Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251 Vedanta Excel in Mathematics - Book 8 ∴ BD = DC = a 2 By using Pythagoras Theorem in right-angled D ABD, AD = AB2 – BD2 = a2 – a2 4 = 3 2 a Now, area of D ABC = 1 2 base × height = 1 2 BC × AD = 1 2 a × 3 2 a = 3 4 a2 Thus, area of equilateral triangle = 3 4 (side)2 (iii) Area of an isosceles triangle In the given isosceles triangle ABC, AB = AC = a and BC = b (suppose). AD perpendicular to BC is drawn. We know that, in an isosceles triangle, perpendicular drawn from the vertical angle to the base bisects the base. Therefore, BD = DC = b 2 By using Pythagoras Theorem in right-angled D ABD, AD = AB2 – BD2 = a2 – b2 4 = 4a2 – b2 2 Now, area of D ABC = 1 2 base × height = 1 2 BC × AD = 1 2 b × 4a2 – b2 2 = 1 4 b 4a2 – b2 Thus, area of an isosceles triangle = 1 4 b 4a2 – b2 17.3 Area of quadrilateral In the adjoining figure, ABCD is a quadrilateral in which AC is the diagonal. Let's draw BX ⊥ AC and DY ⊥ AC. Here, BX (p1 ) and DY (p2 ) are the heights of DABC and DACD respectively. Now, area of DABC = 1 2 b × h = 1 2 AC × p1 Also, area of DACD = 1 2 b × h = 1 2 AC × p2 ∴ Area of quad. ABCD = area of (DABC + DACD) = 1 2 AC × p1 + 1 2 AC × p2 =1 2 AC (p1 + p2 ) Thus, area of a quadrilateral = 1 2 diagonal (p1 + p2 ). A h a B D C a a a 2 a 2 A a a B b D C 2 b 2 www.geogebra.org/classroom/jgvcbcvy Classroom code: JGVC BCVY Vedanta ICT Corner Please! Scan this QR code or browse the link given below: www.geogebra.org/classroom/sznb44qv Classroom code: SZNB 44QV Vedanta ICT Corner Please! Scan this QR code or browse the link given below: p1 p2 A C B D Area and Volume
Vedanta Excel in Mathematics - Book 8 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (i) Area of parallelogram Let's draw a parallelogram ABCD in a cardboard paper. Fold the base AB through D along DE such that DE ⊥ AB. Mark DE as h which is the height of the parallelogram. Now, cut the paper along DE and place the edge AD along BC so that a rectangle AEDC is formed. Here, AB = AE = b and DE = AC = h Now, area of parallelogram ABCD = area of rectangle AEDC = length × breadth = AE × AC = b × h Thus, area of parallelogram = base × height = b × h. (ii) Area of rhombus The figure alongside is a rhombus ABCD in which AC and BD are the diagonals. Let BD be d1 and AC be d2 . The diagonals of a rhombus bisect each other perpendicularly. So, BE ⊥ AC and CE ⊥ BD. Let's cut out the rhombus along the diagonal BD. Again, cut out the triangle BCD along CE. Now, fit the sides BC and CD on AB and AD in such way that a rectangle ECBD is formed. Now, area of rhombus ABCD = Area of rectangle ECBD = length × breadth = d1 × d2 2 = 1 2 d1 × d2 Thus, area of rhombus = 1 2 d1 × d2 . h b A E B C D h h E b B A D C www.geogebra.org/classroom/pu6ssewh Classroom code: PU6S SEWH Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A E B C D d2 2 d2 2 d1 A E B C D d2 2 d2 2 d2 2 d1 A B E C D d2 2 d2 2 d1 www.geogebra.org/classroom/fp82y8be Classroom code: FP82 Y8BE Vedanta ICT Corner Please! Scan this QR code or browse the link given below: A E B d1 d2 C D Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 253 Vedanta Excel in Mathematics - Book 8 (iii) Area of kite In the case of a kite, the diagonals intersect each other perpendicularly and the longer diagonal bisects the shorter one. Let's cut out the kite along the diagonal BD. Again, cut out the triangle BCD along CE. Now, fit the sides BC and CD on AB and AD in such way that a rectangle ECBD is formed. Now, area of kite ABCD = Area of rectangle ECBD = length × breadth = d1 × d2 2 = 1 2 d1 × d2 Thus, area of kite = 1 2 d1 × d2 . (iv) Area of trapezium The figure given alongside is a trapezium ABCD in which AB // CD and h is the height of the trapezium. Let's fold the trapezium in a such a way that the side CD is just fitted on the side AB. In this case, the height (h) of the trapezium is folded into half through EF. Now, cut out the trapezium through EF. Again, fold the uppermost coloured part of the trapezium into half and cut it out. Then, fit CF on BF and DE on AE in such a way that a rectangle DCFE is formed. Now, area of trapezium = Area of rectangle DCFE = length × breadth = (b 2 + a + b 2 ) × h 2 = 1 2 h(a + b) Thus, area of trapezium= 1 2 height (Sum of the two parallel sides) d2 d1 A B C D E A B C D E d1 d2 2 d2 2 A B C D E d1 d2 2 A B C D E d1 d2 2 d2 2 b a A B C h D b a A E F B D C h 2 h 2 E F D b C h 2 a A B h 2 E F D C h 2 a A B h 2 b 2 b 2 E F D A a B C h 2 h 2 b 2 b 2 Area and Volume
Vedanta Excel in Mathematics - Book 8 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Facts to remember Name of plane figure Shape of plane figure Formula of area Triangle Area = 1 2 base × height A = 1 2 bh Rectangle Area = length × breadth A = l × b h b b l Parallelogram Area = base × height A = b × h Rhombus Area = 1 2 d1 × d2 Kite Area = 1 2 (Product of diagonals) A = 1 2 d1 × d2 Trapezium Area = 1 2 (Sum of parallel sides) × h A = 1 2 h (l 1 + l 2 ) Quadrilateral Area = 1 2 diagonal (Sum of perpendiculars) A = 1 2 d (p1 + p2 ) d1 d2 h b d1 d2 h l 2 l 1 p2 p1 d Square Area = (side)2 or, Area = 1 2 (diagonal)2 A = l 2 or, A = 1 2 d2 l d Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 255 Vedanta Excel in Mathematics - Book 8 Worked-out Examples Example 1: Find the area of the following figures. Solution: a) Area of triangle = 1 2 base × height A = 1 2 × 7 cm × 6 cm = 21 cm2 b) Area of rhombus = 1 2 × product of diagonals = 1 2 × 8 cm × 6 cm = 24 cm2 c) Area of trapezium = 1 2 height (Sum of parallel sides) = 1 2 × 8 cm (12 cm + 7 cm) = 4 cm × 19 cm = 76 cm2 Example 2: Find the area of the given plane figures. Solution: Area of the rectangle = l × b = 24 cm × 6 cm = 144 cm2 Area of 2 triangles at two ends = 2 × 1 2 × b × h = 6 cm × 8 cm = 48 cm2 ∴ Area of the figure = 144 cm2 + 48 cm2 = 192 cm2 . 24 cm 6 cm 8 cm 8 cm a) 7 cm 6 cm b) 6cm 8cm c) 12 cm 7 cm 8 cm Example 3: Find the area of the shaded regions in the following figures. a) 15 cm 20 cm A D E C c) 25 cm 16 cm D A C B H E G F 10 cm 8 cm b) 14 cm S R T P Q 10 cm B 12 cm Area and Volume
Vedanta Excel in Mathematics - Book 8 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Solution: a) Area of the trapezium ABCD = 1 2 h (l 1 + l 2 ) = 1 2 × 12 cm (15 cm + 20 cm) = 6 cm × 35 cm = 210 cm2 Also, the area of ∆AEB = 1 2 × b × h = 1 2 × 15 cm × 12 cm = 90 cm2 ∴ Area of the shaded region = Area of trapezium ABCD – Area of ∆AEB = 210 cm2 – 90 cm2 = 120 cm2 b) Area of the parallelogram PQRS = b × h = 14 cm × 10 cm = 140 cm2 Also, Area of ∆PST = 1 2 × b × h = 1 2 × 14 cm × 10 cm = 70 cm2 ∴ Area of the shaded region = Area of parallelogram ABCD – Area of ∆PST = 140 cm2 – 70 cm2 = 70 cm2 c) Area of rectangle ABCD = l × b = 25 cm × 16 cm = 400 cm2 Also, area of rectangle EFGH = l × b = 10 cm × 8 cm = 80 cm2 ∴ Area of the shaded region = Area of rectangle ABCD – Area of rectangle EFGH = 400 cm2 – 80 cm2 = 320 cm2 Example 4: The perimeter of a squared floor of a room is 48 ft. a)Find its length b) Find its area c)Find the cost of carpeting the floor at Rs 44 per sq. ft. Solution: a) The perimeter of the squared floor = 48 ft or, 4l = 48 ft or, l = 48 4 ft = 12 m ∴ Length of the floor is 12 ft. b) Area of the floor = l 2 = (12 ft)2 = 144 sq. ft. c) The cost of carpeting in the floor = Area of floor × Rate of cost = 144 × Rs 44 = Rs 6,336 Example 5: A rectangular garden is twice as long as its breadth and its perimeter is 66 m. a) Find its length and breadth b) Find its area c) Find the cost of growing grass in the garden at Rs 40.50 per sq. metre. Solution: a) Let the breadth of the garden (b) = x m ∴ The length of the garden (l) = 2x m Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 257 Vedanta Excel in Mathematics - Book 8 The perimeter of the garden = 66 m or, 2 (l + b) = 66 m or, 2 (2x + x) = 66 m or, 6x = 66 m or, x = 66 6 m = 11 m ∴ Breadth of the garden = x = 11 m Length of the garden = 2x = 2 × 11 m = 22 m b) Area of the garden = l × b = 22 m × 11 m = 242 m2 c) The cost of growing grass in the garden = Area × Rate of cost = 424 × Rs 40.50 = Rs 9,801 EXERCISE 17.1 General Section – Classwork 1. Let’s say and write the answers as quickly as possible. a) Base of a triangle is a cm and height is b cm, area is ....................................... b) Base of a parallelogram is x cm and height is y cm, area is ............................ c) Diagonals of a rhombus are a cm and b cm, area is ........................................ d) Diagonals of a kite are x cm and y cm, area is ............................................... e) Length of two parallel sides of a trapezium are a cm and b cm, and height is h cm , area is ................................................ f) Diagonal of a quadrilateral is x cm, two perpendiculars from the opposite vertex to the diagonal are a cm and b cm, its area is ....................................... Creative Section - A 2. Find the area of the following figures: a) b) c) 7 cm A B D C 10 cm 8 cm P Q T 6 cmR I J K 10.5 cm 12 cm Area and Volume
Vedanta Excel in Mathematics - Book 8 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 3. Find the area of the following figures: d) e) f) g) A B C 14.5 cm 8 cm D B E S 9 cm T R A J 3 2 cm U C O 6 cm W h) A i) j) k) B C 4 2 cm 12cm A M T H 4cm U J V M P 10cm A B E C D 15 cm 5 cm 16 cm 9 cm b) 18 cm 15 cm P Q R S T a) 10 cm 10 cm 12 cm 8 cm A B C D l) m) n) o) 10.5cm4.5cm W X Y Z T 8cm 6cm A B C D 8cm 6.5cm P Q R S K I T E 15cm 6cm 8cm p) q) r) 24cm 8.5cm P Q R S 7cm 7.5cm 11 cm A B C D 6.5cm 4.5cm 10cm Q U A D M N 10 cm 25 cm c) 18 cm A B C D E d) 30 cm 5cm 5cm 15 cm A B D C E F Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 259 Vedanta Excel in Mathematics - Book 8 5. a) The perimeter of a square room is 40 m. (i) Find its length. (ii) Find the area of its floor. (iii) Find the cost of carpeting the floor at Rs 70 per sq. metre. b) The perimeter of a square ground is 100 m. Find the cost of plastering the ground at Rs 80 per sq. metre. 6. a) A rectangular park is 40 m long and 28 m broad. (i) Find its area. (ii) Find the cost of paving tiles all over it at Rs 90 per sq. metre. b) A rectangular field is 32 m long and its perimeter is 114 m. (i) Find the breadth of the field. (ii) Find the area of the field. (iii) Find the cost of plastering the field at Rs 75 per sq. metre. c) A rectangular room is 12 m broad and its perimeter is 54 m. Find the cost of carpeting its floor at Rs 50 per sq. metre. d) The length of the floor of room is two times its breadth and the perimeter of the floor is 48 m. Find the cost of paving marbles on the floor at Rs 125 per sq. metre. e) The length of a rectangular ground is three times its breadth. If the perimeter of the ground is 96 m, find the cost of growing grass on it at Rs 65 per sq. metre. 4. Find the area of shaded regions in the following figures: Creative Section -B 6 cm 8 cm h) 11 cm 5 cm e) 24 cm 16 cm 6 cm 4 cm 15 cm 10 cm g) 9 cm 4cm 3cm 18 cm 16 cm f) 8 cm6 cm i) 4 cm 12 cm 4 cm 4 cm 10 cm k) 14 cm 7 cm 5 cm l) 4 cm 12 cm 4 cm 5 cm 5 cm j) 15 cm 12 cm 5cm 5cm 4cm 4cm a) 10 cm 16 cm 5 cm 14 cm 17 cm 9 cm d) 18 cm15 cm b) 8 cm 16 cm 12 cm c) 5 cm 10 cm 15 cm Area and Volume
Vedanta Excel in Mathematics - Book 8 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 7. a) A rectangular park of length 60 m and breadth 50 m encloses a volleyball court of length 18 m and breadth 10 m. Find the area of the park excluding the volleyball court. Also, find the cost of paving stone in the park excluding the court at the rate of Rs 110 per square meter. b) Mr. Thapa built a circular pond of diameter 28 ft. for fish farming in his rectangular field of length 70 ft. and breadth 40 ft. Find the area of the field excluding the pond. Also, estimate the cost of planting vegetable at Rs 11 per square feet. It’s your time - Project Work and Activity Section 8. a) Take a rectangular sheet of photocopy paper. Measure its length and breadth. Draw a circle by using a compass with an arc of 3.5 cm radius in the middle of the paper. Lay the circle out by the help of scissors. Find the area of paper remained in the sheet. b) Let’s take a measuring tape and measure the outer length and breadth of white board of your class. Also measure the length and breadth of white part of the board and calculate the area of its wooden frame. 17.4 Perimeter of circle - Looking back Classroom - Exercise Study the given figures. Tell and write the answers as quickly as possible. 1. In the circle given alongside, a) O is called the ............................ b) OA is called the ............................ c) AB is called the ............................ d) PQ is called the ............................ e) If OB = 2 cm, OC = ............................, AB = ............................ 2. In the adjoining circle, a) The shaded region AOB is called ............................ b) The shaded region PRQ is called ............................ A B C P Q O A B P Q O R Draw three circles with radii 2 cm, 3 cm and 4 cm respectively. Place three pieces of thread along the circumference of each circle. 2 cm 3 cm 4 cm Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 261 Vedanta Excel in Mathematics - Book 8 Now, measure the length of each thread separately by placing them on a ruler. The length of each thread gives the circumference of the circle on which it is placed. Now, find the ratios of the length of a circumference of each circle to its diameter. You will find that the ratio circumference diameter is almost the same for every circle. This constant ratio is represented by the Greek letter ‘π’ (pie). Thus, if c be the circumference and d be the diameter of a circle, then, circumference diameter = π or, c d = π or, c = πd Again, diameter of a circle (d) = 2 × radius (r). So, circumference (or perimeter) of a circle (c) = 2πr ∴Perimeter of circle = πd or 2πr The approximate value of π is 3.142159..., which is approximately equivalent to 22 7 . www.geogebra.org/classroom/a5kptqqs Classroom code: A5KP TQQS Vedanta ICT Corner Please! Scan this QR code or browse the link given below: 17.5 Area of circle Let’s do the following activities to find the formula of area of circles. Take a circular piece of paper. Cut it into equal pieces as small as possible and arrange the pieces as shown in the figure. Thus, a rectangle ABCD is formed by this arrangement. The length of rectangle ABCD (l) = 1 2 × circumference = 1 2 × 2πr = πr The breadth of rectangle ABCD (b) = r Here, area of circle = Area of the rectangle ABCD. = length × breadth = πr × r = πr2 Thus, area of circle = πr2 Again, radius of cirlce (r) = d 2 So, area of circle = πr2 = π ( d 2 ) 2 = πd2 4 D (length) C A B (breadth) r of 2πr 1 2 Area and Volume
Vedanta Excel in Mathematics - Book 8 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Worked-out Examples Example 1: If the diameter of a circle is 7 cm find its (i) perimeter and (ii) area. Solution: Here, the radius of the circle (r) = diameter (d) 2 = 7 2 cm = 3.5 cm (i) Now, the perimeter of the circle = 2pr = 2 × 22 7 × 3.5 = 22 cm (ii) Also, the area of the circle = pr2 = 22 7 × 3.5 × 3.5 = 38.5 cm2 Example 2: When a boy completes 5 rounds around a circular pond, he covers a distance of 440 m. (i) Find the perimeter of the pond (ii) Find the diameter of the pond. (π = 22 7 ) Solution: (i) Here, the distance covered in 5 rounds = 440 m The distance covered in 1 round = 440 5 m = 88 m ∴ The perimeter of the pond = 88 m (ii) We know that the perimeter of a circle = πd ∴ πd = 88 m or, 22 7 × d = 88 m or, d = 88 × 7 22 m = 28 m So, the perimeter of the pond is 88 m and its diameter is 28 m. Example 3: If the perimeter of the base of a cylindrical tank is 308 cm, find the diameter and area of the base of the tank. Solution: Here, the perimeter of the base = 308 cm or, pd = 308 cm or, 22 7 × d = 308 cm or, d = 98 cm So, the diameter of the base of the tank is 98 cm. Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 263 Vedanta Excel in Mathematics - Book 8 Again, area of the base = pd2 4 = 22 7 × 1 4 × 98 × 98 = 7546 cm2 . So, the area of the base of the tank is 7546 cm2 . Example 4: Find the area of the shaded region in the following figures. a) b) c) Solution: a) Here, the radius of the circle (r) = 7 cm ∴ Area of the circle (A1 ) = pr2 = 22 7 × 7 × 7 cm2 = 154 cm2 The length of the rectangle (l) = 10 cm The breadth of the rectangle (b) = 8 cm ∴ Area of the rectangle (A2 ) = l × b = 10 cm × 8 cm = 80 cm2 Now, the area of the shaded region = A1 – A2 = 154 cm2 – 80 cm2 = 74 cm2 b) Here, the length of the side of the square (l) = 14 cm ∴ Area of the square (A1 ) = l2 = (14 cm)2 = 196 cm2 Also, the diameter of the semi-circle = 14 cm Then, the radius of the semi-circle (r) = 14 2 cm = 7 cm. ∴ Area of a semi-circle = 1 2 pr2 = 1 2 × 22 7 × 7 × 7 cm2 = 77 cm2 . Area of two semi-circles (A2 ) = 2 × 77 cm2 = 154 cm2 . Now, the area of the shaded region = A1 – A2 = 196 cm2 – 154 cm2 = 42 cm2 c) We know that 360° makes 1 (whole) circle. 1° makes 1 360 part of a circle. 45° makes 1 360 × 45 = 1 8 part of a circle. So, the area of the sector (A) = 1 8 of the area of the circle. Here, the radius of the circle (r) = 28 cm Angle at the centre of a shaded sector (q) = 45° Area (A) = ? Now, Area of sector (A) = 1 8 pr2 = 1 8 × 22 7 × 28 × 28 cm2 = 308 cm2 Hence, the area of shaded sector is 308 cm2 . O 7 cm 10 cm 8 cm 14 cm 14 cm 45°28cm Area and Volume
Vedanta Excel in Mathematics - Book 8 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 17.2 General Section – Classwork Let’s say and write the answers as quickly as possible. 1. a) The ratio of circumference of any circle to its diameter is represented by a symbol ................................ b) The approximate value of p is ................................ c) If r be the radius of a circle, its perimeter = ............................... d) If x be the radius of a circle, its area = ................................ e) If d be the diameter of a circle, its perimeter = ................................ f) If y be the diameter of a circle, its area = ................................ 2. a) In the figure, the area of the bigger circle is 510 cm2 and that of the smaller circle is 350 cm2 . The area of the shaded region is ................................ b) In the given figure, the area of the square is 225 cm2 and that of the circle is 154 cm2 . The area of the shaded region is ................................ Creative Section - A 3. Find the perimeter and area of the following circles. a) b) 4. a) Find the perimeter of a circle with radius 3.5 cm. b) Find the area of a circle with radius 14 cm. c) Find the perimeter and area of a circle with diameter 14 m. 5. a) If the radius of a circular pond is 70 m, find its (i) perimeter and (ii) area b) If the diameter of a circular floor of a room is 35 m find its (i) perimeter (ii) area 7 cm 28 cm Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265 Vedanta Excel in Mathematics - Book 8 6. a) Find the length of wire required to fence a circular park of diameter 280 m with 3 rounds. b) The radius of a circular play ground is 77 m. Calculate the distance covered by a runner in 5 complete rounds around the ground. c) The radius of a vegetable garden is 63 m. If 1188 m of wire is given to fence around it, how many rounds of fencing can be made? Creative Section - B 7. a) When a marathon runner completes 3 round around a circular track he covers a distance of 66 km. Find the radius of the track. b) If 3520 ft of wire is required to fence a circular garden with 4 rounds, find the diameter of the garden. c) If a wheel of a bicycle covers 440 m in 200 rotations, find the diameter of the wheel. 8. a) The perimeter of a circle is 132 cm, calculate (i) its radius and (ii) the area. b) The perimeter of a circular park is 880 m. Find the area of the park. c) The area of the base of a cylindrical tank is 38.5 m2 . Find the perimeter of the base. d) The area of a circular pond is 1386 m2 . Find the length of wire required to fence around it with 1 round. 9. a) Samriddhi bent a rope to form a circle. If she got the circumference of the circle is 30 cm more than its diameter. What was the length of radius? b) The rope required to measure the diameter of a circular pond is 90 m shorter than measuring the circumference. Find the diameter of the pond. 10. Find the area of the shaded regions in the following figures. a) b) c) d) e) f) g) h) i) 21 cm 60° 14 cm 14 cm 8 cm 9 cm 7 cm 14 cm 14 cm 12 cm 12 cm 28 cm 28 cm 42 cm r=3cm10cm 14 cm Area and Volume
Vedanta Excel in Mathematics - Book 8 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur It’s your time: Project Work and Activity Section 11. a) Let’s take a piece of wire and measure its length. Then bent it to form a circle and measure its diameter. Calculate the circumference by using formula C = pd. Write the conclusion on length of wire you had before and the circumference of the circle that you just calculated using formula. b) Let’s draw three circles of different length of radii in a chart paper. Then use thread to measure their circumference and complete the table given below. Fig. no. Diameter (d) Circumference (C) C d Remarks (i) (ii) (iii) Draw a conclusion and present in your class and weekly seminar of your school. 17.6 Solids and their nets Let’s trace the outline of a 10 rupee note, a square shaped biscuit and a 2 rupee coin. The shapes so formed are rectangle, square and a circle. Rectangle and square have flat surfaces. A circle has flat curved surface. Rectangle, square, circle, triangle etc. are the examples of plane figures. The plane figures are also called 2-dimensional figures. Let’s place some 10 rupee notes, square shaped biscuits and 2 rupee coins and observe the solid figures so formed. The solid shapes so formed are a cuboid, a cube and a cylinder. These shapes do not lie on the plane completely and have the third dimension namely the height along with the dimensions length and breadth. Thus, the solid shapes which have three dimensions namely length, breadth and height are called three dimensional shapes i.e., 3-D shapes. Cube, cuboid, cylinder, sphere, cone, pyramid, triangular prism, etc. are a few examples of solids. Solids have three measurable parts that we call the dimensions. So, solids are also called 3D shapes. Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 267 Vedanta Excel in Mathematics - Book 8 Let’s observe a few real life examples of solid objects. Solids have faces (sides), edges and vertices (corners). The exception is the sphere which has no edges or vertices. Prisms A prism is a polyhedron for which the top and bottom faces are congruent polygons and these congruent faces are called the base of the prism. All the other faces of a prism are rectangles which are called the lateral faces. But, a cylinder has a curved lateral face. Such prisms are also called the right prisms. A prism is described by the shape of its base. For example, a cube has square base, so it is a square prism. A cylinder has circular base; so, it is a circular prism. Triangular prism Cylinder (Circular prism) Cube (Square prism) Cuboid (Rectangular prism) (i) Cube and its net A cube is a square prism. It has six square faces. When we fold up the net of a square prism, a cube is formed. www.geogebra.org/classroom/qx67fxyw Classroom code: QX67 FXYW Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Net of cube Cube vertex edge face Area and Volume
Vedanta Excel in Mathematics - Book 8 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (ii) Cuboid and its net A cuboid is a rectangular prism. It has six rectangular faces. When the net of a rectangular prism is folded up a cuboid is formed. Net of cuboid Cuboid (iii) Triangular prism and its net A triangular prism has two bases which are triangles. A triangular prism has three rectangular lateral faces. Triangular prism Net of triangular prism (iv) Cylinder and its net A cylinder is also like a prism. It has two circular bases and a curved lateral face. Pyramids A pyramid is a polyhedron for which the base is a polygon and all lateral faces are triangles. A pyramid is described by the shape of its base. For example, a triangular pyramid has a triangular base, square pyramid has a square base. (i) Square pyramid and its net A square pyramid has a square base. Its four lateral faces are the isosceles triangles. www.geogebra.org/classroom/msphzkjw Classroom code: MSPH ZKJW Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Net of cylinder Cylinder www.geogebra.org/classroom/pkkc6ddn Classroom code: PKKC 6DDN Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Square pyramid Net of a square pyramid A square pyramid Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269 Vedanta Excel in Mathematics - Book 8 (ii) Triangular pyramid and its net A triangular pyramid has a triangular base. It has three congruent triangular faces. The tetrahedron is a triangular pyramid having congruent equilateral triangles for base and for each of its faces. (iii) Pentagonal pyramid and its net A pentagonal pyramid has a pentagon base and five congruent triangular faces. (iv) Hexagonal pyramid and its net A hexagonal pyramid has a hexagon base and six congruent triangular faces. (v) Cone and its net A cone is also like a pyramid. But it has a circular base and a curved face. Net of a pentagonal pyramid A pentagonal pyramid Net of a hexagonal pyramid A hexagonal pyramid Net of a cone A cone EXERCISE 17.3 General Section -Classwork 1. Name the following solids: a) b) c) Net of a triangular pyramid A triangular pyramid Area and Volume
Vedanta Excel in Mathematics - Book 8 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) e) f) g) h) i) a) b) C A B E D C A B F D E 2. Let’s write the names of the base and the lateral faces of the following solids: a) b) c) 3. Let’s study the nets of solids carefully. Then name the solids formed by each of these nets. d) e) f) g) h) i) Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 271 Vedanta Excel in Mathematics - Book 8 4. Trace the nets of the following solids: a) Cube/hexahedron b) cuboid c) cone d) cylinder e) tetrahedron f) triangular prism g) square based pyramid It’s your time - Project Work and Activity Section 5. Make the groups of your friends. Take the hard paper and cut and make the nets of cube, cuboid, cone, cylinder, etc. and paste with glue. Then submit the solids you or your group prepared to your subject teacher. 17.7 Area of solids For any three dimensional figure (3D solids), we can calculate the three types of surface area. They are area of the base, lateral surface area, and total surface area. The top and buttom faces of a solid which are parallel to each other are the bases. The remaining faces of the solid are the lateral surfaces. So, the combination of bases and the lateral surfaces form the total surface of the solid. (i) Area of cube A cube is bounded by six regular plane square faces. In the given figure, ABCD, EFGH, EFBA, HGCD, EHDA, and FGCB are the six faces of the cube. Here, ABCD and EFGH are two bases and the remaining faces are the lateral surfaces of the cube. Let, l be the length of each square face of the cube. Then, the area of each face = l 2 Now, the area of the 4 lateral faces = 4l 2 And, the area of 6 faces = 6l 2 Thus, the total surface area of a cube = 6l 2 and lateral surface area of a cube = 4l 2 . (ii) Area of cuboid A cuboid has 6 rectangular faces. In the given figure, The faces ABCD and EFGH are the bases. The faces EFBA and HGCD are two opposite lateral surfaces. Also, the faces EHDA and FGCB are the remaining two opposite lateral surfaces. Here, the area of 2 bases = area of (ABCD + EFGH) = 2lb The area of 2 opposite lateral surfaces = Area of (EFBA + HGCD) = 2lh The area of 2 other opposite lateral surfaces = Area of (EHDA + FGCB) = 2bh Thus, the total surface area of the cuboid = 2lb + 2lh + 2bh = 2 (lb + lh + bh) A D C B H E F G l l l A D G B C H E F l b h Area and Volume
Vedanta Excel in Mathematics - Book 8 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 17.8 Volume of solids The space occupied by any object is called its volume. In the case of a solid, obviously, its base covers the area of surface and its height occupies the region above the surface. So, its area of the base times height gives its volume. Thus, volume of a solid = Area of its base × its height (i) Volume of cube The area of the base of a cube is l 2 , where l is the length of the sides of the base. The height of the cube is also equal to the length of the base. ∴ Volume of a cube = area of the base × height = l 2 × l = l 3 If the cube is of unit (say 1 cm) length, breadth and height, then its volume is 1 cm3 . So, cm3 , m3 , etc. are the unit of the measurement of volume. (ii) Volume of cuboid Let, l, b and h be the length, breadth and height of a cuboid respectively. Then, area of the base of the cuboid = l × b ∴ Volume of the cuboid = area of base × height = l × b × h The base of a cuboid is covering the surface area of water which is equal to the area of the base of the cuboid. The base of the cuboid occupies its height times the space of water inside the water which is the volume of a cuboid. l b h l b h l l l l b h Worked-out Examples Example 1: Find the total surface area of the following solids: 4 cm 4 cm 4 cm 10 cm 5 cm 3 cm a) b) Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273 Vedanta Excel in Mathematics - Book 8 Solution: a) Here, the length of the side of the cube (l) = 4 cm ∴ Area of the cube = 6l 2 = 6 × 42 = 96 cm2 . b) Here, the length of the cuboid (l) = 10 cm the breadth of the cuboid (b) = 3 cm the height of the cuboid (h) = 5 cm Now, area of the cuboid = 2 (lb + lh + bh) = 2 (10 × 3 + 10 × 5 + 3 × 5) = 190 cm2 Example 1: Find the volume of the following solids: a) b) Solution: a) The length of the sides of the cube (l) = 6 cm ∴ Volume of the cube = l 3 = (6 cm)3 = 216 cm3 . b) The length of the cuboid (l) = 15 cm breadth of the cuboid (b) = 4 cm height of the cuboid (h) = 6 cm ∴ Volume of the cuboid = l × b × h = 15 cm × 4 cm × 6 cm = 360 cm3 Example 2: The base area of a cubical box is 2.25 m2 . Find the volume of the box. Solution: Here, area of the base of the cubical box = 2.25 m2 or, l 2 = 2.25 m2 or, l = 2.25 m2 = 1.5 m Now, the volume of the box = l 3 = (1.5)3 = 3.375 m3 . Example 3: A rectangular water tank is 3 m long, 2.5 m broad, and 2 m high. How much water does it hold? Solution: Here, length of the tank (l) = 3 m breadth of the tank (b) = 2.5 m height of the tank (h) = 2 m Now, volume of the tank = l × b × h = 3 m × 2.5 m × 2 m = 15 m3 The capacity of the tank is 15 m3 . So, it holds 15 m3 of water. 6 cm 6 cm 6 cm 15 cm 6 cm 4 cm Area and Volume
Vedanta Excel in Mathematics - Book 8 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 4: A carton is 50 cm long, 20 cm broad, and 30 cm high. How many cubical boxes each of 10 cm long can be put inside the carton? Solution: Here, length of the carton (l) = 50 cm breadth of the carton (b) = 20 cm height of the carton (h) = 30 cm Now, volume of the carton = l × b × h = 50 cm × 20 cm × 30 cm = 30000 cm3 . Also, the volume of each cubical box = (10 cm)3 = 1000 cm3 Again, the number of cubical boxes = Volume of carton Volume of each box = 30000 cm3 1000 cm3 = 30 Hence, 30 cubical boxes can be put inside the carton. Example 5: The length of a box is twice the breadth and thrice the height. If the volume of the box is 288 cm3 , find the length breadth and height of the box. Solution: Let the length of the box (l) = x cm Then, the breadth of the box (b) = x 2 cm and the height of the box (h) = x 3 cm the volume of the box (V) = 288 cm3 Now, volume of the box (V) = x × x 2 × x 3 or, 288 cm3 = x3 6 or, 1728 = x3 or, x3 = 1728 cm3 or, x = 1728 cm3 = 12 cm ∴ The length of the box (l) = 12 cm the breadth of the box (b) = x 2 = 12 2 = 6 cm the height of the box (h) = x 3 = 12 3 = 4 cm Example 6: Find the volume of the given solid. Solution: Here, volume of the whole solid is the sum of the volume of 3 smaller solids. Volume of the lowermost one cuboid = l × b × h = 12 cm × 5 cm × 4 cm = 240 cm3 . 10 cm 12 cm 5 cm 4 cm 4 cm 4 cm Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 275 Vedanta Excel in Mathematics - Book 8 Also, length of each uppermost cuboid = 4 cm breadth of each uppermost cuboid = 5 cm height of each uppermost cuboid = (10 – 4) cm = 6 cm ∴ Volume of 2 uppermost cuboids = 2 (4 cm × 5 cm × 6 cm) = 2 × 120 cm3 = 240 cm3 Now, volume of the whole cuboid = 240 cm3 + 240 cm3 = 480 cm3 EXERCISE 17.4 General Section – Classwork 1. Let’s say and write the answers as quickly as possible. a) The parallel faces of a solid are called .............................. b) If a be the length of each face of a cube, the area of the bases = .............................. c) If l be the length of each face of a cube, the lateral surface area = .............................. d) If l, b, and h be the length, breadth and height of a cuboid, (i) its area of base = .............................. (ii) its lateral surface area = .............................. (iii) its total surface area = .............................. 2. Let’s say and write the answers as quickly as possible. a) If the area of the base of a cube is a2 cm2 , its volume is ................................... b) If the length of a cube is 2 cm, its volume is ................................... c) If the area of the base of a cuboid is (x × y) cm2 and height is z cm, its volume is ................................... d) If the area of the base of a cuboid of height 3 cm is 10 cm2 , its volume is ................................... e) If the length, breadth and height of a box is 5 cm, 4 cm and 3 cm respectively, its volume is ................................... 3. a) If the volume of a cube is 8 cm3 , its length is ................................... and area of the base is ................................... b) If the volume of a cuboid is 60 cm3 and height is 3 cm, its area of the base is ................................... c) If the volume of a cuboid is 60 cm3 and its area of the base is 20 cm2 , its height is ................................... Area and Volume
Vedanta Excel in Mathematics - Book 8 276 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Creative Section - A 4. Find the total surface area of the following solids: a) b) c) 5. a) A cubical block 10 cm long is placed on the table. How much area does it cover on the surface of the table? b) Find the total surface area and lateral surface area of a chalk box whose height, length and breadth are 25 cm, 20 cm and 15 cm respectively. 6. Find the volume of the following solids: a) b) c) 7. a) Find the volume of the cubes of the given lengths. (i) 5 cm (ii) 4.5 cm (iii) 8 cm (iv) 2.5 m b) Find the volume of the cuboids of the following measurements. Length breadth height (i) 5 cm 4 cm 3 cm (ii) 1.5 m 20 cm 50 cm (iii) 42 cm 30 cm 2.5 m (iv) 1.2 m 1 m 80 cm 8. a) The area of the base of a cubical box is 100 cm2 , (i) Write the formula to find the area of the base of a cube. (ii) Find the length of the cubical box. (iii) Find the volume of the box. b) If the volume of a cube is 3375 cm3 , find its height and area of the base. c) A cubical water tank is 1.2 m high. How many litres of water does it hold? (1000 cm3 = 1 litre) 3 cm 3 cm 3 cm 3.5 cm 15 cm 2 cm 12 cm 6 cm 4 cm 4 cm 4 cm 4 cm 6 cm 10 cm 5 cm 15.5 cm 7 cm 8 cm Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 277 Vedanta Excel in Mathematics - Book 8 9. a) The measurement of a rectangular vessel is 40 cm × 30 cm × 50 cm. (i) Write the formula to find the volume of a cuboid. (ii) Find the volume of the rectangular vessel. (iii) How many litres of oil can be put inside it to fill it completely? b) A rectangular water tank is 1.5 m × 1.4 m × 2 m. How many litres of water does it hold when it is completely filled? 10. a) A cuboidal wooden block is 12 cm broad, 10 cm high and its volume is 1800 cm3 . (i) Find the length of the block . (ii) Find the total surface area of the block. (iii) Find the cost of painting the block at 50 paisa per sq. cm. b) A rectangular box is 1 m long and 75 cm high. If the volume of the box is 375000 cm3 , find the width of the box. c) The area of the base of a rectangular tank is 2.4 m2 . If the capacity of the tank is 3.6 m3 , find the height of the tank. d) A rectangular vessel can hold 40 litres of milk when it is full. (1 l = 1000 cm3 ) (i) Find the volume of the vessel in cubic centimeter. (ii) If the area of the base of the vessel is 2000 cm2 , find the height of the vessel. Creative Section - B 11. a) A carton is 60 cm long, 40 cm broad, and 50 cm high. It is tightly filled with cubical boxes each of 20 cm long. (i) Find the volume of the carton. (ii) Find the volume of each cubical box. (iii) How many cubical boxes can be put inside the carton? b) A cubical box is 75 cm high. How many rectangular packets each of 25 cm × 15 cm × 5 cm are required to fill the box? c) A rectangular box is 5 cm long and 6 cm wide. How many cubical packets each of 1 cm3 can be placed on the floor of its base? If 10 layers of such packets are placed on the floor, how many packets are there? d) A wall is 15 m long, 0.2 m wide and 2 m high. How many bricks of size 15 cm × 8 cm × 5 cm each are required to construct the wall? Area and Volume 12. Find the volume of the following solids. 7 cm 8 cm 3 cm 2cm 12 cm 5 cm 3 cm 10 cm 15 cm 4 cm a) b)
Vedanta Excel in Mathematics - Book 8 278 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur e) 3 cm 3 cm 3 cm 2 cm 8 cm 7 cm 3 cm f) 12 cm 12 cm 10 cm 3cm 4 cm 4 cm 6 cm 6 cm 6 cm 5 cm 5 cm 5 cm 10 cm 7 cm 3 cm 3 cm 3 cm 3 cm 6 cm 6 cm 13. a) The ratio of length, breadth and height of a cuboid is 3:2:1. If the volume of the cuboid is 48 cm3 , find the base area of the cuboid. b) The length, breadth and height of a rectangular room are in the ratio 4:3:2. If the room contains 192 m3 of air, find the area of floor of the room. c) The length of a room is three times the height and breadth is twice of its height. If the volume of the room is 384 m3 , find its length, breadth, and height. d) The length of a rectangular carton is twice the breadth and thrice the height. If the volume of the carton is 4,500 cm3 , find its length, breadth, and height. It’s your time- Project Work and Activity Section 14. a) Let’s measure the length, breadth and thickness (or height) of the following objects and estimate the total surface area of the objects. (i) Mathematics book (ii) Instrument box b) Let’s take a rectangular sheet of a chart paper and measure it’s length and breadth to find its surface area. Now roll it to form a cylinder and measure its diameter and radius. Then, find the curved surface area. Is the curved surface area is same as the area of rectangle? 15. a) Let’s measure the length, breadth and thickness of your mathematics book and find its volume. b) Let’s measure the length, breadth and height of a cuboidal soap and calculate the volume of such 10 soaps. c) Let’s measure the length, breadth and height of cubical or cuboidal objects found in your home or in your classroom. Find the volume of each object. Area and Volume
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 279 Vedanta Excel in Mathematics - Book 8 18.1 Symmetrical and asymmetrical shapes - Looking back Let's draw dotted line to make the following shapes exactly halves. a) b) c) d) e) Can you make the figure (b) and (d) exactly halves by drawing dotted lines? A geometric shape or object is said to be symmetrical if it can be divided into two or more identical parts. However, if a shape cannot be divided into the identical parts, it is called asymmetrical. Symmetrical Asymmetrical Symmetrical Asymmetrical 18.2 Line or axis of symmetry In the following figures, one or more dotted line segments divide the figures into two identical halves. The dotted lines are called the line of symmetry or axis of symmetry. l 1 and l 2 and l 3 and l 4 are lines of symmetry l 1 and l 2 and l 3 are lines of symmetry l 1 is a line of symmetry l 1 l 1 l 1 l 1 l 2 l 2 l 2 l 3 l 3 l 4 l 1 and l 2 is a lines of symmetry Thus, the line of symmetry is the line (or imaginary line) that passes through the centre of a symmetrical shape and divides it into identical halves. The line of symmetry is also called the mirror line or axis of symmetry. Unit 18 Symmetry, Design and Tessellation
Vedanta Excel in Mathematics - Book 8 280 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 18.1 General Section - Classwork 1. Let's say and write, how many lines of symmetry you can find for each of the shapes below. If in doubt, trace and fold them. 2. Let's draw the lines of symmetry of the following geometrical figures by using dotted lines. 18.3 Tessellations Let's complete the patterns and make designs. Colour the patterns and make the design attractive. A tessellation is a covering of the plane with congruent geometrical shapes in a repeating pattern without leaving any gaps and without overlapping each other. Tessellation is also known as tiling. Symmetry, Design and Tessellation
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 281 Vedanta Excel in Mathematics - Book 8 In a tessellation, the shapes are often polygons. The polygon may be either an equilateral triangle, a square, or a regular hexagon. Tessellation is done on the surface of carpets and on the surface of floor or wall to make the surface more attractive. Thus, to make a tessellation (or tiling) – (i) Use the sets of congruent figures (tiles). (ii) Do not leave any gaps. (iii) Do not have any overlaps. 18.4 Types of tessellations (i) Regular tessellations In this case, we use the same types of regular polygons. The polygons that we use may be equilateral triangle, square, regular hexagon, etc. (ii) Semi-regular tessellations In this case, we use two or more regular polygons. In the adjoining tessellation, regular octagons and squares are used. Symmetry, Design and Tessellation
Vedanta Excel in Mathematics - Book 8 282 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iii) Irregular tessellations In this case, we use irregular types of polygons. EXERCISE 18.2 1. Define the following terms. a) Tessellation b) Regular tessellation c) Semi-regular tessellation d) Irregular tessellation 2. What are three important rules while making a tessellation? 3. Let's use graph paper to copy the following tessellation patterns and complete them. Symmetry, Design and Tessellation
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 283 Vedanta Excel in Mathematics - Book 8 4. Let's use graph paper and draw the following patterns of design. Colour your designs and make them attractive. 5. Let's complete the following patterns by tessellating a few more patterns. Colour the design and make them attractive. It's your time - Project Work and Activity Section 6. a) Let's draw your own patterns of designs in graph papers. Colour the patterns and demonstrate in your class. b) Let's observe the various patterns of carpet designs. Copy the patterns in graph papers and make attractive design by colouring. Symmetry, Design and Tessellation
Vedanta Excel in Mathematics - Book 8 284 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 19.1 Bearing – Looking back Classroom - Exercise Let’s say and write the answers as quickly as possible. 1. a) How many directions are there in compass bearing ? What are they ? ................................................................................................................................. ................................................................................................................................. b) Which is the base line direction to find the bearing of any place ? ................................................................................................................................. 2. a) Tell and write the directions represented by (i) NE ......................... (ii) SE ................................ (iii) SW ..................... (iv) NW ............................ b) Tell and write the bearing of these direction (i) East ......................... (ii) South .......................... (iii) West ............................. In the adjoining compass NOS represents NorthSouth and EOW represents East-West directions. The angle between N and E is 90°. The direction NE lies exactly in between N and E. So, the angle between N and NE is 45°. Let’s take O as the point of reference and ON (the North line) as base line, then, the direction of NE from O in clockwise direction is 045°. It is called bearing of NE from O. Similarly, the bearing of E from O is 090°, the bearing of SW from, O is 225° and so on. Thus, the direction in which an object (or place) is sighted may be specified by giving angle in degrees that the direction makes with north line in clockwise is called the bearing of the object (or place). Of course, bearing of a place from another place with north line as the base tells the distance between these two places in terms of angles written in 3 digits. So, it is also called three-digit bearing (or three-figure bearing). North (N) South (S) North West (NW) North East (NE) South West (SW) West (W) East (E) South East (SE) N S W NW SE NE SW 045° 255° E Unit 19 Bearing and Scale Drawing
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 285 Vedanta Excel in Mathematics - Book 8 Worked-out examples Example 1: Write down the bearing of the places, A, B, C, and D from O. c) C 140° d) D 35° a) O A 70° N b) O B 162° N N N O O Solution: a) Bearing of A from O is 070° b) Bearing of B from O is 162° c) Bearing of C from O is (360° – 140°) = 220° d) Bearing of D from O is (360° – 35°) = 325°. Example 2: In the given diagram, if the bearing of A from B is 065°, find the bearing of B from A. Solution: Here, bearing of A from B is 060°. Here, BN // AN1 . So, ∠NBA and ∠N1 AB are co-interior angles. ∴ ∠N1 AB = 180° – 60° [Sum of co-interior angles is 180°] = 120° ∴ Bearing of B from A = 360° – 120° = 240°. Example 3: An aeroplane was flying in the bearing of 085° from Kathmandu. After flying over a certain distance, it changed its direction and flew in the bearing of 240°. By what angle did the aeroplane change its direction? Solution: Here, bearing of P1 from K = 085° Bearing of P2 from K = 240° ∴ The direction changed by the plane from P1 to P2 = 240° – 085° = 155° N N1 B A 060° N K P1 P2 c) 085° 240° Bearing and Scale Drawing
Vedanta Excel in Mathematics - Book 8 286 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 4: If the bearing from Birganj to Dharan is 084°, find the bearing from Dharan to Birgunj. Solution: Let B represents Birganj and D represents Dharan. Here, bearing from B to D = ∠NBD = 084° Now, ∠NBD + obtuse ∠N1 DB = 180° [NB // N1 D and the sum of co-interior angles] or, 084° + obt. ∠N1 DB = 180° ∴ Obt. ∠N1 DB = 180° – 084° = 096° Again, bearing from D to B = 360° – obt.∠N1 DB = 360° – 096° = 264° So, bearing from Dharan to Birganj is 264°. EXERCISE 19.1 General Section – Classwork 1. Let’s tell and write the bearing of the following directions. a) NNE ............... b) NOE ............... c) NSE ............... d) NOS ............... e) NSW ............... f) NOW ............... g) NNW ............... 2. Let’s tell and write the bearing of A from O. N S O W NW SE NE SW E a) b) 40° 150° 60° A O A O N c) N A N N N1 D B 084° Bearing and Scale Drawing
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 287 Vedanta Excel in Mathematics - Book 8 3. Study the map of Nepal given below and answer the following questions. a) If Kathmandu is the point of reference, write the compass direction of: (i) Dharan .................................... (ii) Birgunj ..................................... (iii) Dhangadhi .............................. (iv) Jomsom .................................. b) If Nepalgunj is the point of reference, write the compass direction of : (i) Jumla .................................... (ii) Pokhara ..................................... (iii) Biratnagar ............................ (iv) Baitadi ...................................... c) If Dharan is the point of reference, write the compass direction of: (i) Birgunj ............................ (ii) Pokhara ............................ (iii) Biratnagar ............................ (iv) Kathmandu ............................ Baitadi Pokhara Kathmandu Birgunj Dharan Biratnagar Jumla Dhangadhhi Nepalgunj Jomsom Creative Section - A 4. Write the bearings of the following compass directions. E.g., Bearing of NE = 090° Bearing of NNW = 270° + 45° = 315° a) N and NE (NNE) b) N and SE (NSE) c) N and S (NS) d) N and SW (NSW) e) N and W (NW) f) N and NW (NNW) Bearing and Scale Drawing
Vedanta Excel in Mathematics - Book 8 288 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur N A A A A N a) b) c) N d) N 60° 90° 125° 160° A A A A N N N N e) f) g) h) 150° 130° 90° 42° 6. In the adjoining figure, calculate the bearings of A, B, C, D and E from O. 7. The bearings of A from B are given in the figures. Find the bearings of B from A. N S A B E W O D C 35° 40° 60° 10°30° 20° 80° N B A A A A B B B N N1 N1 N N N1 N1 a) b) c) d) 040° 078° 115° 150° 5. Write down the bearings of the aeroplanes from A shown below. N1 N N N A B B B B A A A N N1 N1 N1 e) f) g) h) 200° 270° 285° 330° Bearing and Scale Drawing
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 289 Vedanta Excel in Mathematics - Book 8 Creative Section -B 8. a) If the bearing of a place A from the place B is 055°, find the bearing of B from A. b) If the bearing of a place P from the place Q is 170°, find the bearing of Q from P. c) If the bearing of Kathmandu from Pokhara is 095°, what is the bearing of Pokhara from Kathmandu? d) If the bearing of your school from your house is 082°, what is the bearing of your house from the school? e) An aeroplane was flying in the bearing of 075° from Kathmandu. After flying over a certain distance it changed its direction and flew in the bearing of 310°. By what angle did the aeroplane change its direction? It’s your time - Project Work and Activity Section 9. a) Let’s estimate the bearing of the following places from your home. (i) your school (ii) nearby health post or hospital (iii) nearby police station b) Let’s estimate the bearing of the following places from your school. (i) your home (ii) nearby health post or hospital (iii) nearby police station 19.2 Scale drawing – Review It is not possible to draw the actual distance between two places in drawing. In such cases, we take a convenient scale and reduce the distance between the places in the drawing. It is known as scale drawing. The scale which is taken to reduce (or enlarge) the actual size is scale factor. Scale factor is the ratio of the size of the drawing to the actual size of the object. For example, (i) A distance of 5 km (ii) A length of 80 m Scale: 1 cm to 1 km (or 1:100000) Scale 1 cm to 20 m (or 1:2000) Scale drawing Scale drawing 0 1 2 3 4 5 km 0 20 40 60 80 m Worked-out Examples Example 1: What is the actual distance between two places which is represented by 5.4 cm on a map which is drawn to the scale 1:1000000? Solution: Here, the scale 1:1000000 means, 1 cm represents 1000000 cm = 1000000 100 × 1000 km = 10 km Now, 1 cm represents 10 km ∴ 5.4 cm represents 5.4 × 10 km = 54 km So, the distance between two places is 54 km. Bearing and Scale Drawing
Vedanta Excel in Mathematics - Book 8 290 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: An aeroplane flies 500 km in the bearing of 040° and 600 km in the bearing of 110°. Find the distance between the two places and the bearing of the starting place from the last place. (Scale 1 cm = 100 km) Solution: In the figure, O is the starting place and B is the last place. Here, 100 km = 1 cm ∴ 500 km = 5 cm Also, 600 km = 6 cm By measurement, OB = 9 cm Now, the distance between two places= 9 × 100 km = 900 km Again, by measurement, ∠N2 BO = 360° - 101° = 259° So, the bearing of the starting place from the last place is 259°. N O N1 N2 110° 040° 6 cm 5 cm A B EXERCISE 19.2 General Section – Classwork 1. Let’s say and write the answers as quickly as possible. a) When the scale is 1 : 100, the actual length of 5 cm is .......................................... b) When the scale is 1 : 200, the actual length of 3 cm is .......................................... c) When the scale is 1 : 500, the actual length of 2 cm is .......................................... 2. a) When the scale is 1 : 100, the map / drawing length of 4 m is ............................. b) When the scale is 1 : 200, the map / drawing length of 8 m is .............................. c) When the scale is 1 : 500, the map / drawing length of 10 m is ............................ Bearing and Scale Drawing
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 291 Vedanta Excel in Mathematics - Book 8 Creative Section 3. Copy and complete the table. Length of map/drawing Scale Actual length a) b) c) d) 2 cm 3 cm 4.2 cm 5.5 cm 1:5000 1:10000 1:100000 1:2000000 .......................... m .......................... m ..........................km .......................... km 4. Copy the complete the table. Actual length Scale Length on map/drawing a) b) c) d) 24 m 650 m 40 km 80 km 1:200 1:1000 1:500000 1:1000000 .......................... cm .......................... cm .......................... cm .......................... mm 5. a) What is the actual length which is represented by 4.5 cm on a scale drawing with scale of 1 cm to 5 m? b) What is the actual distance between two places which is represented by 3.2 cm on a scale drawing with scale of 1:1000000? c) The scale drawing distance between Kathmandu and Biratnagar is 4.7 cm. If the scale factor is 1:10000000, find the actual distance between these two places. 6. a) An aeroplane flies 60 km from A to B in the bearing of 070° and then 90 km from B to C in the bearing of 105°. Represent this information by drawing in a scale of 1 cm = 10 km and answer the following questions. (i) How far is C from A? (ii) What is the bearing of C from A? (iii) What is the bearing of A from C? b) A bus is at P, a place 250 km east from Q. Traveling a distance of 400 km to due west of P, it reaches at R. Express this information by scale drawing and find (i) the distance from Q to R (ii) the bearing of Q from R. It’s your time - Project Work and Activity Section 7. a) Let’s choose an appropriate scale and draw a road map of your school from your home. b) Let’s sketch the layout of ground floor of a house with an appropriate scale to the following structures. sitting room - 12 m by 10 m, a bed-room - 10 m by 8 m, a kitchen - 8 m by 6 m, a dinning room - 8 m by 6 m, a washroom - 4 m by 3 m, guest room - 10 m by 8 m. Bearing and Scale Drawing
Vedanta Excel in Mathematics - Book 8 292 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Assessment - IV 1. In the given figure, AB // CD, ∠AGH = (3x + 5)o , ∠GHD = (x + 45)o , ∠EGB = y and ∠FHD = z. Study the figure and answer the following questions. (a) Which is the alternate angle of ∠AGH? (b) Write down a pair of co-interior angles. (c) Find the value of x, y and z. 2. Study the given figure and answer the following questions. (a) Name the corresponding angle of ∠ABC? (b) What is the value of x? (c) What is the value of y? (d) Is triangle CDE an isosceles triangle? Give reason. 3. In the given triangle PQR, PQ = PR and the ratio of ∠QPR and ∠PQR is 1:2. (a) Write the relation between PQR and ∠PRQ. (b) Find the measure of ∠PQR. (c) Construct a triangle PQR in which PQ = PR = 5 cm and QR = 4 cm. 4. In a parallelogram ABCD, AB = 5 cm, ∠BAD = 60o and AD = 3.5 cm. (a) Write down any two properties of parallelogram. (b) If CD = (3x – 1) cm, find the value of x. (c) Construct the parallelogram. (d) If the adjacent sides AB and BC of the parallelogram were equal, name the special type of the parallelogram? 5. In the given kite ABCD, BD is a diagonal. (a) Define a kite. (b) Prove that: ∆BAD ≅ ∆BCD. (c) If ∠BDC = 300 , what would be the measure of ∠BDA? (d) If the diagonals AC and BD interest at O, length of diagonal AC = 12 cm, AD = 10 cm and OB=7 cm, what would be the length of the diagonal BD? 6. In the given map; ABCD is the compound, PQRS is the house, FEC is the garden and OM is the radius of circular swimming pool. (a) Find the area of the compound. (b) Find the area of the land covered by the house. (c) Which is more spacious, the garden or the swimming pool? 7. A (0, 4), B (3, -1) and C (-2, 0) are the vertices of ∆ABC. Answer the following questions. (a) Find the coordinates of its image under the reflection about x-axis. Also, present both the triangles on the same graph paper. (b) Find the lengths of each side of the triangle ABC by using distance formula. E G H F z B (3x + 5)° (x + 45)° y C D A A E B C D (2x+5)° 65° y° 50° P Q R A B C D 30 m 20 m 25 m 40 m 16 m 14 m A D E C F B O M Q R P S
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 293 Vedanta Excel in Mathematics - Book 8 20.1 Review Statistics is a branch of mathematics in which data of facts and information are collected, sorted, displayed, and analysed. Statistics are used to make decision and prediction about the future plans and policies. Statistical information helps to understand the economic problems and formulation of economic policies. The word ‘statistics’ comes from the word ‘state’, largely because it was the job of the state to keep records and make decisions based on census result. 20.2 Collection of data The marks obtained by 24 students of class 8 in Mathematics out of 20 full marks are given below. 15, 10, 17, 14, 19, 12, 15, 10, 16, 20 14, 18, 15, 10, 17, 16, 18, 17, 20, 13 Such numerical figures are called data. Data should be presented in a proper order so that it is easier to get the necessary information for which they are collected. When the data are arranged either in ascending or in descending order, they are said to be in proper order. The properly arranged data are called arrayed data; otherwise, they are said to be raw data. 20.3 Frequency table The data given below are the hourly wages (in Rs) of 20 workers of a factory: 75, 90, 65, 80, 70, 65, 75, 75, 80, 70, 75, 80, 90, 80, 70, 70, 75, 75, 70, 75 → Raw data 65, 65, 70, 70, 70, 70, 70, 75, 75, 75, 75, 75, 75, 75, 80, 80, 80, 80, 90, 90 → Arrayed data Here, Rs 65 is repeated 2 times. So, its frequency is 2. Rs 70 is repeated 5 times. So, its frequency is 5. Rs 75 is repeated 7 times. So, its frequency is 7. Rs 80 is repeated 4 times. So, its frequency is 4. Rs 90 is repeated 2 times. So, its frequency is 2. In this way, a frequency is the number of times a value occurs. Data and their frequencies can be presented in a table called frequency table. Unit 20 Statistics
Vedanta Excel in Mathematics - Book 8 294 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Now, let’s present the above wages of 20 workers in a frequency table. Wages (in Rs) Tally marks Frequency 65 70 75 80 90 || |||| |||| || ||| || 2 5 7 4 2 Total 20 Tallying is a system of showing frequencies using diagonal lines grouped in fives. Each time five is reached, a horizontal line is drawn through the tally marks to make a group of five. The next line starts a new group. For example, 1 → | 2 → || 3 → ||| 4 → |||| 5 → |||| 6 → |||| | 7 → |||| || 8 → |||| ||| 9 → |||| |||| 10 → |||| |||| 20.4 Grouped and continuous data Let following are the marks obtained by 20 students in a Mathematics exam of full marks 50. 27, 38, 25, 18, 9, 24, 48, 15, 27, 35 23, 45, 32, 16, 26, 39, 20, 33, 40, 37 The above mentioned data are called individual or discrete data. Another way of organizing data is to present them in a grouped form. For grouping the given data, we should first see the smallest value and the largest value, then we have to divide the data into an appropriate class–interval. The numbers of values falling within each class– interval give the frequency. For example, Marks Tally marks Frequency 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | ||| |||| || |||| | ||| 1 3 7 6 3 Total 20 In the above series, 9 is the smallest value and 48 is the largest value. So, the data are grouped into the interval of 0 – 10, 10 – 20, etc., so that the smallest and the largest values should fall in the lowest and the highest class–interval respectively. Statistics
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 295 Vedanta Excel in Mathematics - Book 8 Let’s consider a class–interval 10 – 20. Here, 10 is called the lower limit and 20 is the upper limit of the class–interval. The difference between two limit is called the length or height of each class–interval. For example, in 10 – 20 the length of the class–interval is 10. Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, … Here, the upper limit of a pervious class–interval has repeated as the lower limit of the consecutive next class–interval. Such an arrangement of data is known as grouped and continuous data. 20.5 Cumulative frequency table The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile up’. The table given below shows the marks obtained by 20 students in a mathematics test and the corresponding cumulative frequency of each class–interval Marks Frequency (f) Cumulative frequency (c.f) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 2 5 8 4 1 2 5 + 2 = 7 8 + 7 = 15 15 + 4 = 19 19 + 1 = 20 Total 20 Thus, cumulative frequency corresponding to a class–interval is the sum of all frequencies up to and including that class–interval. 20.6 Graphical representation of data We have already discussed to present data in frequency distribution tables. Alternatively, we can also present data graphically. Different types of diagrams are used for this purpose. Here, we shall discuss two types of diagrams: line graph and pie chart. (i) Line graphs Data can also be represented by plotting the corresponding frequencies in the graph paper. The line so obtained by joining the points is called the line graph. While constructing a line graph, the frequencies of the items are plotted along y–axis. The line graph given below represents the daily wags of the workers of a company. Wages in Rs. 40 50 60 70 80 90 100 No. of workers 15 8 22 18 10 8 5 → 2 students obtained marks less than 10. → 7 students obtained mark less than 20. → 15 students obtained marks less than 30. → 19 students obtained marks less then 40 → 20 students obtained marks less than 50. Statistics
Vedanta Excel in Mathematics - Book 8 296 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur EXERCISE 20.1 General Section – Classwork Lets say and write the answers as quickly as possible. 1.a) A branch of mathematics which deals with the collection, presentation, analysis and interpretation of data is known as ............................................ b) The properly arranged data are called ............................................ c) The number of times that a particular observation occurs in the data is the ........................................... d) In a class interval 40 - 50, the lower limit is ....................................... and the upper limit is .................................... 2. The adjoining line graph shows the average rain fall in mm during 6 months of the year 2020. (i) In which month was the minimum average rainfall? On which month was it maximum? ................................................ (ii) How much was the average rainfall recorded in August? ................................................ (iii) Write a paragraph about the common trend of rainfall during these months in your exercise book. Y X 25 40 50 60 70 80 90 100 20 15 10 No. of workers 5 0 wages (in Rs) 100 120 Apr May June July Aug Sep 80 60 40 20 0 Rainfall (mm) Months Statistics
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 297 Vedanta Excel in Mathematics - Book 8 Creative Section 3. a) From the marks given below obtained by 20 students in Mathematics, construct a frequency distribution table with tally marks. 15, 18, 12, 16, 18, 10, 15, 16, 15, 12 10, 12, 15, 12, 16, 18, 12, 15, 12, 16 b) The marks obtained by 40 students in mathematics in SEE examination are given below. Group the data into the class intervals of length 10 and construct a cumulative frequency distribution table. 42, 68, 80, 45, 92, 36, 8, 17, 49, 30 5, 26, 98, 74, 53, 65, 72, 28, 55, 46 86, 70, 62, 27, 16, 44, 85, 59, 51, 73 66, 78, 38, 81, 97, 77, 69, 45, 33, 67 c) The marks obtained by 50 students of class 8 in Mathematics are shown in the table given below. Construct a cumulative frequency table to represent the data. Marks 10 20 30 40 50 No. of students 4 9 20 15 2 d) The hourly wages of 40 workers in a factory are shown in the table given below. Show their wages in a cumulative frequency table. Wages (in Rs.) 50 60 70 80 90 No. of workers 7 14 11 5 3 4. a) Class 8 students conducted a survey and recorded the number of children in primary level in their school in the following table. Construct a line graph to show the numbers in different classes. Classes I II III IV V No. of children 25 20 40 50 45 b) The table given below shows the velocity of a bus at different interval of time. Draw a line graph to show the velocity–time graph. Time (in second) 5 10 15 20 25 30 Velocity (in m/s) 5 15 20 30 10 25 It’s your time - Project Work and Activity Section 5. a) Let’s collect and write the marks obtained by your friends in the recently conducted mathematics exam. Group the data into the class interval of length 10 and show them in a cumulative frequency distribution table. b) Let’s collect and write the number of students from class 1 to 8 in your school. Show the data in a line graph. (ii) Pie chart A pie chart is a circular statistical graphic which is divided into sectors and the angles of the sectors represent the frequency. Constructing pie–charts Follow these steps to draw a pie chart. Statistics
Vedanta Excel in Mathematics - Book 8 298 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur – Add all the frequencies and write each frequency as a fraction of the total frequency. – Change each fraction into a number of degrees multiplying by 360°. (There are 360° in a circle) i.e. number of degrees of each fraction = corresponding frequency Total frequency × 360° – Tabulate the angles in ascending or descending order. – Draw a circle of convenient size. Then draw a radius as a starting point. – Use a protractor to construct the angles at the centre corresponding to each sector. Worked-out Examples Example 1: The table given below shows the number of students in classes 1 to 5 of a school. Draw a pie chart to represent the numbers. Class I II III IV V No. of students 40 50 30 45 15 Solution: Here, total number of students = 40 + 50 + 30 + 45 + 15 = 180 Number of degrees of the sector of class I 40 180 = × 360° = 80° Number of degrees of the sector of class II 50 180 = × 360° = 100° Number of degrees of the sector of class III 30 180 = × 360° = 60° Number of degrees of the sector of class IV 45 180 = × 360° = 90° Number of degrees of the sector of class V 15 180 = × 360° = 30° Example 2: The table given below shows the monthly income of a family from different sources. Represent the data in a pie chart. Source House rent Business Agriculture Salary Others Income (in Rs) 1500 1700 1200 2500 300 Solution: Here, total monthly income = Rs 7200 Number of degrees of the sector of house rent Number of degrees of the sector of business 1500 7200 = × 360° = 75° Number of degrees of the sector of agriculture 1700 7200 = × 360° = 85° Number of degrees of the sector of salary 1200 7200 = × 360° = 60° Number of degrees of the sector of others 2500 7200 = × 360° = 125° 300 7200 = × 360° = 15° Class II Class III Class IV Class V Class I 100° 80° 60° 90° 30° Salary 125° Business 85° House rent 75° Agriculture 60° Other15° Statistics
Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 299 Vedanta Excel in Mathematics - Book 8 Now, tabulating the angles in descending order: Source Salary Business House rent Agriculture Others Angle 125° 85° 75° 60° 15° Example 3: The monthly budget of a family is shown in the given pie chart. If the total expenditure is Rs 10800, calculate the expenditure on each item and show in a table. Solution: Here, total expenditure = Rs 10800 Food 120° Rent 75° Fuel 40° Miscellaneous 30° Education 95° Now, the table given below shows the actual expenditure. Items Food Education Rent Fuel Miscellaneous Expenditure (Rs) 3600 2850 2250 1200 900 Expenditure on food Expenditure on miscellaneous Expenditure on fuel Expenditure on rent Expenditure on education 10800 360 = Rs × 120 = Rs 3600 10800 360 = Rs × 95 = Rs 2850 10800 360 = Rs × 75 = Rs 2250 10800 360 = Rs × 40 = Rs 1200 10800 360 = Rs × 30 = Rs 900 Whole circle represents the total expenditure. i.e. 360° represents Rs 10800 1° represents Rs 10800 360 120° represents Rs 10800 360 × 120 EXERCISE 20.2 General Section – Classwork 1. The given pie-chart shows the income of a family from three different sources. Answer the following questions. a) How many degrees represent the total income of the family? .............................. b) From which source does the family have maximum income? .............................. c) From which source does the family have minimum income? .............................. d) If the annual income of the family is Rs 4,50,000, how much is the income from farming? .............................. Farming 180° Business 120° Services 60° Statistics
Vedanta Excel in Mathematics - Book 8 300 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 2. a) The table given below shows the number of students of a school from class 4 to 8. Represent the data in a pie chart. Class IV V VI VII VIII No. of students 50 45 40 35 10 b) The monthly budget of a family is given below. Food – Rs 2100 Clothing – Rs 3300 Miscellaneous – Rs 2250 Saving – Rs 3150 Represent the above data in a pie chart. c) Mr. Limbu’s spends Rs 10,800 in a month. His monthly expenditure on different headings are given below: Food – Rs 2,400 Education – Rs 3,600 Rent – Rs 4,200 Miscellaneous – Rs 600 Show the above data in a pie chart. d) Represent the following data in a pie chart. Monthly expenditure of a family. Item Education House rent Food Others Amount (Rs) 8000 4000 10000 14000 e) World consumption of energy is given below: Natural gas – 20 % Nuclear – 5 % Coal – 25 % Hydroelectric – 10 % Oil – 40 % Represent the above data in a pie chart. 3. a) Mrs. Rai’s monthly expenditure is Rs 10,800. The diagram on the right is a pie chart showing her expenditure on different headings. Work out how much was spent under each heading. b) The diagram on the right is a pie chart showing the expenses of a small manufacturing firm. The total expenses were Rs 1,44,000 in a month. Calculate the expenditure on each heading. c) The pie chart given alongside shows the votes secured by three candidates A, B and C in an election. If A secured 5,760 votes, i) how many votes did C secure? ii) who secured the least number of votes? How many votes did he secure? Food 150° 60° Study Rent 70° 50° 30° Transportation Miscellaneous Wages 150° Fuel 40° Rent 50° Raw materials 120° X 120° Y Z 140° Creative Section Statistics