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# 11 Mensuration

## 11.1 Introduction

We have learnt that for a closed plane figure, the perimeter is the distance around its boundary and its area is the region covered by it. We found the area and perimeter of various plane figures such as triangles, rectangles, circles etc. We have also learnt to find the area of pathways or borders in rectangular shapes.

In this chapter, we will try to solve problems related to perimeter and area of other plane closed figures like quadrilaterals.

We will also learn about surface area and volume of solids such as cube, cuboid and cylinder.

## 11.2 Let us Recall

Let us take an example to review our previous knowledge.

This is a figure of a rectangular park (Fig 11.1) whose length is 30 m and width is 20 m.

- What is the total length of the fence surrounding it? To find the length of the fence we need to find the perimeter of this park, which is 100 m. (Check it)
- How much land is occupied by the park? To find the land occupied by this park we need to find the area of this park which is 600 square meters (m2) (How?).
- There is a path of one metre width running inside along the perimeter of the park that has to be cemented. If 1 bag of cement is required to cement 4 m2 area, how many bags of cement would be required to construct the cemented path?
- There are two rectangular flower beds of size 1.5 m × 2 m each in the park as shown in the diagram (Fig 11.1) and the rest has grass on it. Find the area covered by grass.

**Fig 11.1**

We can say that the number of cement bags used = area of the path/area cemented by 1 bag

Area of cemented path = Area of park - Area of park not cemented. Path is 1 m wide, so the rectangular area not
cemented is (30 - 2) × (20 - 2) m^{2}. That is 28 × 18 m^{2}.

Hence number of cement bags used = ------------------

Area of rectangular beds = ------------------

Area of park left after cementing the path = ------------------

Area covered by the grass = ------------------

We can find areas of geometrical shapes other than rectangles also if certain
measurements are given to us . Try to recall and match the following:

## 11.3 Area of Trapezium

Nazma owns a plot near a main road (Fig 11.2). Unlike some other rectangular plots in her neighbourhood, the plot has only one pair of parallel opposite sides. So, it is nearly a trapezium in shape. Can you find out its area? Let us name the vertices of this plot as shown in Fig 11.3.

By drawing EC || AB, we can divide it into two parts, one of rectangular shape and the other of triangular shape, (which is right angled at C), as shown in Fig 11.3.

**Fig 11.2**

**Fig 11.3**

Area of triangle ECD = ½h × c = ½×12×10 = 60 m^{2}.

Area of rectangle ABCE = h × a = 12 × 20 = 240 m^{2}.

Area of trapezium ABDE = area of triangle ECD + Area of rectangle ABCE = 60 + 240 = 300 m^{2}.

We can write the area by combining the two areas and write the area of trapezium as

area of ABDE = ½h × c + h × a = h(c/2 + a)

= h((c+2a)/2) = h((c+a+a)/2)

=h((b+a)/2)=height(sum of parallel sides/2)

By substituting the values of h, b and a in this expression, we find h (b+a)/2 = 300m^{2}

So to find the area of a trapezium we need to know the length of the parallel sides and the perpendicular distance between these two parallel sides. Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the area of trapezium.

We know that all congruent figures are equal in area. Can we say figures equal in area need to be congruent too? Are these figures congruent?

Draw at least three trapeziums which have different areas but equal perimeters on a squared sheet.

## 11.4 Area of a General Quadrilateral

A general quadrilateral can be split into two triangles by drawing one of its diagonals. This “triangulation” helps us to find a formula for any general quadrilateral. Study the Fig 11.10.

Area of quadrilateral ABCD = (area of Δ ABC) + (area of Δ ADC)

= (½AC × h_{1}) + (½AC × h_{2})

**Fig 11.10**

= ½ AC × ( h1 + h2)

= ½ d ( h1 + h2) where d denotes the length of diagonal AC.

Find the area of quadrilateral PQRS shown in Fig 11.11.

In this case, d = 5.5 cm, h1 = 2.5cm, h2 = 1.5 cm,

Area = ½ d ( h1 + h2)

= ½ × 5.5 × (2.5 + 1.5) cm^{2}

= ½ × 5.5 × 4 cm^{2} = 11 cm^{2}

**Fig 11.11**

### 11.4.1 Area of special quadrilaterals

We can use the same method of splitting into triangles (which we called “triangulation”) to find a formula for the area of a rhombus. In Fig 11.13 ABCD is a rhombus. Therefore, its diagonals are perpendicular bisectors of each other.

Area of rhombus ABCD = (area of ΔACD) + (area of ΔABC)

= (½× AC × OD) +(½× AC × OB) =½AC × (OD + OB)

= AC × BD =d_{1} × d_{2} where AC = d_{1} and BD = d_{2}

**Fig 11.13**

In other words, area of a rhombus is half the product of its diagonals.

Find the area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm.

Area of the rhombus =½ d_{1} d_{2} where d_{1}, d_{2} are lengths of diagonals.

=½ × 10 × 8.2 cm2 = 41 cm2.

A parallelogram is divided into two congruent triangles by drawing a diagonal across it. Can we divide a trapezium into two congruent triangles?

## 11.5 Area of a Polygon

We split a quadrilateral into triangles and find its area. Similar methods can be used to find the area of a polygon. Observe the following for a pentagon: (Fig 11.15, 11.16)

**Fig 11.15**

By constructing two diagonals AC and AD the pentagon ABCDE is divided into three parts. So, area ABCDE = area of Δ ABC + area of Δ ACD + area of Δ AED.

**Fig 11.16**

By constructing one diagonal AD and two perpendiculars BF and CG on it, pentagon ABCDE is divided into four parts. So, area of ABCDE = area of right angled Δ AFB + area of trapezium BFGC + area of right angled Δ CGD + area of Δ AED. (Identify the parallel sides of trapezium BFGC.)

The area of a trapezium shaped field is 480 m2, the distance between two parallel sides is 15 m and one of the parallel side is 20 m. Find the other parallel side.

One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.
The given area of trapezium = 480 m^{2}.

Area of a trapezium =½ h (a + b)

So 480 =× 15 × ½(20 + b) or (480 x 2)/15 = 20 + b

or 64 = 20 + b or b = 44 m

Hence the other parallel side of the trapezium is 44 m.

The area of a rhombus is 240 cm2 and one of the diagonals is 16 cm. Find the other diagonal.

Let length of one diagonal d_{1} = 16 cm
and length of the other diagonal = d_{2}

Area of the rhombus = ½ d_{1} . d_{2} = 240

So, ½ 16.d_{2} = 240

Therefore, d_{2} = 30 cm

Hence the length of the second diagonal is 30 cm.

There is a regular hexagon MNOPQR of side 5 cm (Fig 11.20).Aman and Ridhima divided it in two different ways (Fig 11.21). Find the area of this hexagon using both ways.

**Fig 11.20**

**Fig 11.21**

**Aman's Method**

Since it is a regular hexagon so NQ divides the hexagon into two congruent trapeziums.

You can verify it by paper folding (Fig 11.22).

Now area of trapezium MNQR = 4× [11 + 5] / 2 = 2 × 16 = 32 cm^{2}.

**Fig 11.22**

So the area of hexagon MNOPQR = 2 × 32 = 64 cm^{2}.

**Ridhima’s method:**

ΔMNO and ΔRPQ are congruent triangles with altitude 3 cm (Fig 11.23).

You can verify this by cutting off these two triangles and placing them on one another.

**Fig 11.23**

Area of ΔMNO = ½ × 8 × 3 = 12 cm^{2} = Area of ΔRPQ

Area of rectangle MOPR = 8 × 5 = 40 cm^{2}.

Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm^{2}.

## 11.6 Solid Shapes

In your earlier classes you have studied that two dimensional figures can be identified as the faces of three dimensional shapes. Observe the solids which we have discussed so far (Fig 11.24).

**Fig 11.24**

Observe that some shapes have two or more than two identical (congruent) faces. Name them. Which solid has all congruent faces?

All six faces are rectangular, and opposites faces are identical. So there are three pairs of identical faces.

All six faces are squares and identical.

One curved surface and two circular faces which are identical.

**Circular base and top are identical**

Now take one type of box at a time. Cut out all the faces it has. Observe the shape of each face and find the number of faces of the box that are identical by placing them on each other. Write down your observations.

Did you notice the following:

The cylinder has congruent circular faces that are parallel to each other (Fig 11.26). Observe that the line segment joining the center of circular faces is perpendicular to the base. Such cylinders are known as **right circular cylinders**. We are only going to study this type of cylinders, though there are other types of cylinders as well (Fig 11.27).

**Fig 11.26**

(This is right circular cynlinder)

**Fig 11.27**

(This is not a right circular cynlinder)

Why is it incorrect to call the solid shown here a cylinder?

## 11.7 Surface Area of Cube, Cuboid and Cylinder

Imran, Monica and Jaspal are painting a cuboidal, cubical and a cylindrical box respectively of same height (Fig 11.28).

**Fig 11.28**

They try to determine who has painted more area. Hari suggested that finding the surface area of each box would help them find it out.

To find the total surface area, find the area of each face and then add. The surface area of a solid is the sum of the areas of its faces. To clarify further, we take each shape one by one.

### 11.7.1 Cuboid

Suppose you cut open a cuboidal box and lay it flat (Fig 11.29).We can see a net as shown below (Fig 11.30).

Write the dimension of each side. You know that a cuboid has three pairs of identical faces. What expression can you use to find the area of each face?

Find the total area of all the faces of the box. We see that the total surface area of a cuboid is area I + area II + area III + area IV +area V + area VI

= h × l + b × l + b × h + l × h + b × h + l × b

**Fig 11.29**

**Fig 11.30**

So total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl ) where h, l and b are the height, length and width of the cuboid respectively.

Suppose the height, length and width of the box shown above are 20 cm, 15 cm and 10 cm respectively.

Then the total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)

= 2 ( 300 + 200 + 150) = 1300 m^{2}.

The side walls (the faces excluding the top and bottom) make the lateral surface area of the cuboid. For example, the total area of all the four walls of the cuboidal room in which you are sitting is the lateral surface area of this room (Fig 11.32).Hence, the lateral surface area of a cuboid is given by 2(h × l + b × h) or 2h (l + b).

**Fig 11.32**

- Can we say that the total surface area of cuboid = lateral surface area + 2 × area of base?
- If we interchange the lengths of the base and the height of a cuboid (Fig 11.33(i)) to get another cuboid (Fig 11.33(ii)), will its lateral surface area change?

(i)

(ii)

**Fig 11.33**

### 11.7.2 Cube

- Two cubes each with side b are joined to form a cuboid (Fig 11.37). What is the surface area of this cuboid? Is it 12b2? Is the surface area of cuboid formed by joining three such cubes, 18b2? Why?
- How will you arrange 12 cubes of equal length to form a cuboid of smallest surface area?
- After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions (Fig 11.38). How many have no face painted? 1 face painted? 2 faces painted? 3 faces painted?

**Fig 11.37**

### 11.7.3 Cylinders

Most of the cylinders we observe are right circular cylinders. For example, a tin, round pillars, tube lights, water pipes etc.

Note that lateral surface area of a cylinder is the circumference of base × height of cylinder. Can we write lateral surface area of a cuboid as perimeter of base × height of cuboid?

An aquarium is in the form of a cuboid whose external measures are 80 cm × 30 cm × 40 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed?

The length of the aquarium = l = 80 cm

Width of the aquarium = b = 30 cm

Height of the aquarium = h = 40 cm

Area of the base = l × b = 80 × 30 = 2400 cm^{2}

Area of the side face = b × h = 30 × 40 = 1200 cm^{2}

Area of the back face = l × h = 80 × 40 = 3200 cm^{2}

Required area = Area of the base + area of the back face + (2 × area of a side face)

= 2400 + 3200 + (2 × 1200) = 8000 cm^{2}

Hence the area of the coloured paper required is 8000 cm^{2}.

The internal measures of a cuboidal room are 12 m × 8 m × 4 m. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is Rs 5 per m^{2}. What will be the cost of white washing if the ceiling of the room is also white washed.

Let the length of the room = l = 12 m

Width of the room = b = 8 m

Height of the room = h = 4 m

Area of the four walls of the room = Perimeter of the base × Height of the room

= 2 (l + b) × h = 2 (12 + 8) × 4

= 2 × 20 × 4 = 160 m^{2}.

Cost of white washing per m^{2}= Rs 5

Hence the total cost of white washing four walls of the room = Rs (160 × 5)

Area of ceiling is 12 × 8 = 96 m^{2}

Cost of white washing the ceiling = Rs (96 × 5) = Rs 480

So the total cost of white washing = Rs (800 + 480) = Rs 1280

In a building there are 24 cylindrical pillars. The radius of each pillar is 28 cm and height is 4 m. Find the total cost of painting the curved surface area of all pillars at the rate of Rs 8 per m^{2}.

Radius of cylindrical pillar, r = 28 cm = 0.28 m

height, h = 4 m

curved surface area of a cylinder = 2πrh

curved surface area of a pillar =2 × 22/7 × 0.28 × 4=7.04 m^{2}

curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m^{2}

cost of painting an area of 1 m^{2} = Rs 8

Therefore, cost of painting 1689.6 m^{2} = 168.96 × 8 = Rs 1351.68

Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm^{2}.

Let height of the cylinder = h, radius = r = 7cm

Total surface area = 2πr (h + r)

i.e., 2 × 22/7 × 7 × (7 + h) = 968

h = 15 cm

Hence, the height of the cylinder is 15 cm.

## 11.8 Volume of Cube, Cuboid and Cylinder

Amount of space occupied by a three dimensional object is called its volume. Try to compare the volume of objects surrounding you. For example, volume of a room is greater than the volume of an almirah kept inside it. Similarly, volume of your pencil box is greater than the volume of the pen and the eraser kept inside it. Can you measure volume of either of these objects?

Remember, we use square units to find the area of a region. Here we will use cubic units to find the volume of a solid, as cube is the most convenient solid shape (just as square is the most convenient shape to measure area of a region).

For finding the area we divide the region into square units, similarly, to find the volume of a solid we need to divide it into cubical units. Observe that the volume of each of the adjoining solids is 8 cubic units (Fig 11.42 ).

**Fig 11.42**

We can say that the volume of a solid is measured by counting the number of unit cubes it contains. Cubic units which we generally use to measure volume are

1 cubic cm = 1 cm × 1 cm × 1 cm = 1 cm^{3}

= 10 mm × 10 mm × 10 mm = .........mm^{3}

1 cubic m = 1 m × 1 m × 1 m = 1 m3

= ............................... cm^{3}

1 cubic mm = 1 mm × 1 mm × 1 mm = 1 mm^{3}

= 0.1 cm × 0.1 cm × 0.1 cm = ...................... cm^{3}

We nowfind some expressions to find volume of a cuboid, cube and cylinder. Let us take each solid one by one.

### 11.8.1 Cuboid

Take 36 cubes of equal size (i.e., length of each cube is same). Arrange them to form a cuboid. You can arrange them in many ways. Observe the following table and fill in the blanks.

Sr.no | cuboid | length | breadth | height | l × b × h = V |

(i) | 12 | 3 | 1 | 12 × 3 × 1 = 36 | |

(ii) | .... | .... | .... | .... | |

(iii) | .... | .... | .... | .... | |

(iv) | .... | .... | .... | .... |

What do you observe?

Since we have used 36 cubes to form these cuboids, volume of each cuboid is 36 cubic units. Also volume of each cuboid is equal to the product of length, breadth and height of the cuboid. From the above example we can say volume of cuboid = l × b × h. Since l × b is the area of its base we can also say that,

Volume of cuboid = area of the base × height

### 11.8.2 Cube

The cube is a special case of a cuboid, where l = b = h.

Hence, volume of cube = l × l × l = l^{3}

A company sells biscuits. For packing purpose they are using cuboidal boxes:

box A →3 cm × 8 cm × 20 cm, box B →4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Can you suggest any other size (dimensions) which has the same volume but is more economical than these?

### 11.8.3 Cylinder

We know that volume of a cuboid can be found by finding the product of area of base and its height. Can we find the volume of a cylinder in the same way?

Just like cuboid, cylinder has got a top and a base which are congruent and parallel to each other. Its lateral surface is also perpendicular to the base, just like cuboid.

So theVolume of a cuboid = area of base × height

= l × b × h = lbh

Volume of cylinder = area of base × height

= πr^{2} × h = πr^{2}h

## 11.9 Volume and Capacity

There is not much difference between these two words.

- Volume refers to the amount of space occupied by an object.
- Capacity refers to the quantity that a container holds.

Note: If a water tin holds 100 cm^{3} of water then the capacity of the water tin is 100 cm^{3}.

Capacity is also measured in terms of litres. The relation between litre and cm3 is,
1 mL = 1 cm^{3},1 L = 1000 cm^{3}. Thus, 1 m3 = 1000000 cm^{3} = 1000 L.

Find the height of a cuboid whose volume is 275 cm^{3} and base area is 25 cm^{2} .

Volume of a cuboid = Base area × Height

Hence height of the cuboid = Volume of cuboid / Base of area

275 / 25 = 11 cm

Height of the cuboid is 11 cm.

A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m^{3} ?

Volume of one box = 0.8 m^{3}

Volume of godown = 60 × 40 × 30 = 72000 m^{3}

Number of boxes that can be stored in the godown = Volume of godown / Volume of one box

= (60 x 40 x 30) / 0.8 = 90,000

Hence the number of cuboidal boxes that can be stored in the godown is 90,000.

A rectangular paper of width 14 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder (Fig 11.45)? (Take 22/7 for π)

A cylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is 20 cm.

**Fig 11.45**

Height of the cylinder = h = 14 cm

Radius = r = 20 cm

Volume of the cylinder = V = πr^{2}h

= 22/7 × 20 × 20 × 14=17600 cm^{3}

Hence, the volume of the cylinder is 17600 cm^{3}.

A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.

Length of the paper becomes the perimeter of the base of the cylinder and width becomes height.

Let radius of the cylinder = r and height = h

Perimeter of the base of the cylinder = 2πr = 11

or 2× 22/7 ×r = 11

Therefore, r= 7/4 cm

Volume of the cylinder = V = πr^{2}h

22/7 × 7/4 × 7/4 × 4cm^{3} = 38.5 cm^{3}

Hence the volume of the cylinder is 38.5 cm^{3}.