Mensuration

Welcome dear students! Today we are going to learn about Mensuration from Class 8 Maths.

We begin with section 9.1, 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 and so on. 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.

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Moving to section 9.2, 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. Let us observe the following for a pentagon. In Figure 9.1, by constructing two diagonals AC and AD, the pentagon ABCDE is divided into three parts. So, the area of ABCDE equals the area of ∆ABC plus the area of ∆ACD plus the area of ∆AED. In Figure 9.2, by constructing one diagonal AD and two perpendiculars BF and CG on it, the pentagon ABCDE is divided into four parts. So, the area of ABCDE equals the area of right angled ∆AFB plus the area of trapezium BFGC plus the area of right angled ∆CGD plus the area of ∆AED. Identify the parallel sides of trapezium BFGC.

Now let us try these exercises. First, divide the polygons in Figure 9.3 into parts like triangles and trapeziums to find out their areas. The first polygon is EFGHI with diagonal FI. Draw perpendiculars from vertices G and H to the diagonal FI. This divides the polygon into two triangles and one trapezium. Calculate the area of each part using 1/2 × base × height for triangles and 1/2 × h × (sum of parallel sides) for the trapezium, then add them. The second polygon is MNOPQR with diagonal NQ. Draw perpendiculars from vertices O, P, and R to the diagonal NQ. This creates triangles and trapeziums. Measure the required lengths from your textbook diagram, apply the area formulas for each part, and sum them to find the total area.

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Second, polygon ABCDE is divided into parts as shown in Figure 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. The area of polygon ABCDE equals area of ∆AFB + area of trapezium FBCH + area of ∆CHD + area of ∆ADE. Area of ∆AFB = 1/2 × AF × BF = 1/2 × 3 × 2 = 3 cm². Area of trapezium FBCH = FH × (BF + CH)/2. FH = AH − AF = 6 − 3 = 3 cm. So area is 3 × (2 + 3)/2 = 3 × 2.5 = 7.5 cm². Area of ∆CHD = 1/2 × HD × CH. HD = AD − AH = 8 − 6 = 2 cm. So 1/2 × 2 × 3 = 3 cm². Area of ∆ADE = 1/2 × AD × GE = 1/2 × 8 × 2.5 = 10 cm². So, the area of polygon ABCDE = 3 + 7.5 + 3 + 10 = 23.5 cm².

Third, find the area of polygon MNOPQR in Figure 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. The perpendiculars are NA = 2.5 cm, RB = 2.5 cm, OC = 3 cm, QD = 2 cm. The area is the sum of areas of triangles and trapeziums formed. Area of ∆MAN = 1/2 × MA × NA = 1/2 × 2 × 2.5 = 2.5 cm². Area of trapezium NABR = 1/2 × (NA + RB) × AB. AB = MB − MA = 4 − 2 = 2 cm. So 1/2 × (2.5 + 2.5) × 2 = 5 cm². Area of trapezium RBCO = 1/2 × (RB + OC) × BC. BC = MC − MB = 6 − 4 = 2 cm. So 1/2 × (2.5 + 3) × 2 = 5.5 cm². Area of trapezium OCDQ = 1/2 × (OC + QD) × CD. CD = MD − MC = 7 − 6 = 1 cm. So 1/2 × (3 + 2) × 1 = 2.5 cm². Area of ∆QDP = 1/2 × DP × QD. DP = MP − MD = 9 − 7 = 2 cm. So 1/2 × 2 × 2 = 2 cm². Total area = 2.5 + 5 + 5.5 + 2.5 + 2 = 17.5 cm².

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Now, 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 = 1/2 h (a + b). So 480 = 1/2 × 15 × (20 + b). Multiply both sides by 2: 960 = 15 × (20 + b). Divide by 15: 64 = 20 + b. Subtract 20: b = 44 m. 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 d_1 = 16 cm and length of the other diagonal = d_2. Area of the rhombus = 1/2 d_1 × d_2 = 240. So, 1/2 × 16 × d_2 = 240. 8 × d_2 = 240. Therefore, d_2 = 30 cm. Hence the length of the second diagonal is 30 cm.

Example 3. There is a hexagon MNOPQR of side 5 cm as shown in Figure 9.6. Aman and Ridhima divided it in two different ways as shown in Figure 9.7. Find the area of this hexagon using both ways. In Figure 9.6, the hexagon has side MR = 11 cm, side OP = 5 cm, and bottom RQ = 8 cm. Aman divides it by diagonal NQ into two congruent trapeziums. You can verify this by paper folding as in Figure 9.8. Now area of trapezium MNQR = 4 × (11 + 5)/2 = 4 × 8 = 32 cm². So the area of hexagon MNOPQR = 2 × 32 = 64 cm². Ridhima divides it into rectangle MOPR and two triangles MNO and RPQ, with altitude 3 cm marked on both triangles, and rectangle height 5 cm as in Figure 9.9. Triangles MNO and RPQ are congruent triangles with altitude 3 cm. You can verify this by cutting off these two triangles and placing them on one another. Area of ∆MNO = 1/2 × 8 × 3 = 12 cm². This equals area of ∆RPQ. Area of rectangle MOPR = 8 × 5 = 40 cm². Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm². Both methods give the same result.

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Let us now solve Exercise 9.1 completely. Question 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. Solution: Area of trapezium = 1/2 × height × sum of parallel sides. So area = 1/2 × 0.8 × (1 + 1.2) = 0.4 × 2.2 = 0.88 m².

Question 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. Solution: Area = 1/2 × h × (a + b). 34 = 1/2 × 4 × (10 + b). 34 = 2 × (10 + b). 17 = 10 + b. So b = 7 cm.

Question 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. Solution: Perimeter = AB + BC + CD + DA. 120 = AB + 48 + 17 + 40. 120 = AB + 105. So AB = 15 m. Since AB is perpendicular to AD and BC, it is the height. Area = 1/2 × AB × (AD + BC) = 1/2 × 15 × (40 + 48) = 7.5 × 88 = 660 m².

Question 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. Solution: Area = area of two triangles sharing the diagonal. Area = 1/2 × diagonal × (h_1 + h_2) = 1/2 × 24 × (8 + 13) = 12 × 21 = 252 m².

Question 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area. Solution: Area = 1/2 × d_1 × d_2 = 1/2 × 7.5 × 12 = 7.5 × 6 = 45 cm².

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Question 6. Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal. Solution: Area of rhombus = base × altitude = 5 × 4.8 = 24 cm². Using diagonal formula, area = 1/2 × d_1 × d_2. 24 = 1/2 × 8 × d_2. 24 = 4 × d_2. So d_2 = 6 cm.

Question 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 ₹ 4. Solution: Area of one tile = 1/2 × 45 × 30 = 675 cm². Total area for 3000 tiles = 3000 × 675 = 2025000 cm². Convert to m²: divide by 10000. So area = 202.5 m². Cost = area × rate = 202.5 × 4 = ₹ 810.

Question 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. Solution: Let side along road be x. Side along river is 2x. Height is 100. Area = 1/2 × 100 × (x + 2x). 10500 = 50 × 3x. 10500 = 150x. x = 70 m. Side along river is 2x = 140 m.

Question 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. Solution: Refer to the figure in your textbook for the exact measurements. The figure shows an octagon divided into two trapeziums and one rectangle. To find the area, calculate the area of one trapezium using 1/2 × height × (sum of parallel sides), multiply by two for both trapeziums, then add the area of the central rectangle (length × breadth). Sum these three values to get the total octagonal surface area.

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Question 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? Solution: Refer to the figure in your textbook for the exact dimensions. Jyoti divides it into two congruent trapeziums. Calculate the area of one trapezium using 1/2 × height × (sum of parallel sides) and multiply by two. Kavita divides it into a triangle and a rectangle. Calculate the triangle area as 1/2 × base × height and the rectangle area as length × breadth, then add them. Both methods will yield the same total area. Another way is to divide the pentagon into three triangles by drawing diagonals from one vertex to the non-adjacent vertices, calculate each triangle's area, and sum them.

Question 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. Solution: The frame has four sections: top, bottom, left, right. The width of the frame is uniform. Outer length 28, inner length 20. Difference is 8, so width on each side is 4 cm. Outer breadth 24, inner breadth 16. Difference is 8, so width on top and bottom is 4 cm. The top and bottom sections are trapeziums with parallel sides 28 cm and 20 cm, height 4 cm. Area of one top or bottom section = 1/2 × 4 × (28 + 20) = 2 × 48 = 96 cm². The left and right sections are trapeziums with parallel sides 24 cm and 16 cm, height 4 cm. Area of one left or right section = 1/2 × 4 × (24 + 16) = 2 × 40 = 80 cm². So the top and bottom sections each have an area of 96 cm², and the left and right sections each have an area of 80 cm².

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Now we move to section 9.3, 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 we have discussed so far in Figure 9.10. The figure shows a cuboid, cylinder, cone, cube, and pyramid. Each has vertices, faces, and edges. Observe that some shapes have two or more than two identical or congruent faces. Name them. Which solid has all congruent faces? The cube has all six faces congruent.

Do this activity. Soaps, toys, pastes, snacks and so on often come in the packing of cuboidal, cubical or cylindrical boxes. Collect such boxes as shown in Figure 9.11. The cuboidal box has all six faces rectangular, and opposite faces are identical. So there are three pairs of identical faces. The cubical box has all six faces as squares and identical. The cylindrical box has one curved surface and two circular faces which are identical. The curved surface connects the circular base and top. 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 as shown in Figure 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 as shown in Figure 9.13. Figure 9.12 shows a right circular cylinder. Figure 9.13 shows a cylinder that is slanted. Think, discuss and write: Why is it incorrect to call the solid shown here a cylinder? It is because the cross section is not uniform and the sides are not perpendicular to the base, so it does not meet the definition of a right circular cylinder.

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Section 9.4 covers Surface Area of Cube, Cuboid and Cylinder. Imran, Monica and Jaspal are painting a cuboidal, cubical and a cylindrical box respectively of same height as in Figure 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.

Subsection 9.4.1, Cuboid. Suppose you cut open a cuboidal box and lay it flat as in Figure 9.15. We can see a net as shown below in Figure 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) = 2 × 650 = 1300 cm².

Try these. Find the total surface area of the following cuboids in Figure 9.17. First cuboid has dimensions 6 cm, 4 cm, 2 cm. Surface area = 2(6 × 4 + 4 × 2 + 2 × 6) = 2(24 + 8 + 12) = 2 × 44 = 88 cm². Second cuboid has dimensions 4 cm, 4 cm, 10 cm. Surface area = 2(4 × 4 + 4 × 10 + 10 × 4) = 2(16 + 40 + 40) = 2 × 96 = 192 cm².

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The side walls, meaning 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 as shown in Figure 9.18. The figure shows a room with height h, length l, breadth b, roof and base. Hence, the lateral surface area of a cuboid is given by 2(h × l + b × h), or 2h(l + b). Do this activity. Cover the lateral surface of a cuboidal duster using a strip of brown sheet of paper, such that it just fits around the surface. Remove the paper. Measure the area of the paper. Is it the lateral surface area of the duster? Yes, it is. Second, measure length, width and height of your classroom and find the total surface area of the room, ignoring windows and doors. Find the lateral surface area of this room. Find the total area of the room which is to be white washed. This means lateral area plus ceiling area. Think, discuss and write. First, can we say that the total surface area of cuboid = lateral surface area + 2 × area of base? Yes, because total surface area includes top, bottom, and four walls. Second, if we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change? Yes, because the lateral surface area depends on the perimeter of the base times height. Changing dimensions changes the perimeter and height, so the lateral surface area will change.

Subsection 9.4.2, Cube. Do this activity. Draw the pattern shown on a squared paper and cut it out as in Figure 9.20. You know that this pattern is a net of a cube. Fold it along the lines and tape the edges to form a cube as in Figure 9.21. 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. Write the area of each of the faces. Are they equal? Yes. Write the total surface area of this cube. If each side of the cube is l, what will be the area of each face? The area is l². Can we say that the total surface area of a cube of side l is 6l²? Yes. Try these. Find the surface area of cube A and lateral surface area of cube B in Figure 9.22. Cube A has side 10 cm. Surface area = 6 × 10² = 600 cm². Cube B has side 7 cm. Lateral surface area = 4 × 7² = 4 × 49 = 196 cm².

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Think, discuss and write. First, two cubes each with side b are joined to form a cuboid. What is the surface area of this cuboid? Is it 12b²? No. When joined, two faces are hidden. The new dimensions are 2b, b, b. Surface area = 2(2b × b + b × b + b × 2b) = 2(2b² + b² + 2b²) = 2 × 5b² = 10b². Is the surface area of cuboid formed by joining three such cubes, 18b²? No. Dimensions are 3b, b, b. Surface area = 2(3b × b + b × b + b × 3b) = 2 × 7b² = 14b². Why? Because joining cubes hides faces, reducing total surface area. Second, how will you arrange 12 cubes of equal length to form a cuboid of smallest surface area? Arrange them to be as close to a cube as possible, like 3 × 2 × 2. Third, after the surface area of a cube is painted, the cube is cut into 64 smaller cubes of same dimensions. How many have no face painted? One face painted? Two faces painted? Three faces painted? The cube is 4 × 4 × 4. No face painted: inner cubes, which is 2 × 2 × 2 = 8. One face painted: center of each face, 6 faces × 2 × 2 = 24. Two faces painted: edges excluding corners, 12 edges × 2 = 24. Three faces painted: corners = 8.

Subsection 9.4.3, Cylinders. Most of the cylinders we observe are right circular cylinders. For example, a tin, round pillars, tube lights, water pipes and so on. Do this activity. Take a cylindrical can or box and trace the base of the can on graph paper and cut it. 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. Tape the pieces together to form a cylinder. 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 equal to length of the strip? Yes. 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).

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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 in Figure 9.26. The figure shows a cylinder with height h, cut vertically, then unrolled to show a top circle with area πr², a middle rectangle with length 2πr and width h, area 2πrh, and a bottom circle with area πr². Note: We take π to be 22/7 unless otherwise stated. The lateral or curved surface area of a cylinder is 2πrh. The total surface area of a cylinder = πr² + 2πrh + πr² = 2πr² + 2πrh = 2πr(r + h). Try these. Find total surface area of the following cylinders in Figure 9.27. First cylinder has radius 14 cm and height 8 cm. Total surface area = 2πr(r + h) = 2 × 22/7 × 14 × (14 + 8) = 88 × 22 = 1936 cm². Second cylinder has diameter 2 m, so radius 1 m, and length 2 m. Total surface area = 2 × 22/7 × 1 × (1 + 2) = 44/7 × 3 = 132/7 ≈ 18.86 m². Think, discuss and write. 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? Yes, because perimeter of base is 2(l + b), multiplied by height gives 2h(l + b), which matches the formula.

Example 4. 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 = b = 30 cm. Height = h = 40 cm. Area of the base = l × b = 80 × 30 = 2400 cm². Area of one 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 = 2400 + 3200 + 2400 = 8000 cm². Hence the area of the coloured paper required is 8000 cm².

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Example 5. 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 = b = 8 m. Height = 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 6. 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 × 22/7 × 0.28 × 4 = 7.04 m². Curved surface area of 24 such pillars = 7.04 × 24 = 168.96 m². Cost of painting an area of 1 m² = ₹ 8. Therefore, cost of painting 168.96 m² = 168.96 × 8 = ₹ 1351.68.

Example 7. 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 = 7 cm. Total surface area = 2πr(h + r). That is, 2 × 22/7 × 7 × (7 + h) = 968. The sevens cancel, leaving 44 × (7 + h) = 968. Divide by 44: 7 + h = 22. So h = 15 cm. Hence, the height of the cylinder is 15 cm.

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Now we solve Exercise 9.2. Question 1. There are two cuboidal boxes. Box a has length 60 cm, breadth 40 cm, height 50 cm. Box b is a cube with side 50 cm. Which box requires the lesser amount of material to make? Solution: Material needed = surface area. Box a surface area = 2(60 × 40 + 40 × 50 + 50 × 60) = 2(2400 + 2000 + 3000) = 2 × 7400 = 14800 cm². Box b surface area = 6 × 50² = 6 × 2500 = 15000 cm². Box a requires lesser material.

Question 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? Solution: Surface area of one suitcase = 2(80 × 48 + 48 × 24 + 24 × 80) = 2(3840 + 1152 + 1920) = 2 × 6912 = 13824 cm². For 100 suitcases, area = 1,382,400 cm². Tarpaulin width is 96 cm. Length required = area divided by width = 1,382,400 ÷ 96 = 14400 cm. Convert to metres: divide by 100. So 144 m of tarpaulin is required.

Question 3. Find the side of a cube whose surface area is 600 cm². Solution: Surface area = 6 × side². 600 = 6 × side². side² = 100. side = 10 cm.

Question 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. Solution: Dimensions are l = 2, b = 1, h = 1.5. Total surface area = 2(2 × 1 + 1 × 1.5 + 1.5 × 2) = 2(2 + 1.5 + 3) = 2 × 6.5 = 13 m². Subtract bottom area, which is l × b = 2 × 1 = 2 m². Painted area = 13 − 2 = 11 m².

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Question 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? Solution: Area to be painted = lateral surface area + ceiling area. Lateral area = 2(15 + 10) × 7 = 2 × 25 × 7 = 350 m². Ceiling area = 15 × 10 = 150 m². Total area = 350 + 150 = 500 m². Each can covers 100 m². Number of cans = 500 ÷ 100 = 5 cans.

Question 6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area? Figure shows a cylinder with diameter 7 cm and height 7 cm, and a cube with side 7 cm. Solution: Both have the same height of 7 cm. The cylinder has a circular base, the cube has a square base. Lateral surface area of cylinder = 2πrh. Radius is 3.5. So 2 × 22/7 × 3.5 × 7 = 154 cm². Lateral surface area of cube = 4 × side² = 4 × 49 = 196 cm². The cube has larger lateral surface area.

Question 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? Solution: Total surface area = 2πr(r + h) = 2 × 22/7 × 7 × (7 + 3) = 44 × 10 = 440 m².

Question 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? Solution: Lateral surface area = area of rectangular sheet. Area = length × width. 4224 = length × 33. Length = 4224 ÷ 33 = 128 cm. Perimeter = 2 × (length + width) = 2 × (128 + 33) = 2 × 161 = 322 cm.

Question 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. Solution: Diameter 84 cm means radius 42 cm = 0.42 m. Length = 1 m. Curved surface area of one revolution = 2πrh = 2 × 22/7 × 0.42 × 1 = 2.64 m². For 750 revolutions, area = 750 × 2.64 = 1980 m².

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Question 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. If the label is placed 2 cm from top and bottom, what is the area of the label. Solution: Diameter 14 means radius 7 cm. Label height = total height − top and bottom margins = 20 − 2 − 2 = 16 cm. Label area = lateral surface area of cylinder with height 16. Area = 2πrh = 2 × 22/7 × 7 × 16 = 44 × 16 = 704 cm².

Now section 9.5 covers 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. 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 in Figure 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 cm³ = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm³. 1 m³ = 1 m × 1 m × 1 m = 1000000 cm³. 1 mm³ = 1 mm × 1 mm × 1 mm = 0.1 cm × 0.1 cm × 0.1 cm = 0.001 cm³. We now find some expressions to find volume of a cuboid, cube and cylinder.

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Subsection 9.5.1, Cuboid. Take 36 cubes of equal size. Arrange them to form a cuboid. You can arrange them in many ways. Observe the following arrangements. First cuboid has length 12, breadth 3, height 1. Volume = 12 × 3 × 1 = 36. Second cuboid has length 6, breadth 3, height 2. Volume = 6 × 3 × 2 = 36. Third cuboid has length 9, breadth 1, height 4. Volume = 9 × 1 × 4 = 36. Fourth cuboid has length 6, breadth 6, height 1. Volume = 6 × 6 × 1 = 36. 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 activity. Take a sheet of paper. Measure its area. Pile up such sheets of paper of same size to make a cuboid. 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? Yes, like a stack of books, a deck of cards, or a pile of coins. Try these. Find the volume of the following cuboids in Figure 9.30. First has dimensions 8 cm, 3 cm, 2 cm. Volume = 8 × 3 × 2 = 48 cm³. Second has base area 24 m² and height 3 m. Volume = base area × height = 24 × 3 = 72 m³.

Subsection 9.5.2, Cube. The cube is a special case of a cuboid, where l = b = h. Hence, volume of cube = l × l × l = l³. Try these. Find the volume of the following cubes. First with side 4 cm. Volume = 4³ = 64 cm³. Second with side 1.5 m. Volume = 1.5³ = 3.375 m³. Do this activity. Arrange 64 cubes of equal size in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid shapes of same volume have same surface area? No, they usually have different surface areas. For example, 1 × 1 × 64 has huge surface area, while 4 × 4 × 4 has minimum surface area. Think, discuss and write. A company sells biscuits. For packing purpose they are using cuboidal boxes: box A is 3 cm × 8 cm × 20 cm, box B is 4 cm × 12 cm × 10 cm. What size of the box will be economical for the company? Why? Both have volume 480 cm³. Calculate surface areas. Box A surface area = 2(3 × 8 + 8 × 20 + 20 × 3) = 2(24 + 160 + 60) = 2 × 244 = 488 cm². Box B surface area = 2(4 × 12 + 12 × 10 + 10 × 4) = 2(48 + 120 + 40) = 2 × 208 = 416 cm². Box B uses less material, so it is more economical. Can you suggest any other size which has the same volume but is more economical than these? Yes, a cube shape or closer to cube, like 8 × 6 × 10, would have even less surface area.

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Subsection 9.5.3, 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 the Volume of a cuboid = area of base × height = l × b × h. Volume of cylinder = area of base × height = πr²h. Try these. Find the volume of the following cylinders. First has 7 cm diameter, so radius 3.5 cm, and 10 cm length. Volume = πr²h = 22/7 × 3.5² × 10 = 22/7 × 12.25 × 10 = 385 cm³. Second has base area 250 m², height 2 m. Volume = 250 × 2 = 500 m³.

Section 9.6 covers 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 cm³ of water then the capacity of the water tin is 100 cm³. Capacity is also measured in terms of litres. The relation between litre and cubic centimetre is, 1 mL = 1 cm³, 1 L = 1000 cm³. Thus, 1 m³ = 1000000 cm³ = 1000 L.

Example 8. Find the height of a cuboid whose volume is 275 cm³ and base area is 25 cm². Solution: Volume of a cuboid = Base area × Height. Hence height of the cuboid = Volume ÷ Base area = 275 ÷ 25 = 11 cm. Height of the cuboid is 11 cm.

Example 9. 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 m³? Solution: Volume of one box = 0.8 m³. Volume of godown = 60 × 40 × 30 = 72000 m³. Number of boxes that can be stored = Volume of godown ÷ Volume of one box = 72000 ÷ 0.8 = 90000. Hence the number of cuboidal boxes that can be stored in the godown is 90000.

Example 10. 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. 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 = πr²h = 22/7 × 20 × 20 × 14 = 22/7 × 400 × 14. The seven cancels with fourteen to give two. So 22 × 400 × 2 = 17600 cm³. Hence, the volume of the cylinder is 17600 cm³.

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Example 11. 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. So 2 × 22/7 × r = 11. Therefore, r = 7/4 cm. Volume of the cylinder = V = πr²h = 22/7 × (7/4)² × 4 = 22/7 × 49/16 × 4. The sevens cancel. 22 × 7/16 × 4. Four cancels with sixteen to give four. So 22 × 7 / 4 = 154 / 4 = 38.5 cm³. Hence the volume of the cylinder is 38.5 cm³.

Now we solve Exercise 9.3 completely. Question 1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume. First, to find how much it can hold. This requires volume. Second, number of cement bags required to plaster it. This requires surface area. Third, to find the number of smaller tanks that can be filled with water from it. This requires volume.

Question 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? Solution: Volume depends on radius squared times height. Cylinder B has double the radius, so its volume will be greater because radius is squared. Verification: Cylinder A radius 3.5, height 14. Volume = 22/7 × 3.5² × 14 = 22/7 × 12.25 × 14 = 539 cm³. Cylinder B radius 7, height 7. Volume = 22/7 × 49 × 7 = 1078 cm³. Cylinder B has greater volume. Surface area of A = 2πr(r + h) = 2 × 22/7 × 3.5 × (3.5 + 14) = 22 × 17.5 = 385 cm². Surface area of B = 2 × 22/7 × 7 × (7 + 7) = 44 × 14 = 616 cm². Yes, cylinder B also has greater surface area.

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Question 3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³? Solution: Volume = base area × height. 900 = 180 × height. Height = 900 ÷ 180 = 5 cm.

Question 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? Solution: Volume of cuboid = 60 × 54 × 30 = 97200 cm³. Volume of one small cube = 6³ = 216 cm³. Number of cubes = 97200 ÷ 216 = 450.

Question 5. Find the height of the cylinder whose volume is 1.54 m³ and diameter of the base is 140 cm? Solution: Diameter 140 cm means radius 70 cm = 0.7 m. Volume = πr²h. 1.54 = 22/7 × 0.7² × h. 1.54 = 22/7 × 0.49 × h. 1.54 = 1.54 × h. So h = 1 m.

Question 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? Solution: Volume = πr²h = 22/7 × 1.5² × 7. The sevens cancel. So 22 × 2.25 = 49.5 m³. Convert to litres: 1 m³ = 1000 L. So 49.5 × 1000 = 49500 L.

Question 7. If each edge of a cube is doubled, first, how many times will its surface area increase? Second, how many times will its volume increase? Solution: Let original edge be a. Original surface area = 6a². New edge is 2a. New surface area = 6 × (2a)² = 24a². Ratio is 24a² ÷ 6a² = 4. So surface area increases 4 times. Original volume = a³. New volume = (2a)³ = 8a³. Ratio is 8. So volume increases 8 times.

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Question 8. Water is pouring into a cuboidal 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. Solution: Volume of reservoir = 108 m³. Convert to litres: 108 × 1000 = 108000 L. Rate is 60 L/min. Time in minutes = 108000 ÷ 60 = 1800 minutes. Convert to hours: divide by 60. 1800 ÷ 60 = 30 hours.

Finally, let us review what we have discussed. Surface area of a solid is the sum of the areas of its faces. Surface area of a cuboid = 2(lb + bh + hl). Surface area of a cube = 6l². Surface area of a cylinder = 2πr(r + h). Amount of region occupied by a solid is called its volume. Volume of a cuboid = l × b × h. Volume of a cube = l³. Volume of a cylinder = πr²h. 1 cm³ = 1 mL. 1 L = 1000 cm³. 1 m³ = 1000000 cm³ = 1000 L.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]