Mensuration

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

Mensuration

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 9.1, 9.2)

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

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 ∆ AFB + area of trapezium BFGC + area of right angled ∆ CGD + area of ∆ AED. (Identify the parallel sides of trapezium BFGC.)

TRY THESE

(i) Divide the following polygons (Fig 9.3) into parts (triangles and trapezium) to find out its area

FI is a diagonal of polygon EFGHI                NQ is a diagonal of polygon MNOPQR

(ii) Polygon ABCDE is divided into parts as shown below (Fig 9.4). Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Area of Polygon ABCDE = area of ∆AFB + ....

Area of ∆ AFB = \(\frac{1}{2}\)× AF × BF = \(\frac{1}{2}\) × 3 × 2 = ....

Area of trapezium FBCH = FH × (BF CH) 2 + = 3 × \(\frac{\left( 2+3 \right)}{2}\) [FH = AH – AF]

Area of ∆CHD =\(\frac{1}{2}\)× HD× CH = ....; Area of ∆ADE = \[\frac{1}{2}\] × AD × GE = ....

So, the area of polygon ABCDE = ....

(iii) Find the area of polygon MNOPQR (Fig 9.5) if MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm, MA = 2 cm NA, OC, QD and RB are perpendiculars to diagonal MP

Example 1: 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.

Solution: 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 m² .

Area of a trapezium = \(\frac{1}{2}\) h (a + b)

So 480 = \(\frac{1}{2}\)× 15 × (20 + b)

or \(\frac{480\times 2}{15}\)= 20 + b

or 64 = 20 + b or b = 4

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

Example 2: The area of a rhombus is 240 cm² and one of the diagonals is 16 cm. Find the other diagonal. Solution: Let length of one diagonal d₁ = 16 cm

and length of the other diagonal = d₂

Area of the rhombus = \(\frac{1}{2}\)d₁ . d₂ = 240

So, \(\frac{1}{2}\)16 d₂ = 240

 Therefore, d₂ = 30 cm

Hence the length of the second diagonal is 30 cm

Example 3: There is a hexagon MNOPQR of side 5 cm (Fig 9.6). Aman and Ridhima divided it in two different ways (Fig 9.7).

 Find the area of this hexagon using both ways

Solution: Aman’s method:

Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding (Fig).

Now area of trapezium MNQR = \(4\times \frac{\left( 11+5 \right)}{2}\) = 2 × 16 = 32 cm² .

So the area of hexagon MNOPQR = 2 × 32 = 64 cm² .

 Ridhima’s method:

 ∆ MNO and ∆ RPQ are congruent triangles with altitude 3 cm (Fig ).

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

Area of ∆ MNO = \(\frac{1}{2}\)× 8 × 3 = 12 cm² = Area of ∆ RPQ

Area of rectangle MOPR = 8 × 5 = 40 cm² .

Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm² .

Mensuration

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 9.10).

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

DO THIS

Soaps, toys, pastes, snacks etc. often come in the packing of cuboidal, cubical or cylindrical boxes. Collect, such boxes (Fig 9.11).

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 9.12).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 9.13).

Mensuration

Surface Area of Cube

Imran, Monica and Jaspal are painting a cuboidal, cubical and a cylindrical box respectively of same height (Fig 9.4).

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.

DO THIS

Draw the pattern shown on a squared paper and cut it out [Fig 9.20(i)]. (You know that this pattern is a net of a cube. Fold it along the lines [Fig 9.20(ii)] and tape the edges to form a cube [Fig 9.20(iii)].

(a) What is the length, width and height of the cube? Observe that all the faces of a cube are square in shape. This makes length, height and width of a cube equal (Fig 9.21(i)).

(b) Write the area of each of the faces. Are they equal?

(c) Write the total surface area of this cube.

(d) If each side of the cube is l, what will be the area of each face? (Fig 9.21(ii)). Can we say that the total surface area of a cube of side l is 6l ² ?

Mensuration

Surface Area of Cuboid

Suppose you cut open a cuboidal box and lay it flat (Fig 9.15). We can see a net as shown below (Fig 9.16).

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

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 m² .

• 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 9.18).

Hence, the lateral surface area of a cuboid is given by  2(h × l + b × h) or 2h (l + b).

Mensuration

Cylinders

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

DO THIS

(i) Take a cylindrical can or box and trace the base of the can on graph paper and cut it [Fig 9.25(i)]. Take another graph paper in such a way that its width is equal to the height of the can. Wrap the strip around the can such that it just fits around the can (remove the excess paper) [Fig 9.25(ii)]. Tape the pieces [Fig 9.25(iii)] together to form a cylinder [Fig 9.25(iv)]. What is the shape of the paper that goes around the can?

Of course it is rectangular in shape. When you tape the parts of this cylinder together, the length of the rectangular strip is equal to the circumference of the circle. Record the radius (r) of the circular base, length (l) and width (h) of the rectangular strip. Is 2πr = length of the strip. Check if the area of rectangular strip is 2πrh. Count how many square units of the squared paper are used to form the cylinder. Check if this count is approximately equal to 2πr (r + h).

 (ii) We can deduce the relation 2πr (r + h) as the surface area of a cylinder in another way. Imagine cutting up a cylinder as shown below (Fig 9.26).

The lateral (or curved) surface area of a cylinder is 2πrh.

The total surface area of a cylinder = πr ² + 2πrh + πr ²

= 2πr ² + 2πrh or 2πr (r + h)

TRY THESE

Find total surface area of the following cylinders (Fig 9.27)

Example :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?

Solution: 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 cm²

Area of the side face = b × h = 30 × 40 = 1200 cm²

Area of the back face = l × h = 80 × 40 = 3200 cm²

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

= 2400 + 3200 + (2 × 1200) = 8000 cm²

 Hence the area of the coloured paper required is 8000 cm² .

Example : 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 ` 5 per m2 . What will be the cost of white washing if the ceiling of the room is also whitewashed.

Solution: 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 m² .

 Cost of white washing per m² = ₹ 5

Hence the total cost of white washing four walls of the room = ₹ (160 × 5)  = ₹800

Area of ceiling is 12 × 8 = 96 m²

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

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

Example : 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 ₹ 8 per m² .

Solution: 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 = \(2\times \frac{22}{7}\)× 0.28 ×4  = 7.04 m²

 curved surface area of 24 such pillar = 7.04 × 24 = 168.96 m²

cost of painting an area of 1 m² = ₹8

Therefore, cost of painting 1689.6 m² = 168.96 × 8 = ₹ 1351.68

Example : Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm² .

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

Total surface area = 2πr (h + r)

i.e., \(2\times \frac{22}{7}\)× 7 × (7 + h) = 968

h = 15 cm

 Hence, the height of the cylinder is 15 cm.

Mensuration

Volume of Cube

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 9.28 ).

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

Cube

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

 Hence, volume of cube = l × l × l = l\(^3\)

Mensuration

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.

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

DO THIS

Take a sheet of paper. Measure its area. Pile up such sheets of paper of same size to make a cuboid (Fig 9.29).

Measure the height of this pile. Find the volume of the cuboid by finding the product of the area of the sheet and the height of this pile of sheets. This activity illustrates the idea that volume of a solid can be deduced by this method also (if the base and top of the solid are congruent and parallel to each other and its edges are perpendicular to the base). Can you think of such objects whose volume can be found by using this method?
TRY THESE

Find the volume of the following cuboids (Fig 9.30).

Mensuration

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 the Volume of a cuboid = area of base × height

= l × b × h = lbh

 Volume of cylinder = area of base × height

= πr ² × h = πr ²h

TRY THESE

Find the volume of the following cylinders.

Mensuration

Volume and Capacity

There is not much difference between these two words.

(a) Volume refers to the amount of space occupied by an object.

(b) Capacity refers to the quantity that a container holds.

Note: If a water tin holds 100 cm3 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 cm\(^3\) is, 1 mL = 1 cm\(^3\) ,1 L = 1000 cm\(^3\) .

 Thus, 1 m\(^3\) = 1000000 cm\(^3\) = 1000 L.

Example : Find the height of a cuboid whose volume is 275 cm\(^3\) and base area is 25 cm² .

Solution: Volume of a cuboid = Base area × Height

\(\text{Hence height of the cuboid }=\frac{\text{Volume of cuboid }}{\text{Base area}}\)

=\(\frac{275}{25}=11cm\)

Height of the cuboid is 11 cm.

Example : 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\(^3\) ?

Solution: 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 = \(\frac{\text{Volume of the godown}~}{\text{Volume of one box}}\)= \(\frac{60\times 40\times 30}{0.8}\) = 90,000

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

Example : 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 9.31). (Take 22/ 7 for π)

Solution: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.

Height of the cylinder = h = 14 cm

 Radius = r = 20 cm

Volume of the cylinder = V = π r ² h

= \(\frac{22}{7}\)× 20× 20× 14= 17600 cm\(^3\)

Hence, the volume of the cylinder is 17600 cm\(^3\) .

Example : 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.

 Solution: 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 \(\frac{22}{7}\)× 2× r = 11

Therefore, r = \(\frac{7}{4}\) cm

 Volume of the cylinder = V = πr ²h = \(\frac{22}{7}\)×  \(\frac{7}{4}\)×  \(\frac{7}{4}\)×4 cm\(^3\) = 38.5 cm\(^3\).

 Hence the volume of the cylinder is 38.5 cm\(^3\)