Mensuration — Questions and Answers
Welcome to HSLC Guru. This page gives the complete, step-by-step solutions of ASSEB Class 8 General Mathematics Chapter 11 Mensuration — every question of Exercise 11.1, 11.2 and 11.3 with formulas, labelled figures and full working. It covers the area of trapeziums, rhombuses and polygons, and the surface area and volume of cuboids, cubes and cylinders.
Summary
Mensuration is the branch of mathematics that deals with measuring the area, surface area and volume of geometric shapes. For two-dimensional figures we measure area (in square units such as cm² or m²), and for three-dimensional solids we measure surface area (square units) and volume (cubic units).
The chapter introduces the area formulas for a trapezium, a rhombus and a general quadrilateral. The area of any polygon can be found by dividing it into familiar shapes — triangles, rectangles, trapeziums — and adding their areas.
For solids, it develops the surface area and volume of a cuboid, a cube and a right circular cylinder, together with the relation between volume and capacity (litres). The main formulas are collected in the table below.
| Shape | Area / Surface area / Volume |
|---|---|
| Area of a trapezium | $\frac{1}{2}(a+b)\times h$ |
| Area of a rhombus | $\frac{1}{2}\times d_1\times d_2$ |
| Area of a quadrilateral (offsets $h_1,h_2$, diagonal $d$) | $\frac{1}{2}\times d\times(h_1+h_2)$ |
| Area of a circle | $\pi r^2$ |
| Total surface area of a cuboid | $2(lb+bh+lh)$ |
| Lateral surface area of a cuboid | $2h(l+b)$ |
| Total surface area of a cube | $6l^2$ |
| Curved surface area of a cylinder | $2\pi r h$ |
| Total surface area of a cylinder (closed) | $2\pi r(r+h)$ |
| Volume of a cuboid | $l\times b\times h$ |
| Volume of a cube | $l^3$ |
| Volume of a cylinder | $\pi r^2 h$ |
Summary: This ASSEB Class 8 General Mathematics Chapter 11 Mensuration solution covers the areas of trapezium, rhombus and general quadrilaterals, the area of polygons by splitting into known shapes, and the surface area and volume of cuboids, cubes and right circular cylinders. Every question of Exercise 11.1, 11.2 and 11.3 is solved with formulas, labelled figures and step-by-step working, along with the conversion between volume and capacity (litres).
Textbook Questions and Answers
Exercise 11.1
1. Find the area of the following figures.
Answer: (a) This is a trapezium with parallel sides $6$ cm and $12$ cm and perpendicular distance $4$ cm.
$$\text{Area} = \frac{1}{2}(6+12)\times 4 = \frac{1}{2}\times 18 \times 4 = 36 \text{ sq cm}$$
(b) Split the figure into a rectangle ($10\times 10$) and a triangle. The triangle has base $=10$ cm and height $=16-10=6$ cm.
$$\text{Area} = (10\times 10) + \frac{1}{2}\times 10 \times 6 = 100 + 30 = 130 \text{ sq cm}$$
2. The area of a trapezium is 34 sq cm. If the length of one of the parallel sides is 10 cm and the perpendicular distance between them is 4 cm, find the other parallel side.
Answer: Let the other parallel side $=b$ cm.
$$\frac{1}{2}(10+b)\times 4 = 34 \Rightarrow 2(10+b)=34 \Rightarrow 10+b=17 \Rightarrow b = 7$$
The other parallel side $=7$ cm.
3. The length of a diagonal of a quadrilateral is 20 m. If the perpendiculars upon it from the opposite vertices are of lengths 8.5 m and 11 m, find the area of the quadrilateral.
Answer: Diagonal $d=20$ m, offsets $h_1=8.5$ m, $h_2=11$ m.
$$\text{Area} = \frac{1}{2}\times d\times(h_1+h_2) = \frac{1}{2}\times 20 \times (8.5+11) = 10 \times 19.5 = 195 \text{ sq m}$$
4. Find the area of a rhombus whose diagonals are of length 10 cm and 14 cm.
Answer: $d_1=10$ cm, $d_2=14$ cm.
$$\text{Area} = \frac{1}{2}\times d_1\times d_2 = \frac{1}{2}\times 10 \times 14 = 70 \text{ sq cm}$$
5. The area of a quadrilateral is 100 sq cm. If the offsets are of length 6 cm and 4 cm, find the length of the diagonal.
Answer: Let the diagonal $=d$ cm.
$$\frac{1}{2}\times d\times(6+4) = 100 \Rightarrow \frac{1}{2}\times d\times 10 = 100 \Rightarrow 5d = 100 \Rightarrow d = 20$$
The length of the diagonal $=20$ cm.
6. The length of each side of a regular hexagon ABCDEF is 6 cm. If CF = 10 cm and AE = 8 cm, find the area of the hexagon.
Answer: The diagonal CF divides the hexagon into two trapeziums — FEDC (top) and FABC (bottom). AE = 8 cm is the total perpendicular distance between the top side ED and the bottom side AB, so each trapezium has height $=\frac{8}{2}=4$ cm.
$$\text{Area of one trapezium} = \frac{1}{2}(6+10)\times 4 = \frac{1}{2}\times 16 \times 4 = 32 \text{ sq cm}$$
$$\text{Area of the hexagon} = 2 \times 32 = 64 \text{ sq cm}$$
7. A rhombus and a triangle are equal in areas. If the base and height of the triangle are 24.8 cm and 5.5 cm respectively and the length of one diagonal of the rhombus is 22 cm, find the length of the other diagonal of the rhombus.
Answer: Area of the triangle $=\frac{1}{2}\times 24.8 \times 5.5 = 68.2$ sq cm. Let the other diagonal $=d_2$.
$$\frac{1}{2}\times 22 \times d_2 = 68.2 \Rightarrow 11\,d_2 = 68.2 \Rightarrow d_2 = 6.2$$
The other diagonal of the rhombus $=6.2$ cm.
8. The area of a trapezium shaped garden is 480 sq m and its height is 15 m. If the length of one of the parallel sides is 20 m, find the length of the other parallel side.
Answer: Let the other side $=b$ m.
$$\frac{1}{2}(20+b)\times 15 = 480 \Rightarrow 20+b = \frac{480\times 2}{15} = 64 \Rightarrow b = 44$$
The other parallel side $=44$ m.
9. In the polygon ABCDE, BE = 8 cm, CE = 10 cm, AF = 5 cm, BG = 4 cm, DH = 3 cm. Find the area of the polygon.
Answer: The diagonals EB and EC divide the polygon into three triangles — △ABE, △BCE and △CDE. AF is the perpendicular from A to EB, BG the perpendicular from B to EC and DH the perpendicular from D to EC.
$$\text{Area} = \tfrac{1}{2}\!\cdot\! BE\!\cdot\! AF + \tfrac{1}{2}\!\cdot\! CE\!\cdot\! BG + \tfrac{1}{2}\!\cdot\! CE\!\cdot\! DH$$
$$= \tfrac{1}{2}(8)(5) + \tfrac{1}{2}(10)(4) + \tfrac{1}{2}(10)(3) = 20 + 20 + 15 = 55 \text{ sq cm}$$
10. The area of a trapezium is 68 sq cm and the lengths of the parallel sides are respectively 13 cm and 21 cm. Find the perpendicular distance between the parallel sides.
Answer: Let the perpendicular distance (height) $=h$ cm.
$$\frac{1}{2}(13+21)\times h = 68 \Rightarrow \frac{1}{2}\times 34 \times h = 68 \Rightarrow 17h = 68 \Rightarrow h = 4$$
The perpendicular distance $=4$ cm.
11. Find the area of the given figure.
Answer: Split the figure into a middle rectangle and two trapeziums (top and bottom). The rectangle $=10\times 50$; each trapezium has height $=\frac{70-50}{2}=10$ cm with parallel sides $30$ cm and $10$ cm.
$$\text{Area} = (10\times 50) + 2\times \frac{1}{2}(30+10)\times 10 = 500 + 2(200) = 900 \text{ sq cm}$$
12. Find the area of the shaded portion of the following figures. $\left(\pi=\frac{22}{7}\right)$
Answer: (a) Shaded area $=$ area of the rectangle $-$ area of the semicircle. Rectangle $=7\times 9=63$; radius of the semicircle $=\frac{7}{2}=3.5$.
$$= 63 – \frac{1}{2}\times\frac{22}{7}\times (3.5)^2 = 63 – 19.25 = 43.75 \text{ sq cm}$$
(b) Shaded area $=$ large rectangle $-$ inner square $=(15\times 10) – (7\times 7) = 150 – 49 = 101$ sq cm.
(c) Shaded area (ring) $=$ outer circle $-$ inner circle; outer radius $14$, inner radius $7$.
$$= \frac{22}{7}(14^2 – 7^2) = \frac{22}{7}(196-49) = \frac{22}{7}\times 147 = 462 \text{ sq cm}$$
(d) Shaded area $=$ trapezium $-$ circle. The trapezium has parallel sides $10$ and $16$ with height $10$; the circle has radius $3.5$.
$$= \frac{1}{2}(10+16)\times 10 – \frac{22}{7}(3.5)^2 = 130 – 38.5 = 91.5 \text{ sq cm}$$
(e) Shaded area $=$ large circle $-$ small circle; large radius $21$, small radius $14$.
$$= \frac{22}{7}(21^2 – 14^2) = \frac{22}{7}(441-196) = \frac{22}{7}\times 245 = 770 \text{ sq cm}$$
13. If the lengths of two parallel sides of a trapezium are 7 cm and 5 cm and they are 4 cm apart, find the area of the trapezium.
Answer: $$\text{Area} = \frac{1}{2}(7+5)\times 4 = \frac{1}{2}\times 12 \times 4 = 24 \text{ sq cm}$$
14. The lengths of a side and a diagonal of a rhombus are 13 cm and 24 cm respectively. Find the area of the rhombus.
Answer: The diagonals of a rhombus bisect each other at right angles. Half of one diagonal $=\frac{24}{2}=12$. With side $=13$, half of the other diagonal $=\sqrt{13^2 – 12^2}=\sqrt{169-144}=\sqrt{25}=5$, so the other diagonal $=10$.
$$\text{Area} = \frac{1}{2}\times 24 \times 10 = 120 \text{ sq cm}$$
15. In the parallelogram ABCD, DE and DF are perpendiculars. If AB = 12 cm, DE = 7 cm and DF = 14 cm, find the length of BC.
Answer: Area of a parallelogram $=$ base $\times$ height. DE is the perpendicular to AB and DF is the perpendicular to BC, so the two expressions for the area are equal.
$$AB\times DE = BC\times DF \Rightarrow 12\times 7 = BC\times 14 \Rightarrow BC = \frac{84}{14} = 6$$
The length of BC $=6$ cm.
16. The parallel sides of a trapezium are in the ratio 1:2. If the distance between the parallel sides is 12 cm and the area of the trapezium is 180 sq cm, find the lengths of the parallel sides.
Answer: Let the parallel sides be $x$ and $2x$.
$$\frac{1}{2}(x+2x)\times 12 = 180 \Rightarrow \frac{1}{2}\times 3x \times 12 = 180 \Rightarrow 18x = 180 \Rightarrow x = 10$$
The parallel sides are $10$ cm and $2\times 10 = 20$ cm.
Exercise 11.2
1. The length of each edge of a cube is 27 cm. Find the surface area of the cube.
Answer: $$\text{Surface area} = 6l^2 = 6\times 27^2 = 6\times 729 = 4374 \text{ sq cm}$$
2. The perimeter of the floor of a room is 30 m and the height of the room is $\frac{1}{10}$ th of its perimeter. Find the area of the four walls of the room.
Answer: Height $h=\frac{1}{10}\times 30 = 3$ m. Area of four walls $=$ perimeter $\times$ height.
$$= 30 \times 3 = 90 \text{ sq m}$$
3. Find the area of the base surface of a cuboid whose total surface area and lateral surface area are 50 sq m and 30 sq m respectively.
Answer: Total surface area $=2(lb+bh+lh)$ and lateral surface area $=2(bh+lh)$. Subtracting —
$$2lb = 50 – 30 = 20 \Rightarrow lb = 10 \text{ sq m}$$
Area of the base surface $=lb=10$ sq m.
4. A box of dimensions 80 cm × 48 cm × 24 cm is to be covered with a cloth. How many metres of cloth of width 96 cm is required to cover 100 such boxes?
Answer: Total surface area of one box —
$$2(lb+bh+lh) = 2(80\times 48 + 48\times 24 + 80\times 24) = 2(3840+1152+1920) = 13824 \text{ sq cm}$$
Total area for 100 boxes $=13824\times 100 = 1382400$ sq cm. Width of cloth $=96$ cm.
$$\text{Length of cloth} = \frac{1382400}{96} = 14400 \text{ cm} = 144 \text{ m}$$
5. A roller takes 500 complete revolutions to level a playground. Find the cost of levelling the ground at the rate of 75 paise per sq m if the diameter and length of the roller are 84 cm and 120 cm respectively.
Answer: Radius $r=\frac{84}{2}=42$ cm, length $h=120$ cm. Area covered in one revolution $=$ curved surface area $=2\pi r h$.
$$2\pi r h = 2\times \frac{22}{7}\times 42 \times 120 = 31680 \text{ sq cm}$$
Area in 500 revolutions $=31680\times 500 = 15840000$ sq cm $=1584$ sq m. Cost $=1584\times 0.75 = ₹1188$.
6. Find the curved surface area of a cylindrical curd container with one end open whose height and radius are 30 cm and 14 cm respectively.
Answer: $$\text{Curved surface area} = 2\pi r h = 2\times \frac{22}{7}\times 14 \times 30 = 2640 \text{ sq cm}$$
7. 3 cubes each of side length 5 cm are joined end to end. Find the total surface area of the cuboid so formed.
Answer: Joining three cubes end to end gives a cuboid with length $l=3\times 5 = 15$, breadth $b=5$, height $h=5$ cm.
$$\text{Total surface area} = 2(lb+bh+lh) = 2(15\times 5 + 5\times 5 + 15\times 5) = 2(75+25+75) = 350 \text{ sq cm}$$
8. Find the diameter of the base of a cylinder of height 14 cm and curved surface area 88 sq cm.
Answer: $2\pi r h = 88$, so —
$$2\times \frac{22}{7}\times r \times 14 = 88 \Rightarrow 88r = 88 \Rightarrow r = 1 \text{ cm}$$
Diameter of the base $=2r=2$ cm.
9. There are 25 right circular cylindrical posts in a temple. Each post is of height 4 m and radius 28 cm. Find the cost of painting the curved surface of the posts at the rate of Rs 8 per square metre.
Answer: $r=28$ cm $=0.28$ m, $h=4$ m. Curved surface area of one post —
$$2\pi r h = 2\times \frac{22}{7}\times 0.28 \times 4 = 7.04 \text{ sq m}$$
Total area of 25 posts $=7.04\times 25 = 176$ sq m. Cost $=176\times 8 = ₹1408$.
10. Find the lateral surface area and the area of the circular base of a cylindrical glass of height 12 cm and diameter 7 cm.
Answer: Radius $r=\frac{7}{2}=3.5$ cm, $h=12$ cm.
$$\text{Lateral surface area} = 2\pi r h = 2\times \frac{22}{7}\times 3.5 \times 12 = 264 \text{ sq cm}$$
$$\text{Area of circular base} = \pi r^2 = \frac{22}{7}\times (3.5)^2 = 38.5 \text{ sq cm}$$
11. The cross-sectional diameter and length of a hume pipe are 1400 mm and 2500 mm respectively. Find the cost of painting the exterior side of 22 such pipes at the rate of Rs 8 per square metre.
Answer: Diameter $=1400$ mm $=1.4$ m, so $r=0.7$ m; length $h=2500$ mm $=2.5$ m. A pipe is open at both ends, so its exterior surface area $=2\pi r h$.
$$2\pi r h = 2\times \frac{22}{7}\times 0.7 \times 2.5 = 11 \text{ sq m}$$
Total area of 22 pipes $=11\times 22 = 242$ sq m. Cost $=242\times 8 = ₹1936$.
Exercise 11.3
1. A cuboidal container can hold 105 litres of water. If the inner base is of dimensions 15 m × 3.5 m, find the height of the container.
Answer: Volume $=105$ litres $=\frac{105}{1000}=0.105$ cu m. Base area $=15\times 3.5 = 52.5$ sq m.
$$h = \frac{\text{Volume}}{\text{Base area}} = \frac{0.105}{52.5} = 0.002 \text{ m} = 0.2 \text{ cm}$$
2. A metallic cube of side 12 cm is melted to form three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm respectively, find the edge of the third cube.
Answer: The total volume of metal is unchanged. Let the edge of the third cube $=x$.
$$12^3 = 6^3 + 8^3 + x^3 \Rightarrow 1728 = 216 + 512 + x^3 \Rightarrow x^3 = 1000 \Rightarrow x = 10$$
The edge of the third cube $=10$ cm.
3. Find the height of a cuboid whose volume and area of the base are 900 cu cm and 180 sq cm respectively.
Answer: $$h = \frac{\text{Volume}}{\text{Base area}} = \frac{900}{180} = 5 \text{ cm}$$
4. How many cubes of side 6 cm can be put inside a cuboid, open at one end, of dimensions 60 cm × 54 cm × 30 cm?
Answer: Volume of the cuboid $=60\times 54 \times 30 = 97200$ cu cm; volume of one cube $=6^3 = 216$ cu cm.
$$\text{Number of cubes} = \frac{97200}{216} = 450$$
5. What is the amount of soil dug out from a circular pit of 21 metres deep and 6 metre diameter?
Answer: The pit is a cylinder. Radius $r=\frac{6}{2}=3$ m, depth $h=21$ m.
$$\text{Volume} = \pi r^2 h = \frac{22}{7}\times 3^2 \times 21 = \frac{22}{7}\times 9 \times 21 = 594 \text{ cu m}$$
6. Water is filled at the rate of 40 litres per minute into a cuboidal tank. Find the total time to fill the tank if its volume is 54 cu m.
Answer: Volume $=54$ cu m $=54\times 1000 = 54000$ litres.
$$\text{Time} = \frac{54000}{40} = 1350 \text{ minutes} = 22.5 \text{ hours}$$
7. A metallic solid of volume 2200 cu cm is melted to form a uniform wire of cross-sectional diameter 0.5 cm. Find the length of the wire.
Answer: The wire is a cylinder. Radius $r=\frac{0.5}{2}=0.25$ cm. The volume stays the same —
$$\pi r^2 h = 2200 \Rightarrow \frac{22}{7}\times (0.25)^2 \times h = 2200 \Rightarrow h = \frac{2200\times 7}{22\times 0.0625} = 11200 \text{ cm}$$
Length of the wire $=11200$ cm $=112$ m.
8. Find the height of a cuboid whose volume and area of base are 440 cu cm and 88 sq cm respectively.
Answer: $$h = \frac{440}{88} = 5 \text{ cm}$$
9. Find the height of a cuboid whose volume and area of the base are 168 cu m and 2800 sq cm respectively.
Answer: The units must be made consistent. Base area $=2800$ sq cm $=\frac{2800}{10000}=0.28$ sq m.
$$h = \frac{168}{0.28} = 600 \text{ m}$$
10. Find the amount of water required to fill a rectangular water reservoir whose inner dimensions are 6 m × 2 m × 1 m.
Answer: Volume $=6\times 2 \times 1 = 12$ cu m. Since $1$ cu m $=1000$ litres —
$$\text{Capacity} = 12\times 1000 = 12000 \text{ litres}$$
11. The circumference of the base of a cylinder is 132 cm and its height is 25 cm. Find the volume of the cylinder.
Answer: $2\pi r = 132$, so —
$$r = \frac{132\times 7}{2\times 22} = 21 \text{ cm}$$
$$\text{Volume} = \pi r^2 h = \frac{22}{7}\times 21^2 \times 25 = 34650 \text{ cu cm}$$
12. The heights of two cylinders having equal volumes are in the ratio 1:4. Find the ratio of the radii of their bases.
Answer: Since the volumes are equal, $\pi r_1^2 h_1 = \pi r_2^2 h_2$.
$$\frac{r_1^2}{r_2^2} = \frac{h_2}{h_1} = \frac{4}{1} \Rightarrow \frac{r_1}{r_2} = \frac{2}{1}$$
The ratio of the radii $=2:1$.
13. A cuboidal water tank measures internally 4.2 m length, 300 cm breadth and 1.8 m height. Find the capacity of the tank in litres.
Answer: $300$ cm $=3$ m. Volume $=4.2\times 3 \times 1.8 = 22.68$ cu m.
$$\text{Capacity} = 22.68\times 1000 = 22680 \text{ litres}$$
14. The total surface area of a solid cylindrical pillar is 924 sq cm. If its curved surface area is two-thirds of its total surface area, find the radius of its base and its volume.
Answer: Total surface area $=2\pi r(r+h) = 924$; curved surface area $=2\pi r h = \frac{2}{3}\times 924 = 616$.
$$2\pi r^2 = 924 – 616 = 308 \Rightarrow \pi r^2 = 154 \Rightarrow r^2 = \frac{154\times 7}{22} = 49 \Rightarrow r = 7 \text{ cm}$$
$2\pi r h = 616 \Rightarrow 44h = 616 \Rightarrow h = 14$ cm.
$$\text{Volume} = \pi r^2 h = \frac{22}{7}\times 49 \times 14 = 2156 \text{ cu cm}$$
Complete the tables (textbook activity)
(a) Complete the table of volumes of cuboids (Volume $=l\times b\times h$).
| Length $l$ (cm) | Breadth $b$ (cm) | Height $h$ (cm) | Volume (cu cm) |
|---|---|---|---|
| 10 | 7 | 6 | 420 |
| 8 | 9 | 12 | 864 |
| 3 | 6 | 11 | 198 |
| 4 | 15 | 20 | 1200 |
| 12 | 8 | 6 | 576 |
| 9 | 7 | 12 | 756 |
(b) Complete the table of volumes of cylinders (base area $=\pi r^2$, volume $=\pi r^2 h$, $\pi=\frac{22}{7}$).
| Height $h$ (cm) | Radius $r$ (cm) | Base area $\pi r^2$ (sq cm) | Volume (cu cm) |
|---|---|---|---|
| 15 | 7 | 154 | 2310 |
| 10 | — | 44 | 440 |
| 20 | 0.7 | 1.54 | 30.8 |
| 2 | — | 250 | 500 |
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. What is the formula for the area of a trapezium? (a) $\frac{1}{2}(a+b)h$ (b) $a\times b$ (c) $\pi r^2$ (d) $l^3$
Answer: (a) $\frac{1}{2}(a+b)h$.
2. What is the total surface area of a cube? (a) $4l^2$ (b) $6l^2$ (c) $l^3$ (d) $2l^2$
Answer: (b) $6l^2$.
3. What is the volume of a right circular cylinder? (a) $2\pi r h$ (b) $\pi r^2$ (c) $\pi r^2 h$ (d) $\frac{1}{3}\pi r^2 h$
Answer: (c) $\pi r^2 h$.
4. 1 cubic metre is equal to how many litres? (a) 10 (b) 100 (c) 1000 (d) 10000
Answer: (c) 1000 litres.
5. What is the area of a rhombus? (a) $d_1\times d_2$ (b) $\frac{1}{2}d_1 d_2$ (c) $4a$ (d) $a^2$
Answer: (b) $\frac{1}{2}d_1 d_2$.
6. What is the volume of a cuboid? (a) $l\times b\times h$ (b) $2(lb+bh+lh)$ (c) $6l^2$ (d) $\pi r^2 h$
Answer: (a) $l\times b\times h$.
7. What is the area of a circle? (a) $2\pi r$ (b) $\pi r^2$ (c) $\pi d$ (d) $\frac{1}{2}\pi r$
Answer: (b) $\pi r^2$.
8. 1 cubic cm is equal to how many ml? (a) 1 (b) 10 (c) 100 (d) 1000
Answer: (a) 1 ml.
9. What is the surface area of a cylinder open at both ends? (a) $2\pi r(r+h)$ (b) $2\pi r h$ (c) $2\pi r h+\pi r^2$ (d) $\pi r^2 h$
Answer: (b) $2\pi r h$.
10. If the volume of a cube is 64 cu cm, what is the length of its edge? (a) 4 cm (b) 8 cm (c) 6 cm (d) 16 cm
Answer: (a) 4 cm, because $\sqrt[3]{64}=4$.
Fill in the Blanks
1. Total surface area of a cuboid $=$ ______.
Answer: $2(lb+bh+lh)$.
2. Volume of a cube $=$ ______.
Answer: $l^3$.
3. 1 litre $=$ ______ cu cm.
Answer: 1000.
4. Total surface area of a closed cylinder $=$ ______.
Answer: $2\pi r(r+h)$.
5. The approximate value of $\pi$ $=$ ______.
Answer: $\frac{22}{7}$.
True or False
1. Two different shapes with equal perimeter always have equal areas.
Answer: False.
2. A cube is a special case of a cuboid where $l=b=h$.
Answer: True.
3. Volume and capacity are always equal.
Answer: False (they are treated as equal only when the wall thickness is negligible).
4. The surface area of a cylinder open at both ends $=2\pi r h$.
Answer: True.
5. The unit of area is cubic cm.
Answer: False (the unit of area is square cm).
Short Answer Questions
1. What is the difference between volume and capacity?
Answer: Volume is the amount of space occupied by an object, while capacity is the quantity a container can hold. When the wall thickness of a container is negligible, its volume and capacity are taken to be equal.
2. What is meant by lateral surface area?
Answer: In any 3-D solid (except a sphere), the sum of the areas of all the faces excluding the top and bottom faces is called its lateral surface area. For a cuboid it is $2h(l+b)$.
3. How is the area of a polygon found?
Answer: By drawing diagonals or perpendiculars, the polygon is divided into familiar shapes such as triangles, rectangles, squares and trapeziums. The sum of the areas of these parts equals the area of the polygon.
4. Find the volume and total surface area of a cube of side 4 cm.
Answer: Volume $=l^3 = 4^3 = 64$ cu cm; total surface area $=6l^2 = 6\times 16 = 96$ sq cm.
Key Terms
| Term | Meaning |
|---|---|
| Mensuration | Measuring the area, surface area and volume of shapes |
| Area | The region enclosed by a two-dimensional surface |
| Perimeter | The total length of the boundary of a shape |
| Trapezium | A quadrilateral with one pair of parallel sides |
| Rhombus | A parallelogram with all four sides equal |
| Polygon | A closed figure bounded by straight line segments |
| Cuboid | A solid with six rectangular faces |
| Cube | A solid with six equal square faces |
| Cylinder | A solid with two circular faces and one curved surface |
| Surface area | The sum of the areas of all faces of a solid |
| Volume | The amount of space occupied by an object |
| Capacity | The quantity a container can hold |
| Curved surface area | The area of the curved face of a cylinder |