Mensuration | ICSE Class 9 Maths Notes

This note covers perimeter and area of triangles and quadrilaterals, Heron’s formula, circumference and area of circles, semicircles and quarter-circles, inner and outer regions, surface area and volume of cubes and cuboids, open containers, internal dimensions, capacity and costs.

What do perimeter, area and volume measure?

Mensuration is the measurement of lengths, areas and volumes of geometric figures. A plane figure lies in a flat surface. Its perimeter is the total length of its boundary, while its area measures the region enclosed by that boundary.

A solid occupies three-dimensional space. Its surface area measures its exposed surfaces, while its volume measures the space it occupies. The capacity of a hollow container is the volume available inside it.

Which units belong to each measurement?

The abbreviations cm, m and mm mean centimetre, metre and millimetre. Perimeter uses units of length. Area uses square units, such as cm², meaning square centimetres. Volume uses cubic units, such as cm³, meaning cubic centimetres.

A unit square has side one unit; a unit cube has length, breadth and height one unit each. Thus area counts square units, whereas volume counts cubic units. This distinction explains why a surface calculation cannot directly give a container’s capacity.

The signs =, ×, + and − mean equals, multiplied by, plus and minus. Use 1 m = 100 cm, so 1 m² = 10,000 cm² and 1 m³ = 1,000,000 cm³. Also, 1 cm³ = 1 mL and 1 L = 1,000 cm³. Convert lengths to a common unit before multiplying them.

Note: Squaring a length means multiplying it by itself; cubing it means multiplying three copies. An area conversion therefore uses two length-conversion factors, while a volume conversion uses three.

How are a triangle’s perimeter and area calculated?

A triangle is a closed figure with three straight sides. A vertex is a corner where sides meet. Let a, b and c denote its side lengths, P its perimeter and A its area. Then P = a + b + c.

Choose one side as the base. Its corresponding altitude, or height, is the perpendicular distance from the opposite vertex to the line containing that base. Perpendicular lines meet at a right angle, which measures 90°, where ° denotes degrees.

Result: Area from a base and its corresponding height

For this formula, let b denote the chosen base length and h the corresponding height. Then A = ½bh. The factor ½ means one-half. A sloping side cannot replace h unless it is perpendicular to the chosen base.

Two congruent copies, meaning copies with the same shape and size, of a triangle can be fitted into a parallelogram. A parallelogram is a quadrilateral, or four-sided figure, with both pairs of opposite sides parallel. Parallel lines lie in the same plane and stay the same distance apart. Its area is base multiplied by perpendicular height.

The two triangular copies occupy equal areas and together fill that parallelogram. Each triangle therefore has half the parallelogram’s area. For an obtuse triangle, containing an angle greater than 90°, the altitude may meet an extension of the base.

Worked example 1. A right-angled triangle has perpendicular sides 3 units and 4 units, and its third side is 5 units. Find its perimeter and area.

Answer: P = 3 + 4 + 5 = 12 units. Taking the perpendicular sides as base and height gives A = ½ × 3 × 4 = 6 square units. The 5-unit side is the hypotenuse, meaning the side opposite the right angle.

When naming a segment, two capital letters identify its endpoints; thus AB names the segment joining vertices A and B. The same letters in the diagram identify points, not the area symbol used in formulas.

The corners are labelled A, B and C. A is above B, while C is to the right of B. The lengths AB, BC and AC are labelled 4, 3 and 5 respectively.

See Fig. 6.25 in your NCERT textbook

How does Heron’s formula use three side lengths?

Heron’s formula gives a triangle’s area when all three side lengths are known. It avoids first calculating an altitude. Let a, b and c be the three side lengths and let s be the semiperimeter, meaning half their sum.

Result: Heron’s formula

s = (a + b + c)/2 and A = √[s(s − a)(s − b)(s − c)]. Here A means area, / means division, and √ means the non-negative square root. Adjacent bracketed factors are multiplied together.

The lengths must form a triangle: the sum of any two must exceed the third. First calculate s, then calculate each difference from that same s. All four factors remain inside the square root. The final area has square units.

Worked example 2. Find the area of a triangle with sides 3 units, 4 units and 5 units using Heron’s formula.

Answer: s = (3 + 4 + 5)/2 = 6 units. Therefore A = √[6(6 − 3)(6 − 4)(6 − 5)] = √(6 × 3 × 2 × 1) = √36 = 6 square units. This agrees with the base-and-height calculation.

How can the perimeter supply a missing side?

Worked example 3. An isosceles triangle, meaning a triangle with two equal sides, has perimeter 40 cm. Each equal side is 15 cm. Find its area.

Answer: The base is 40 − 15 − 15 = 10 cm. The semiperimeter is 40/2 = 20 cm. Hence A = √[20(20 − 15)(20 − 15)(20 − 10)] = √5000 = 50√2 cm².

For an equilateral triangle, all three sides have the same length a. Substituting s = 3a/2 into Heron’s formula gives A = (√3/4)a². Its perimeter is 3a. Keep square-root answers exact unless an approximation is requested.

Which formulas apply to special quadrilaterals?

A polygon is a closed figure made of straight line segments. A quadrilateral is a polygon with four sides. Its perimeter is the sum of all four side lengths. A diagonal joins two opposite vertices. Different types of quadrilateral require different area information.

A rectangle has four right angles. A square is a rectangle whose sides are all equal. A rhombus is a parallelogram with all sides equal. A kite has two distinct pairs of equal adjacent sides, meaning equal sides meeting at a vertex.

How should the symbols in the formulas be read?

In the table, A denotes area and P perimeter. The letters l and b denote a rectangle’s length and breadth. The letter a denotes a square’s or rhombus’s side; for a parallelogram or kite, a and b denote its two side lengths.

For the parallelogram area formula, b is the chosen base and h its corresponding perpendicular height. The symbols d₁ and d₂ are the lengths of the two diagonals; the small numerals distinguish them.

The rhombus and kite diagonal formulas use their perpendicular diagonals. They do not imply that half the product of the diagonals gives the area of every quadrilateral. A rectangle’s length and breadth are perpendicular, which explains its simpler product formula.

Worked example 4. A rhombus has area 240 cm² and one diagonal 16 cm. Find its other diagonal.

Answer: Let d₂ be the unknown diagonal. Then 240 = ½ × 16 × d₂ = 8d₂. Dividing by 8 gives d₂ = 30 cm. This is a length, so its unit is cm rather than cm².

A parallelogram’s two adjacent side lengths alone do not determine its area. The perpendicular height can change as the angle between the sides changes. Distinguish the information needed for its perimeter from the information needed for its area.

How are trapeziums and general quadrilaterals measured?

A trapezium has a pair of opposite parallel sides. Parallel sides lie on lines that stay the same distance apart. Let a and b be their lengths, and h their perpendicular separation. Its area A is A = ½(a + b)h.

Result: Splitting a quadrilateral into triangles

Draw a diagonal inside a convex quadrilateral, meaning one with every interior angle less than 180°. The diagonal divides it into two triangles. For a trapezium, the triangles have bases a and b and the same height h, so their areas add to ½(a + b)h.

To find a trapezium’s perimeter, add all four sides. The sum a + b includes just the parallel pair. It cannot give the complete boundary unless the lengths of the other two sides are also included.

Worked example 5. A trapezium-shaped field has area 480 m². Its parallel sides are 15 m apart, and one parallel side is 20 m. Find the other parallel side.

Answer: Let b be the unknown side. Then 480 = ½ × (20 + b) × 15. Therefore 20 + b = (2 × 480)/15 = 64, so b = 44 m. The given 15 m is the height, not another boundary side.

How does a diagonal help with an irregular quadrilateral?

Let d be the diagonal length, and h₁ and h₂ the perpendicular distances to it from the two remaining vertices on opposite sides. Adding the two triangle areas gives A = ½d(h₁ + h₂). This method does not require equal or parallel opposite sides.

Worked example 6. A convex quadrilateral field has diagonal 24 m. The perpendicular distances from the other two vertices to this diagonal are 8 m and 13 m. Find its area.

Answer: The triangle areas are ½ × 24 × 8 = 96 m² and ½ × 24 × 13 = 156 m². Their sum is 252 m². Equivalently, A = ½ × 24 × (8 + 13) = 252 m².

For a concave quadrilateral, one interior angle exceeds 180°. Choose its diagonal that lies inside the region, then add the two triangle areas. If their side lengths are known, Heron’s formula can find each area.

When dividing a region into pieces, the pieces must cover the required region without overlaps. Add their areas, but do not count an internal diagonal as part of the outer perimeter.

How are a circle’s circumference and area related to its radius?

A circle is the set of points in a plane at a fixed distance from a fixed point called the centre. Its radius, written r, is that distance. Its diameter, written d, passes through the centre and joins two points on the circle.

The diameter contains two radii, so d = 2r. The circumference, written C, is the circle’s perimeter. The symbol π, read as pi, represents the constant ratio of a circle’s circumference to its diameter.

Which formula answers which question?

C = 2πr = πd, whereas the enclosed area A is A = πr². Circumference has length units; area has square units. When the diameter is supplied, halve it before using the radius-based area formula.

The value 22/7 is an approximation to π, not an exact equality. Use the approximation specified in a question. If an exact answer is wanted, an expression containing π can be retained without replacing π by a decimal or fraction.

Worked example 7. A car tyre has diameter 56 cm. How far does it travel in one complete revolution, assuming it rolls without slipping? Use π ≈ 22/7, where ≈ means approximately equal to.

Answer: A revolution is one complete turn. The distance travelled is the circumference, C = πd. Using the given approximation gives C = (22/7) × 56 = 176 cm. The radius would be 28 cm, giving the same result from 2πr.

For a rolling wheel, divide the total distance by the circumference to find the number of revolutions, after expressing both lengths in the same unit. For a circular region to be covered or polished, calculate the area instead.

How do semicircles and quarter-circles differ from whole circles?

An arc is a portion of a circle’s curved boundary. A sector is the region bounded by an arc and the two radii joining its endpoints to the centre. A semicircular region is half a disc; a quarter-circle, or quadrant, is one-quarter.

A disc includes the region enclosed by a circle. For radius r, the two semicircular regions have equal areas, and the four quadrants have equal areas. Their areas are therefore one-half and one-quarter of the full area πr².

Why is arc length different from perimeter?

The semicircle’s straight boundary is its diameter. A quadrant has two straight radius boundaries. Thus both perimeters require adding 2r to the appropriate arc length. Halving or quartering a circumference gives just the curved part.

Figure 6.39 shows two differently coloured halves separated by diameter AB through centre O. Figure 6.40 shows four differently coloured quadrants meeting at O, with A, B, C and D around the circle.

See Figs. 6.39 and 6.40 in your NCERT textbook

Worked example 8. A circle has circumference 44 cm. Find the area of one quadrant. Use π ≈ 22/7.

Answer: From 44 = 2 × (22/7) × r, the radius is r = 7 cm. The quadrant’s area is ¼πr². Using the given approximation, this is ¼ × (22/7) × 7² = 38.5 cm².

In a combined figure, first decide which straight and curved portions actually lie on the outside boundary. A line where two pieces meet internally contributes no exposed boundary length. For area, add adjoining pieces or subtract a removed region.

How are inner and outer areas calculated?

When a smaller region is removed from a larger one, the area left is the outer area minus the inner area. Keep the measurements associated with each boundary separate. Calculate the area of each complete region before subtracting; a difference between two lengths is not itself an area.

How is the area of a circular ring found?

Concentric circles have the same centre. The region between two such circles is an annulus, or circular ring. Let R be the outer radius and r the inner radius, with R greater than r. Its area A is A = πR² − πr² = π(R² − r²).

The ring’s uniform width is R − r. If a question gives the inner radius and the width, add them to obtain the outer radius. If it gives the outer radius and width, subtract the width to obtain the inner radius.

The expression π(R − r)² describes a disc whose radius equals the ring’s width. It does not describe the ring. To measure both boundaries of a complete ring, add the circumferences: total boundary length = 2πR + 2πr.

How does the same subtraction apply to rectangles?

Let L and B denote the outer length and breadth of a rectangular frame, and l and b the corresponding inner dimensions. Its area is LB − lb. The outer rectangle includes both the frame and its opening; subtracting the opening leaves the frame.

For a uniform rectangular border of width t, the inner length is L − 2t and the inner breadth is B − 2t. Each dimension loses the border width at both ends. These formulas assume the inner and outer rectangle sides are parallel.

Note: Area subtraction and boundary addition answer different questions. For a ring, subtract disc areas to measure the material, but add the inner and outer circumferences when both edges are being measured.

How are the surface areas of closed cubes and cuboids found?

A cuboid has six rectangular faces, meaning flat surfaces, arranged in three pairs of equal opposite faces. Let l, b and h denote its length, breadth and height. A cube is a cuboid with all three dimensions equal.

An edge is where two faces meet. A net is a flat arrangement of faces that can be folded into the solid. Counting face areas on a net helps identify the surfaces that must be included in a calculation.

Result: Total surface area of a cuboid

Total surface area, abbreviated TSA, includes all six faces of a closed cuboid. The top and bottom contribute 2lb, the front and back contribute 2lh, and the remaining pair contribute 2bh. Therefore TSA = 2(lb + lh + bh).

The lateral surface area, abbreviated LSA, includes the four side faces relative to the chosen base. It excludes the top and bottom. Therefore LSA = 2h(l + b), which is the base perimeter multiplied by the height.

The cuboid drawing labels length l, width w and height h; w means the breadth called b here. The blue net displays six rectangles with corresponding length, width and height labels.

See Figs. 14.1C and 14.2 in your NCERT textbook

What changes when the solid is a cube?

Let a be the cube’s edge length. Each face has area a², so TSA = 6a². Its four lateral faces have combined area LSA = 4a². These also follow by substituting l = b = h = a into the cuboid formulas.

Worked example 9. Compare the total surface areas of a cube of side 6 cm and a cuboid measuring 9 cm × 6 cm × 4 cm.

Answer: The cube’s surface area is 6 × 6² = 216 cm². The cuboid’s surface area is 2[(9 × 6) + (9 × 4) + (6 × 4)] = 228 cm². The cuboid has the greater surface area.

These are areas of the closed solids’ complete boundaries. If a lid is missing or only selected faces are covered, identify those faces before choosing or modifying the formula.

How do open boxes and covering costs change the calculation?

A cuboid open at the top has its base and four side faces. For length l, breadth b and height h, its area is lb + 2h(l + b). This equals the closed cuboid’s total surface area minus the missing top face lb.

A cube of edge a with one face missing has area 5a². These simple face-area formulas treat the material as having negligible thickness. When thickness matters, inner and outer dimensions describe different surfaces and must be handled separately.

How should selected faces be counted?

Worked example 10. An aquarium has external, meaning outside, length 80 cm, breadth 30 cm and height 40 cm. Coloured paper covers its base, back and two side faces. Find the paper area required.

Answer: The base area is 80 × 30 = 2400 cm². The back area is 80 × 40 = 3200 cm². Each side face has area 30 × 40 = 1200 cm². The required total is 2400 + 3200 + 2 × 1200 = 8000 cm².

The front and top are not among the surfaces specified for paper coverage. A complete surface-area formula would include surfaces that the question does not ask to cover. Listing the required faces makes the calculation match the task.

How is a rate used to find cost?

A rate per unit area is the cost of covering one square unit. Multiply the required area by that rate after matching units. The symbol ₹ denotes rupees. A rate in rupees per m² requires an area expressed in m².

Worked example 11. A room has internal, meaning inside, length 12 m, breadth 8 m and height 4 m. Find the cost of whitewashing all four walls at ₹5 per m², then include the ceiling. Use the full wall areas without deducting openings.

Answer: Wall area = 2 × 4 × (12 + 8) = 160 m²; wall cost = 160 × 5 = ₹800. Ceiling area = 12 × 8 = 96 m²; ceiling cost = 96 × 5 = ₹480. Total cost with ceiling = ₹1280.

If dimensions of doors or windows to be excluded are supplied, subtract their areas from the relevant wall area before applying the rate. Do not assume sizes for unmeasured openings or introduce extra deductions.

How are volume, cross-section and capacity connected?

A cross-section is the flat shape obtained by slicing a solid with a plane. A slice parallel to a cuboid’s rectangular base has the same length and breadth as that base. The cross-sectional area stays constant along its height.

Result: Volume as cross-sectional area multiplied by height

For a cuboid of length l, breadth b and perpendicular height h, the base area is lb. Let V denote volume. Stacking equal rectangular layers through height h gives V = lbh, or V = area of cross-section × height.

Here the cross-section is parallel to the base and the height is perpendicular to it. This argument applies to the cube and cuboid because the cross-sectional area remains constant. It is not a rule for multiplying an arbitrary slice by any length.

For a cube of edge a, V = a³. A container’s capacity uses its inner dimensions. Express capacity in litres by dividing its volume in cm³ by 1000, or multiplying its volume in m³ by 1000.

Worked example 12. A cuboid has base area 180 cm² and volume 900 cm³. Find its height.

Answer: Volume = base area × height, so 900 = 180 × h, where h is the height in centimetres. Therefore h = 900/180 = 5 cm. Dividing cubic units by square units gives a length unit.

Do equal volumes imply equal surface areas?

The cube of side 6 cm has volume 6³ = 216 cm³. The cuboid measuring 9 cm × 6 cm × 4 cm also has volume 216 cm³. Their surface areas, calculated earlier, are 216 cm² and 228 cm² respectively.

Thus equal volumes need not have equal surface areas. The volume measures how much space the solid occupies; the surface area measures how much boundary must be covered. Keeping these purposes separate prevents using one formula to answer the other question.

How are internal dimensions and the volume of material handled?

External dimensions measure a container from its outside surfaces. Internal dimensions measure the empty space within it. When the walls have thickness, the internal dimensions are smaller. The outside volume includes both the container material and the cavity, meaning its hollow interior.

Which dimensions decrease by twice the thickness?

Let L, B and H denote a cuboidal container’s external length, breadth and height. Let t be the uniform thickness of its walls and base. For an open-top container, the internal length is L − 2t and the internal breadth is B − 2t.

Both horizontal dimensions lose t at each of two opposite walls. The internal height is H − t because only the base reduces the depth: there is no lid. Therefore its capacity is (L − 2t)(B − 2t)(H − t) in cubic units.

For a closed hollow cuboid whose lid also has thickness t, the internal height is H − 2t. Its inner volume is then (L − 2t)(B − 2t)(H − 2t). These formulas require positive internal dimensions and the stated uniform thickness.

How is the material volume separated from capacity?

Volume of material = external volume − internal volume. The external volume is LBH. Subtract the appropriate cavity volume for the open or closed container. This measures material, whereas the inner volume alone measures how much the container can hold.

If a cost is given per unit volume of material, multiply the material volume by that rate. This differs from a painting cost per unit area. Read the rate’s unit before deciding which measurement is needed.

1. Identify whether each supplied dimension is internal or external.
2. Identify whether a lid exists and whether thickness is uniform.
3. Write the inner dimensions, accounting for each relevant wall or base.
4. Calculate capacity, material volume or surface area according to the quantity requested.

For a thick open box, painting all exposed surfaces can involve inner walls, outer walls, the base and the top rim, meaning the exposed upper edges of the walls. List the specified surfaces rather than treating the box as a thin five-face model.

Glossary

• Perimeter — The total length of the boundary of a closed plane figure.
• Area — The measure of an enclosed plane region, expressed in square units.
• Altitude — The perpendicular distance from a triangle’s vertex to the line containing its opposite side.
• Semiperimeter — Half the sum of all the side lengths forming a figure’s perimeter.
• Diagonal — A line segment joining two non-adjacent vertices of a polygon.
• Circumference — The perimeter of a circle, measured as a length around its curved boundary.
• Radius — The distance from the centre of a circle to any point on the circle.
• Sector — A circular region enclosed by an arc and the two radii joining its endpoints to the centre.
• Annulus — The ring-shaped region between two circles that have the same centre.
• Total surface area — The sum of the areas of all faces forming a solid’s complete boundary.
• Lateral surface area — The combined area of the side faces, excluding the chosen top and bottom faces.
• Volume — The amount of three-dimensional space occupied by a solid, measured in cubic units.
• Capacity — The volume available inside a hollow container for holding a substance.
• Cross-section — The plane shape formed when a plane slices through a solid.

Common errors and misconceptions

• Misconception: Any side can serve as the height in a triangle’s area formula. Correct: Use the perpendicular height corresponding to the chosen base, extending the base if necessary.
• Misconception: Heron’s formula uses the full perimeter in place of s. Correct: The symbol s represents the semiperimeter, and the entire product of four factors lies under the square root.
• Misconception: Half the product of the diagonals gives every quadrilateral’s area. Correct: The rhombus and kite formulas use perpendicular diagonals; a general quadrilateral requires an appropriate decomposition or other sufficient information.
• Misconception: A semicircle’s perimeter is just half the circumference. Correct: Half the circumference gives its arc length; add its diameter to obtain the complete boundary length.
• Misconception: A circular ring has area π times the square of its width. Correct: Subtract the inner disc’s area from the outer disc’s area.
• Misconception: Every cuboidal box requires all six faces in a covering calculation. Correct: Count the specified faces and omit a missing lid or any face not being covered.
• Misconception: External dimensions directly give a thick container’s capacity. Correct: Use internal dimensions; for uniform thickness, an open container’s depth loses one base thickness.
• Misconception: Area and volume units convert using the same factor as length. Correct: Square or cube the length-conversion factor, according to the quantity being measured.

Exam-style questions with model answers

Q1. A right-angled triangle has perpendicular sides 3 units and 4 units, and third side 5 units. Find its perimeter and area. [2 marks]

1. The perimeter is the sum of its three sides: 3 + 4 + 5 = 12 units.
2. The perpendicular sides give the base and height, so area = ½ × 3 × 4 = 6 square units.

Q2. An isosceles triangle has perimeter 40 cm and equal sides of 15 cm each. Find its area using Heron’s formula. [3 marks]

1. Subtract the two equal sides from the perimeter to find the base: 40 − 15 − 15 = 10 cm.
2. The semiperimeter s is half the perimeter, so s = 40/2 = 20 cm. The three differences are 5 cm, 5 cm and 10 cm.
3. Heron’s formula gives area = √[20 × 5 × 5 × 10] = √5000 = 50√2 cm².

Q3. A trapezium-shaped field has area 480 m², perpendicular distance 15 m between its parallel sides, and one parallel side 20 m. Find the other parallel side. [3 marks]

1. Let b be the unknown parallel side in metres. The area is half the sum of the parallel sides multiplied by their perpendicular separation.
2. Substitute the data: 480 = ½ × (20 + b) × 15. Multiplying by 2 and dividing by 15 gives 20 + b = 64.
3. Subtract 20 from both sides to obtain b = 44 m. The requested quantity is a side length, so the answer uses metres.

Q4. A circle has circumference 44 cm. Find its radius and the area of one quadrant. Use π ≈ 22/7. [3 marks]

1. Let r be the radius in centimetres. Use circumference = 2πr, so 44 = 2 × (22/7) × r. This gives r = 7 cm.
2. A quadrant is one-quarter of the circular region, so its area is ¼πr². The radius must be squared in this formula.
3. Using the stated approximation gives quadrant area = ¼ × (22/7) × 7² = 38.5 cm², expressed in square units.

Q5. Compare the volumes and total surface areas of a cube of side 6 cm and a cuboid of length 9 cm, breadth 6 cm and height 4 cm. Explain what the comparison shows. [5 marks]

1. The cube’s volume is the product of its three equal dimensions: 6 × 6 × 6 = 216 cm³.
2. The cuboid’s volume is length × breadth × height = 9 × 6 × 4 = 216 cm³. Thus both solids occupy the same volume.
3. The cube has six square faces, so its total surface area is 6 × 6² = 216 cm².
4. The cuboid has three pairs of equal rectangular faces. Its total surface area is 2[(9 × 6) + (9 × 4) + (6 × 4)] = 228 cm².
5. The cuboid has the greater surface area despite having the same volume. Equal volumes therefore do not necessarily require equal areas of covering material.

Q6. An aquarium has external length 80 cm, breadth 30 cm and height 40 cm. Find the area of paper needed to cover its base, back and two side faces. [4 marks]

1. The rectangular base measures 80 cm by 30 cm, so its area is 80 × 30 = 2400 cm².
2. The back face measures 80 cm by 40 cm, so its area is 80 × 40 = 3200 cm².
3. Each side face measures 30 cm by 40 cm. Both side faces therefore contribute 2 × 30 × 40 = 2400 cm².
4. Add the specified faces: 2400 + 3200 + 2400 = 8000 cm². The front and top are not included in the requested coverage.

Q7. A room has internal length 12 m, breadth 8 m and height 4 m. Whitewashing costs ₹5 per m². Find the cost for all four walls and then for the walls together with the ceiling, without deducting openings. [5 marks]

1. The four walls form the lateral surface. Their combined area is twice the height multiplied by the sum of length and breadth.
2. Substitute the dimensions: wall area = 2 × 4 × (12 + 8) = 160 m². No door or window area is deducted.
3. Multiply by the stated rate to obtain the wall cost: 160 × ₹5 = ₹800. The area and rate both use square metres.
4. The ceiling has the same rectangular dimensions as the floor. Its area is 12 × 8 = 96 m², giving a ceiling cost of 96 × ₹5 = ₹480.
5. Add the costs for the surfaces requested: ₹800 + ₹480 = ₹1280 for walls and ceiling together. The floor is not being whitewashed.

Q8. A cuboid has base area 180 cm² and volume 900 cm³. Find its perpendicular height. [2 marks]

1. Volume equals base area multiplied by perpendicular height. Let h denote the height in centimetres, giving 900 = 180h.
2. Divide by the base area: h = 900/180 = 5 cm. The height is a length, not an area or volume.

Key takeaways

• Perimeter measures boundary length, area measures a region, and volume measures space; use linear, square and cubic units respectively.
• A triangle’s area uses a base and its perpendicular height, while Heron’s formula uses three sides and their semiperimeter.
• Choose a quadrilateral’s area formula from its properties and supplied measurements; its four boundary lengths give its perimeter.
• A circle’s circumference is 2πr and its area is πr², where r is the radius and π is the circumference-to-diameter ratio.
• Include straight boundaries when finding semicircle and quadrant perimeters; their arc lengths alone do not give the complete perimeter.
• Subtract inner area from outer area to measure a border or ring, keeping the two sets of dimensions separate.
• Count the faces actually present or covered in a surface-area problem, then multiply by an appropriately matched area cost rate.
• Use internal dimensions for capacity and subtract cavity volume from external volume when finding the amount of container material.

Test yourself

What distinguishes a triangle’s height from a sloping side?

The height is perpendicular to the line containing the chosen base. A sloping side can serve as height only when it has that perpendicular relationship.

What does the letter s represent in Heron’s formula?

It represents the semiperimeter, calculated by adding the three side lengths and dividing their sum by two.

What additional length must be added to a semicircular arc to obtain the region’s perimeter?

Add the diameter, which is twice the radius. The complete perimeter includes both the curved boundary and this straight boundary.

Why is π(R − r)² incorrect for a circular ring with outer radius R and inner radius r?

It finds a disc’s area using the ring’s width as radius. The ring requires subtraction of two disc areas: πR² − πr².

Which faces contribute to the lateral surface area of a cuboid?

The four side faces contribute. The chosen top and bottom are excluded, giving twice the height multiplied by the sum of length and breadth.

How is an open-top cuboid’s face area obtained from the closed cuboid formula when thickness is negligible?

Subtract one top-face area from the total surface area. The remaining area includes the base and the four side faces.

Why does an open container’s internal height lose one thickness while its internal length loses two?

The height loses only the base thickness because there is no lid. The length loses one thickness at each of two opposite walls.

Does equal volume guarantee equal surface area for two solids?

No. Volume measures occupied space, while surface area measures the boundary. Solids can occupy the same volume yet have different surface areas.