8 math mensuration

Finish Line & Beyond

Mensuration

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

A

B C

D

E 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 = ABAD ×

Area of ΔAEB and ΔDCF = ABCFABEB ××+××21

21

= )(21 CFEBAB +×

Area of the Trapezium= ABADCFEBAB ×++× )(21

))(21( ADCFEBAB ++=

22ADCFEBAB ++×= )(

21 CFEBADADAB +++×=

)(21 ADEFAB +××=

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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A

B D

C

O

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

= BOAO ×××214

BOAO ××= 2= BOAC ×

Since BO=21

BD So,

BOAC × BDAC ××=21

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 ofa cuboid = 2(lb + bh + hl)a cube = 6l2a cylinder = 2πr(r + h)

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




a cube = l3a 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 fig-ure have the same perimeter. Which field has a larger area?

60 metres 80 metres (a) (b)

Answer: Perimeter of Square = 4× side = 240604 =× metresArea of Square = Side² = 60² = 3600 square metre

Perimeter of Rectangle = 2(length+breadth)Or, 240 = 2(80+breadth)Or, 80+breadth = 120Or, breadth = 120-80=40 metres




Area of rectangle = length × breadth = 32004080 =× square metres

Now it is clear that the area of the square field is greater than the area of the rectan-gular 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 = 3001520 =× square metre

So, area of the garden = 625-300=325 square metre

Cost of developing the garden = Area × Rate= 1650055300 =× rupees

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].




Answer: Area of the rectangular part = length × breadth 140720 =×= square metres.

Area of semicircular portions = π r²

5.3727

27

722 =××= square metres

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 = π2 r

2227

7222 =××= metres

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 cov-er a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill upthe corners).

Answer: Area of the Parallelogram = base× height= 2401024 =× square cms

Required number of tiles = Tiles of AreaFloor of Area

45000240

1001001080 =××= (area of floor is converted into square cms)

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 ris the radius of the circle.




Answer: (a) Perimeter = dd +2

π

2.78.24.48.228.2

722 =+=+×= cms

(b) Perimeter = lengthbreadthd ++ 22

π

2.108.234.4 =++= cms ( perimeter of semicircular part is same as in (a))

(c) Perimeter = slantd 22

+π

4.844.4 =+= 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 = ×21

sum of parallel sides × perpendicular distance

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

8.0)2.11(21 ×+×=

9.18.01.1 =+= 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.



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

So, 4)10(2134 ×+×= x

)10(2 x+×=




x+=⇒ 101771017 =−=⇒ x 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 perpen-dicular 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 22564-2898²-²17 ===

15=⇒ DE m

Area of Trapezium = ×21

Sum of Parallel Sides× Perpendicular Distance

15)4048(21 ×+×=

6601544 =×= sqm

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




Answer: Area of Upper Triangle = baseheight ××21

156241321 =××= square metre

Area of Lower Triangle= baseheight ××21

9624821 =××= sq m

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 = 2121 DiagonalDiagonal ××

45125.721 =××= sqcm

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 diagon-al.

Answer: If one of the Diagonals is 8 cms then other diagonal can be derived as fol-lows using Pythagorean Theorem:

²)221(-Side²²)1

21( DiagonalDiagonal =

)²(Altitude2-Side²²)1( =⇒ Altitude2016-364²-²6 ===

Altitude = 24So, Diagonal 1 = 28

Area = 21628421 =×× sqcm

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.

Answer: Area of Rhombus = 675304521 =××

Cost of Polishing RateTotalArea ×=810000046753000 =××= Rupees




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.



100)21(2110500 ×+×=⇒ SideSide

210100

210500)21( =×=+⇒ SideSide

21013 =×⇒ Side (As one of the parallel sides is double of the other side)701 =⇒ Side1402 =⇒ Side This is the side near the river

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 lar²Perpendicu-²eHypoteneus=

916-254²-²5 ===So, Base = 3Hence 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




Area of Trapezium= ×21


255511

21 =××=

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

55511 =×=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 dis-tance is 7.5 metres

Area of Trapezium= ×21


So, Area of two Trapeziums 5.3375.7)1530( =×+= sq m

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 22515 =×

Area of Triangle 5.112151521 =××=

Total Area =225+112.5 =337.5 sq m




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 = 428 = cms

Sides of both horizontal rectangles =4 cms and 16 cmsSides of both vertical rectangles = 4 cms and 20 cms.

Now, Area of 1 triangle = heightbase ××21

84421 =××= sqcms

Area of the horizontal rectangle = 64164 =×Hence, Area of lower horizontal section of the frame =64+8+8=80 sqcms

Area of the vertical rectangle = 204 × =80Hence, Area of One Vertical Section of the frame = 80+8+8=96




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 = hbl ××120000504060 =××= cubic cms

Volume of Cube = Side³=50³=125000

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 bhlhlb ++=)244824804880(2 ×+×+×=

)115219203840(2 ++= 1382469122 =×=Hence, Surface Area of 100 suitcases = 1382400 sq cmsRequired tarpaulin will have same area

So, required length of tarpaulin = WidthArea

1440096

1382400 == cms = 14.4 metres

3. Find the side of a cube whose surface area is 600 cm2.

Answer: Surface Area of Cube ×= 6 Side²²6600 Side×=⇒

100² =⇒ Side




10=⇒ Side 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 bhlhlb ++=)5.125.1121(2 ×+×+×=

13)35.12(2 =++= sq mArea of Bottom Surface = 221 =× 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 bhlhlb ++=)7107151015(2 ×+×+×=

)70105150(2 ++=6503252 =×=

Area of Floor = 1501015 =×Hence, Painted Area =650-150=500

As, one can of paint is enough for 100 sq m

So, 5100500 = cans are needed to paint the hall.

Alternate Method: Area of Walls of a Room = )(2 blh +××)1015(72 +××=

3502514 =×=Area of Roof = Area of Floor = 150Total Area to be painted = 350+150=500

6. Describe how the two figures given below are alike and how they are dif-ferent. Which box has larger lateral surface area?




Answer: Similarity: Their dimensions are equalDifference: The figure on the left is cylindrical and that on the right is cuboidal.

Curved Surface Area of Cylinder h2 rπ=

154722727

7222 =×=×××= sq cm

Surface Area of Cube ²4 Side×=196494²74 =×=×= sq cms

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 rh2²2 ππ += r )(2 hrr += π

4401044)37(77222 =×=+××= sq m sheet is required

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 perimet-er of rectangular sheet?

Answer: Area of Rectangular Sheet = Surface Area of the CylinderArea of Rectangle bl ×=

4224=×⇒ bl422433 =×⇒ l

12833

4224 ==⇒ l cms

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.




Answer: In one revolution the wheel of the roller will cover an area equal to its sur-face area.

Curved Surface Area of Cylinder h2 rπ=

100427222 ×××= =26400 sq cms = 2.64 sq m

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 cmsCurved Surface Area of Cylinder = rhπ2

7041677222 =×××= sq cms




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 cylin-der 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 cylin-der 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 = ²hrπ

Volume of Cylinder A 5391427

27

722 =×××=

Volume of Cylinder B 1078777722 =×××=

Curved Surface Area of Cylinder rhπ2=

Curved surface Area of Cylinder A 3081427

7222 =×××=

Curved Surface Area of Cylinder B 308777222 =×××=

Total Surface Area of Cylinder )(2 hrr += π

Total Surface Area of Cylinder A = )1427(

27

7222 +×××

3852

3522 =×=




Total Surface Area of Cylinder B )77(77222 +×××=

6161444 =×=

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 = 5180900 ==

BaseAreaVolume

cms

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 Cubes of VolumeCuboid of Volume=

666305460

××××=

4505910 =××=

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 = ²hrπ

Hence. Height = ²r

Volumeπ

10070

10070

722

54.1

××=

1154154

72210054.1 ==

××= metre

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 = ²hrπ

5.4975.15.1722 =×××= cubic metre

As we know, 1 cubic metre = 1000 litresSo, 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?




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 struc-ture

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 litreSo, time = Volume ÷ Rate per minute

utesmin60

108000=

hourshours 306060

108000 =×

=



8 math mensuration

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