO = AC × BO 1 Since BO= BD So, 2 1 AC × BO = × AC × BD 2 =4×

This shows how the formula for area of rhombus is derived. 2. Surface area of a solid is the sum of the areas of its faces. 3. Surface area of a cuboid = 2(lb + bh + hl) a cube = 6l2 a cylinder = 2πr(r + h)

4. Amount of region occupied by a solid is called its volume. 5. Volume of a cuboid = l × b × h

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Finish Line & Beyond a cube = l3 a cylinder = πr2h

6. (i) 1 cm3 = 1 mL (ii) 1L = 1000 cm3 (iii) 1 m3 = 1000000 cm3 = 1000L

Exercise 1 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

60 metres (a)

80 metres (b)

Answer: Perimeter of Square = 4 × side = 4 × 60 = 240 metres Area of Square = Side² = 60² = 3600 square metre Perimeter of Rectangle = 2(length+breadth) Or, 240 = 2(80+breadth) Or, 80+breadth = 120 Or, breadth = 120-80=40 metres

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Finish Line & Beyond Area of rectangle = length

× breadth = 80 × 40 = 3200 square metres

Now it is clear that the area of the square field is greater than the area of the rectangular field. 2. Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m².

Answer: Area of the square plot = Side² = 25² = 625 square metre Area of the house construction part = length

× breadth = 20 × 15 = 300 square metre

So, area of the garden = 625-300=325 square metre Cost of developing the garden = Area = 300 × 55 = 16500 rupees

× Rate

3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].

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Finish Line & Beyond

Answer: Area of the rectangular part = length = 20 × 7 = 140 square metres. Area of semicircular portions =

=

× breadth

π r²

22 7 7 × × = 37.5 square metres 7 2 2

So, total area = 140+37.5=177.5 square metres The perimeter will be equal to the perimeter of circle and twice the length Perimeter of Circle = 2π r

= 2×

22 7 × = 22 metres 7 2

Perimeter of the shape = 22+20+20=62 metres 4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners). Answer: Area of the Parallelogram = base × height = 24 × 10 = 240 square cms Required number of tiles =

=

Area of Floor Area of Tiles

1080 × 100 × 100 = 45000 (area of floor is converted into square cms) 240

5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

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Answer: (a) Perimeter =

πd + d 2

22 2.8 × + 2.8 = 4.4 + 2.8 = 7.2 cms 7 2 πd + 2breadth + length (b) Perimeter = 2 = 4.4 + 3 + 2.8 = 10.2 cms ( perimeter of semicircular part is same as in (a)) =

(c) Perimeter =

πd + 2slant 2

= 4.4 + 4 = 8.4 cms

So, the food shape in (a) requires the ant to cover the least distance.

Exercise 2 1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. Answer: Area of Trapezium =

1 × sum of parallel sides × perpendicular distance 2

Parallel Sides = 1 and 1.2 m and Perpendicular distance = 0.8m

1 × (1 + 1.2) × 0.8 2 = 1.1 + 0.8 = 1.9 square metres =

2. The area of a trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side. Answer: Area of Trapezium =

1 × sum of parallel sides × perpendicular distance 2

Let us assume other parallel side to be x and lengths of one parallel site is 10 cms and that of perpendicular height is 4 cms as per questions

1 × (10 + x) × 4 2 = 2 × (10 + x )

So,

34 =

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Finish Line & Beyond ⇒ 17 = 10 + x ⇒ x = 17 − 10 = 7 cms

3. Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Answer: If a perpendicular from D to BC is drawn then it will be equal to AB, then EC = 48-40 = 8 m. In ΔDEC DE² = DC²-EC² ⇒ DE² = 17² - 8² = 289 - 64 = 225 ⇒ DE = 15 m Area of Trapezium =

1 × Sum of Parallel Sides × Perpendicular Distance 2

1 × (48 + 40) × 15 2 = 44 × 15 = 660 sqm =

4. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

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Finish Line & Beyond Answer: Area of Upper Triangle =

=

1 × 13 × 24 = 156 square metre 2

Area of Lower Triangle=

=

1 × height × base 2

1 × height × base 2

1 × 8 × 24 = 96 sq m 2

Area of the field = 156+96=252 square metre 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Answer: Area of Rhombus =

=

1 × Diagonal1 × Diagonal 2 2

1 × 7.5 × 12 = 45 sqcm 2

6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. Answer: If one of the Diagonals is 8 cms then other diagonal can be derived as follows using Pythagorean Theorem:

1 1 ( Diagonal1)² = Side² - ( Diagonal 2)² 2 2 ⇒ ( Altitude1)² = Side² - (Altitude2)² = 6² - 4² = 36 - 16 = 20 Altitude = 4 2 So, Diagonal 1 = 8 2 Area =

1 × 4 × 8 2 = 16 2 sqcm 2

7. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is Rs 4.

1 × 45 × 30 = 675 2 Cost of Polishing = TotalArea × Rate = 3000 × 675 × 4 = 8100000 Rupees Answer: Area of Rhombus =

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Finish Line & Beyond 8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river. Answer: Area of Trapezium =

1 × sum of parallel sides × perpendicular distance 2

1 × ( Side1 + Side 2) × 100 2 2 ( Side1 + Side 2) = 10500 × = 210 100 3 × Side1 = 210 (As one of the parallel sides is double of the other side) Side1 = 70 Side 2 = 140 This is the side near the river

⇒ 10500 = ⇒ ⇒ ⇒ ⇒

9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Answer: Upper part is in the shape of a trapezium. This is mirrored in the lower part as well. So area of two trapeziums added to the area of middle rectangular portion will be equal to the area of the octagon. The Base of the triangle runs from the base of the perpendicular to the edge of the octagon. This can be calculated using Pythagorean Theorem: Base² = Hypoteneuse² - Perpendicular²

= 5² - 4² = 25 - 16 = 9

So, Base = 3 Hence The Upper Parallel Side of the Trapezium = 11-6=5 cms (As there will be two bases of two triangles adding up to 6 cms) Second point to note is, as it is a regular octagon so the upper parallel side = side of the octagon = 5 cms

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Finish Line & Beyond Area of Trapezium=

=

1 × sum of parallel sides × perpendicular distance 2

1 55 × 11 × 5 = 2 2

So, Area of 2 trapeziums = 55 sq cms Now, Area of Rectangle = Length × Breadth

= 11 × 5 = 55

Hence, Area of the platform = 55+55=110 sq cms 10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?

Answer: Jyoti has divided the park in to congruent trapeziums. The parallel sides of the trapezium are measuring 15 metres and 30 metres and the perpendicular distance is 7.5 metres

1 × sum of parallel sides × perpendicular distance 2 So, Area of two Trapeziums = (30 + 15) × 7.5 = 337.5 sq m Area of Trapezium=

Kavita divided the park into upper triangular portion and lower rectangular portion. Sides of rectangle are 15 metres each. Base and height of triangle are 15 metres each. Area of Square = 15 × 15 = 225 Area of Triangle

=

1 × 15 × 15 = 112.5 2

Total Area =225+112.5 =337.5 sq m

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Finish Line & Beyond

11. Diagram of the adjacent picture frame has outer dimensions =24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Answer: Let us draw a perpendicular from any side of the inner rectangle to the side of the outer rectangle as shown above. This will give us 4 congruent triangles. As 24-16=8 and 28-20= 8 So, height and base of each triangle =

8 = 4 cms 2

Sides of both horizontal rectangles =4 cms and 16 cms Sides of both vertical rectangles = 4 cms and 20 cms. Now, Area of 1 triangle =

=

1 × base × height 2

1 × 4 × 4 = 8 sqcms 2

Area of the horizontal rectangle = 4 × 16 = 64 Hence, Area of lower horizontal section of the frame =64+8+8=80 sqcms Area of the vertical rectangle = 4 × 20 =80 Hence, Area of One Vertical Section of the frame = 80+8+8=96

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Exercise 3 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Answer: Volume of the Cuboid = l × = 60 × 40 × 50 = 120000 cubic cms Volume of Cube = Side³ =50³=125000

b× h

As the volume of the cube is greater than that of the cuboid so it will require more material, and the cuboid will require less material 2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases? Answer: Surface Area of Cuboid

= 2(lb + lh + bh)

= 2(80 × 48 + 80 × 24 + 48 × 24) = 2(3840 + 1920 + 1152) = 2 × 6912 = 13824

Hence, Surface Area of 100 suitcases = 1382400 sq cms Required tarpaulin will have same area So, required length of tarpaulin =

=

Area Width

1382400 = 14400 cms = 14.4 metres 96

3. Find the side of a cube whose surface area is 600 cm2. Answer: Surface Area of Cube = 6 × Side²

⇒ 600 = 6 × Side ² ⇒ Side ² = 100

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Finish Line & Beyond ⇒ Side = 10 cms 4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet. Answer: Surface Area of Cabinet

= 2(lb + lh + bh)

= 2(1 × 2 + 1 × 1.5 + 2 × 1.5) = 2(2 + 1.5 + 3) = 13 sq m Area of Bottom Surface =

1 × 2 = 2 sq m

Hence, Area Covered by painting = 13-2=11 sq m 5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will she need to paint the room? Answer: Area of All walls, ceilings and floor

= 2(lb + lh + bh)

= 2(15 × 10 + 15 × 7 + 10 × 7) = 2(150 + 105 + 70) = 2 × 325 = 650 Area of Floor = 15 × 10 = 150

Hence, Painted Area =650-150=500 As, one can of paint is enough for 100 sq m So,

500 = 5 cans are needed to paint the hall. 100

Alternate Method: Area of Walls of a Room =

2 × h × (l + b)

= 2 × 7 × (15 + 10) = 14 × 25 = 350

Area of Roof = Area of Floor = 150 Total Area to be painted = 350+150=500 6. Describe how the two figures given below are alike and how they are different. Which box has larger lateral surface area?

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Finish Line & Beyond

Answer: Similarity: Their dimensions are equal Difference: The figure on the left is cylindrical and that on the right is cuboidal. Curved Surface Area of Cylinder =

2π rh

22 7 × × 7 = 22 × 7 = 154 sq cm 7 2 Surface Area of Cube = 4 × Side² = 4 × 7² = 4 × 49 = 196 sq cms = 2×

It is clear that the cylinder is having a larger surface area. 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required? Answer: Surface Area of Cylinder = 2π r ² + 2π rh

= 2×

= 2π r (r + h)

22 × 7(7 + 3) = 44 × 10 = 440 sq m sheet is required 7

8. The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet? Answer: Area of Rectangular Sheet = Surface Area of the Cylinder Area of Rectangle = l × b

⇒ l × b = 4224 ⇒ l × 33 = 4224 4224 ⇒ l= = 128 cms 33

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

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Finish Line & Beyond Answer: In one revolution the wheel of the roller will cover an area equal to its surface area. Curved Surface Area of Cylinder

= 2×

= 2π rh

22 × 42 × 100 =26400 sq cms = 2.64 sq m 7

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.

Answer: The cylindrical area covered by the level will have height = 20-4=16 cms Curved Surface Area of Cylinder = 2π rh

= 2×

22 × 7 × 16 = 704 sq cms 7

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Finish Line & Beyond

Exercise 4 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it. Answer: (a) We need to calculate the volume to find the capacity (b) As plastering will cover the surface so we need surface area to know this (c) Volume will give the capacity and that can be compared with capacity of smaller tanks 2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area? Answer: As cylinder A’s radius is half of radius of cylinder B so its volume will be lesser than that of cylinder B. Although Cylinder B’s height is half of height of cylinder A but as you know while calculating the volume we need to square the radius so halving the radius has a greater impact than halving the height. While calculating surface area, the curved surface area in both will be same and the total surface area will be greater in the cylinder with the greater radius. Volume of Cylinder = π r ²h

22 7 7 × × × 14 = 539 7 2 2 22 × 7 × 7 × 7 = 1078 Volume of Cylinder B = 7 Curved Surface Area of Cylinder = 2π rh 22 7 × × 14 = 308 Curved surface Area of Cylinder A = 2 × 7 2 22 × 7 × 7 = 308 Curved Surface Area of Cylinder B = 2 × 7 Volume of Cylinder A

=

2π r (r + h) 22 7 7 × × ( + 14) Total Surface Area of Cylinder A = 2 × 7 2 2 35 = 22 × = 385 2 Total Surface Area of Cylinder =

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Finish Line & Beyond Total Surface Area of Cylinder B

= 2×

= 44 × 14 = 616

22 × 7 × (7 + 7 ) 7

3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3? Answer: Volume = Base Area × Height Hence, Height =

Volume 900 = = 5 cms BaseArea 180

4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid? Answer: Number of Cubes

=

60 × 54 × 30 6× 6× 6 = 10 × 9 × 5 = 450

Volume of Cuboid Volume of Cubes

=

5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm ? Answer: Volume of Cylinder = π r ²h Hence. Height =

Volume π r²

1.54 22 70 70 × × 7 100 100 1.54 × 100 154 = = = 1 metre 22 × 7 154 =

6. A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank? Answer: Volume of Milk Tank =

=

π r ²h

22 × 1.5 × 1.5 × 7 = 49.5 cubic metre 7

As we know, 1 cubic metre = 1000 litres So, 49.5 cubic metre = 49500 litres 7. If each edge of a cube is doubled, (i) how many times will its surface area increase? (ii) how many times will its volume increase?

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Finish Line & Beyond Answer: (i) Whenever sides are doubled in any structure then area becomes 4 times the original structure (ii) Volume becomes 8 times of the original volume if sides are doubled in any structure 8. Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m³, find the number of hours it will take to fill the reservoir. Answer: 108 cubic metre = 108000 litre So, time = Volume ÷ Rate per minute

108000 min utes 60 108000 = hours = 30hours 60 × 60 =

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