MENSURATION 151 COMPETENCY Solution the behavior problems related to perimeter, area and volume. CHAPTERS 9. Perimeter, Area and Volume LEARNING OUTCOMES find the perimeter of triangular surface. find the surface area of cuboid and cube solve the problems related to surface area and volume of cuboid and cube. establish and use the relation of circumference and diameter of a circle. MENSURATION UNIT III Perimeter Area Volume
152 The Leading Mathematics - 7 CHAPTER 9 Perimeter, Area and Volume Lesson Topics Pages 9.1 Perimeter of Triangular Surface 153 9.2 Surface Area of Cube and Cuboid 158 9.3 Volume of Cube and Cuboid 163 9.4 Circumference of Circle 166 Can you read the scales of the ruler or measuring tape in cm and inches? Which materials are used for measuring the length, breadth and height of the compound wall and aquarium? Can you measure their length, breadth and height? Can you add and subtract the numbers 2457 and 1989? Can you multiply the numbers 5479 and 89? Can you divide 598 by 13? Can you add and subtract the decimal numbers 1.4 and 3.28? Can you multiply the numbers 2.54 and 3.28? Can you divide 39.37 by 3.28? Divide 2.54 by 12. Can you divide 39.37 by 3.28? WARM-UP
MENSURATION 153 9.1 Perimeter of Triangular Surface At the end of this topic, the students will be able to: ¾ find the perimeter of rectangle, square and parallelogram. Learning Objectives A point and a line are two basic geometrical objects. No measurement is made in the case of a point. A line segment is measured in terms of what is known as length. At first, we shall be concerned about the measurement of the boundaries of a plane geometrical figure called polygon. Perimeter: The total length of the bounding line segments or curved line of a plane figure is called its perimeter. Perimeter of Plane Geometric Figures Look the following pictures and think about their boundaries: Can youmeasure the boundary sides of all the above pictures using scale orruler ortape? Note the results in your copy separately. Find the sum of boundaries in each picture as Sum of boundaries of set square = .......... + .......... + .......... = ............ Sum of boundaries of Math book = .......... + .......... + .......... + .......... = ............ Sum of boundaries of wooden table = .......... + .......... + .......... + .......... = ............ Sum of boundaries of black board = .......... + .......... + .......... + .......... = ............ Sum of boundaries of door = .......... + .......... + .......... + .......... = ............ The sum so obtained is called the perimeter of respective pictures. By the same process, you can find the perimeter of the classroom, school’ garden, bedroom, field, etc.
154 The Leading Mathematics - 7 Perimeter is the total length of the sides of a closed plane figure. There are different types of triangular surfaces. They have own special characteristics. We have mention the perimeter of the different types of triangular surface according as their lengths of sides or edges in below: Name Figure Sides Perimeter (P) Scalene Triangle a b c a b c P = a + b + c Isosceles Triangle a b a b P = 2a + b Equilateral triangle a a P = 3a Right-angled Triangle b h p p b h P = p + b + h Right-angled Isosceles Triangle b or p h p or b p or b h P = 2p + h = 2b + h CLASSWORK EXAMPLES Example 1 Find the perimeter of a triangular base of the given prism. Solution: Here, Length of sides of the base of prism, a = 7 cm, b = 10 cm, c = 8 cm Perimeter of rectangle (P) = ? Now, we have, P = a + b + c = 7 + 10 + 8 = 25 cm 7 cm 15 cm 10 cm 8 cm
MENSURATION 155 Example 2 Find the total outer edges of the given equilateral triangular bread. Solution: Given, the length of the edge of the given equilateral triangular bread (a) = 9 cm Now, we know that The total outer edges of the bread (P) = 3a = 3 × 9 = 27 cm Example 3 There is triangular pond with edges 13 ft, 15 ft and 20 ft in the ground. A man wants to round on the edges of the pond 5 times by wire to safe from falling babies. (a) Find the perimeter of the pond. (b) How long wire does he need ? (c) Find the cost of wire at Rs. 25/ft. Solution: Given, Length of edges of the triangular pond, a = 13 ft, b = 15 ft, c = 20 ft Total length of wire (T) = ? Now, we have, (a) Length of wire around the pond once (P) = a + b + c = 13 + 15 + 20 = 48 ft (b) Total length of wire around 5 times of the pond (T) = 5P = 5 × 48 = 240 ft (c) Cost of wire at Rs. 25/ft = Rs. 25 × 240 = Rs. 6000. EXERCISE 9.1 Your mastery depends on practice. Practice like you play. 1. Observe the following triangular objects and find the outer perimeter of the following triangular surfaces. (a) 15 cm 9 cm 12 cm (b) 11.3 cm 8 cm 8 cm (c) 10 cm 10 cm 6 cm 9 cm
156 The Leading Mathematics - 7 (d) 11 cm 12 cm 9 cm (e) 7 cm 7 cm 7 cm (f) 14 cm 16 cm 18 cm 2. Find the perimeter of the following triangles: (a) 8 cm 12 cm 11 cm (b) 10 cm 13 cm (c) 14 ft (d) 12 cm 5 cm 13 cm (e) 15.55 cm 11 cm (f) 11.3 cm 9.4 cm 3. (a) Samir makes a triangular garden with edges 25 ft, 34 ft and 38 ft. How long wall does he make around the garden ? Find it. (b) How long thread would be required to make an isosceles triangle having sides 13 cm, 15 cm and 17 cm? (c) Find the length of the railing around the equilateral triangular pond with an edge 11 m. 4. Find the missing sides in the following triangles: (a) 7 cm ? 10 cm (b) ? 14 cm (c) ? Perimeter (P) = 30 cm Perimeter (P) = 39 cm Perimeter (P) = 33 ft (d) 8 cm 6 cm ? (e) (f) ? 9.8 cm Perimeter (P) = 24 cm Perimeter (P) = 34.14 cm Perimeter (P) = 37 cm 14.14 cm ? ?
MENSURATION 157 5. (a) Binod bought a triangular land with boundaries 25 m, 22 m and 28 m. He wants to make a wall around it. i. How long wall would he made ? ii. Find the cost to make it at Rs. 650/m. (b) Sanam makes an isosceles triangular pond with equal edges 30 ft and the third edge 25 ft. She wants to make railing around it. i. How long wall would he need ? ii. Find the cost of railing at Rs. 72.25/ft. 6. (a) There is triangular field with boundaries 17 m, 23 m and 25 m. Its land owner wants to round on its boundaries 6 times by wire to safe from animals. i. Find the total boundaries of the field. ii. How long wire does he need ? iii. Find the cost of wire at Rs. 28.50/ft. (b) There is an equilateral triangular island on the road with an edge 18 ft. A traffic arranges 4 round steel pipes on its edge to safe from vehicles. i. Find its perimeter. ii. How long pipe does he need ? iii. Find the cost of pipe at Rs. 150.75/ft. ANSWERS 1. (a) 36 cm (b) 27.3 cm (c) 26 cm (d) 32 cm (e) 49 cm (f) 48 cm 2. (a) 31 cm (b) 36 cm (c) 42 cm (d) 30cm (e) 37.55 cm (f) 32 cm 3. (a) 97 ft (b) 45 cm (c) 33 m 4. (a) 13 cm (b) 11 cm (c) 11 ft (d) 10 cm (e) 10 cm (f) 13.6 cm 5. (a) i. 13 cm ii. Rs. 48750 (b) i. 85 ft ii. Rs. 6141.25 6. (a) i. 65 m ii. 390 m iii. Rs.11115 (b) i. 54 ft ii. 216 ft iii. Rs. 32562 Project Work 9.1 Measure the length of all the edges of any ten triangular objects. Find their perimeters in tables and present it in the classroom.
158 The Leading Mathematics - 7 9.2 Surface Area of Cuboid and Cube At the end of this topic, the students will be able to: ¾ find the surface area of cuboid and cube. ¾ solve the problems related to surface area of cuboid and cube. Learning Objectives Area of Rectangle Area is measured in square units such as square inches, square feet, square centimeter or square meters. To find the area of a rectangle, multiply the length by the width. i.e., Area of rectangle (A) = length × breadth = l × b sq. units. Area of Square A square is an equilateral and equiangular quadrilateral. So, it has equal length and breadth. To find the area of a square, as follow. i.e., Area of square (A) = length × length = l × l = l2 sq. units. Surface Area of Cuboid Take a cuboid with length l, breadth b and height h. Open the box as shown in the figure: This show that the outer surface of a cuboid is made up of six rectangles but in fact they are the faces of cuboid. So, Surface area of cuboid = Area of all the six rectangles = lb + lb + bh + bh + lh + lh = 2lb + 2bh + 2lh = 2(lb + bh + lh) sq. units. ∴ Total Surface Area of Cuboid (TSA) = 2(lb + bh + lh) sq. units. Length (l) Breadth (b) Length (l) Cuboid h b l Net of Cuboid b b l l l l l l h h h h h b lb b bh lh bh lh lb b b b b
MENSURATION 159 Surface Area of Cube Take a cube with length l. Open the box as shown in the figure: l l l l2 l2 l2 l2 l2 l2 l l l l l l Cuboid Net of Cuboid This shows that the outer surface of a cube is made up of six squares but in fact they are the faces of cube. So, Surface area of cube = Area of all the six squares = 6 × Area of squares = 6 l2 sq. units. ∴ Total Surface Area of Cube (TSA) = 6 l2 sq. units. CLASSWORK EXAMPLES Example 1 (a) Write the formula to find the base area of the cuboid. (b) Find the base area of the solid. (c) Find the total surface area of the given rectangular solid. Solution: (a) Base area of cuboid (A) =l × b (b) In the given rectangular solid, Length (l) = 9 cm, Breadth (b) = 3 cm, Height (h) = 4 cm, Base area (A) = ? Now, we know that Base area (A) = l × b = 9 cm × 3 cm = 27 cm2 4 cm 9 cm 3 cm
160 The Leading Mathematics - 7 (c) TSA of solid = 2(lb + bh + lh) = 2 ( 9 × 3 + 3 × 4 + 9 × 4) = 2(27 + 12 + 36) = 2 × 75 = 150 cm2 = 316 cm2 . Example 2 If the surface area of a cube is 864 m2 , find the length of the cube. Solution: Here, in the given cube, surface area = 864 m2 or, 6l2 = 864 or, l2 = 864 6 or, l2 = 144 or, l2 = 122 ∴ l = 12 cm [l is never negative. Why?] Hence, the length of the cube is 12 cm. Example 3 The surface area of a cuboid with length 14 cm and 10 cm is 664 cm2 . (a) Find its height. (b) Find the lateral surface are of the cuboid. Solution: Here, in the given cuboid Surface area = 664 m2 Length (l) = 14 cm, Breadth (b) = 10 cm, Height (h) = ? Now, we know that (a) TSA of solid = 2(lb + bh + lh) or, 664 = 2(14 × 10 + 10h + 14h) or, 664 2 = 140 + 24h or, 332 – 140 = 24h or, 192 24 = h ∴ h = 8 cm Hence, the height of the cuboid is 8 cm.
MENSURATION 161 (b) Lateral Surface Area (LSA) = 2h(l + b) = 2 × 8 (14 + 10) = 16 × 24 = 384 cm2 Hence, lateral surface area of cuboid is 384 cm2 . EXERCISE 9.2 Your mastery depends on practice. Practice like you play. 1. Observe the following objects and find the total surface area of the following rectangular solids: (a) 6 cm 3 cm 2 cm (b) 6 cm 2 cm 4 cm (c) 2. Find the total surface area of the following cubes: (a) 5 inch (b) 3 cm (c) 5.2 cm 3. Find the surface area of the cuboids with the following length, breadth and height: (a) 15 cm, 12 cm and 11 cm. (b) 18 cm, 21 cm and 15 cm. 4. Calculate the surface area of the cube with the following length: (a) 14 m (b) 17 cm. 5. (a) If the surface area of a cuboid is 636 cm2 and its length and breadth are 12 cm and 9 cm respectively, find its height. (b) If the surface area of a cuboid is 1272 cm2 and its length and breadth are 18 cm and 14 cm respectively, find its height. 11 cm 7 cm 3 cm
162 The Leading Mathematics - 7 6. (a) If the area of a cuboid is 760.80 cm2 and its length and height are 12 cm and 7 cm respectively, find its breadth. (b) If the area of a cuboid is 888 cm2 and its length and height are 16 cm and 12 cm respectively, find its breadth. 7. (a) If the area of a cuboid is 537 cm2 and its breadth and height are 8 cm and 11 cm respectively, find its length. (b) If the area of a cuboid is 2106 cm2 and its breadth and height are 16 cm and 18 cm respectively, find its length. 8. (a) If the area of a cube is 294 cm2 , find its length. (b) If the area of a cube is 433.5 cm2 , find its length. ANSWERS 1. (a) 72 cm2 (b) 88 cm2 (c) 262 cm2 2. (a) 150 inch2 (b) 54 cm2 (c) 162.24 cm2 3. (a) 954 cm2 (b) 1926 cm2 4. (a) 1176 m2 (b) 1734 cm2 5. (a) 10 cm (b) 12 cm 6. (a) 15.6 cm (b) 9 cm 7. (a) 9.5 cm (b) 22.5 cm 8. (a) 7 cm (b) 8.5 cm Project Work 9.2 Measure the length, breadth and thickness of your geometric box, book and notebook. Find the total surface area of the box and present it in your classroom.
MENSURATION 163 9.3 Volume of Cuboid and Cube At the end of this topic, the students will be able to: ¾ find the volume of cuboid and cube. ¾ solve the problems related to volume of cuboid and cube. Learning Objectives How much quantity of water can hold given tiffin box ? For this, we discuss unit cube. Volume of Cuboid Find the volume of the following cuboids by counting unit cubes. (a) (b) (c) What are the volumes of the above cuboid? If you multiply the length, breadth and height of each of the cuboid in the above figure, what would you get? Did you get the same volume as the one you obtained by counting units? Hence we have the formula, Volume of a cuboid (V) = l × b × h cu. units = A × h cu. units [∴ A = l × b] Volume of Cube As the cube is a cuboid of equal length of side, we have Volume of cube (V) = l × b × h [ l = b = h = a] or, V = a × a × a ∴ V = a3 cu. units 1 1 1 l h b l h b
164 The Leading Mathematics - 7 CLASSWORK EXAMPLES Example 1 Find the volume of a cuboid having the given dimensions. Solution: The given cuboid has length (l) = 8 cm, breadth (b) = 5 cm and height (h) = 10 cm. Now, the volume of the cuboid, V = l × b × h = 8 cm × 5 cm × 10 cm = 400 cm3 . Example 2 How many dice of 1 cu. cm volume can be kept in this box in the space of cuboid? Solution: Given, Length of the cuboid (l) = 10 cm, Breadth of the cuboid (b) = 8 cm, Height of the cuboid (h) = 8 cm Now, volume (V) = l × b × h = 10 cm × 8 cm × 8 cm = 640 cm3 Here, Volume of die (V) = 1 cm3 ∴ N = V u = 640 cm3 1 cm3 = 640 Hence, the box holds 640 dice of unit volume in it. EXERCISE 9.3 Your mastery depends on practice. Practice like you play. 1. Find the volume of the following objects in the shape of cuboids: (a) (b) (c) (d) 2. Find the volume of the cuboids if; (a) l = 15 cm, b = 12 cm, h = 8 cm (b) l = 12 cm, b = 6.2 cm, h = 10 cm 3. Find the volume of the cubes having length of side as; (a) 6 cm (b) 8.2 cm (c) 9.1 cm 4. Find the length of side of the cube having the following volumes: (a) 343 cm3 (b) 2197 cm3 (c) 681.472 cm3 10 cm 8 cm 5 cm 8 cm 10 cm 8 cm 8 cm 4 cm 6 cm 10 cm 6 cm 4 cm 10 cm 5 cm 7 cm 4 cm 4 cm 9 cm
MENSURATION 165 5. (a) How much water is there in an underground water tank with dimensions 3 m × 2.5 m × 1.5 m? [Use 1 m3 = 1000l] (b) A rectangular water tank has the length 8 ft, breadth 6 ft. and height 5 ft. Find the capacity of the tank. [Use 1m = 3.28 ft] 6. Find the height of the following objects in the shape of a cuboid: (a) 12 cm 9 cm V = 648 cm3 ? (b) 5 cm 9 cm V = 540cm3 ? (c) 18 cm 7 cm ? V =1260 cm3 7. (a) Find the length of a rectangular box if its volume, breadth and height are 3240 cm3 , 12 cm and 15 cm respectively. (b) Find the breadth of a cuboid if its volume, length and height are 480cm3 , 10cm and 6 cm respectively. (c) Find the height of a rectangular pillar if its volume, length and height are 180000 cm3 , 30 cm and 20 cm respectively. 8. (a) Find the volume of a rectangular container with height 9 ft and base area 150 ft2 . (b) Find the volume of a rectangular box with height 12 cm and base area 36 cm2 . 9. (a) The volumes of a cube with volume 1728 cm3 and a cuboid are the same. If the base area of the cuboid is 288 cm2 , find the height of the cuboid. (b) The volume of a cuboid is double of the cube with side 6 cm. If the height and length of the cuboid are 9 cm and 12 cm, find the breadth of the cuboid. ANSWERS 1. (a) 192 cm3 (b) 240 cm3 (c) 350 cm3 (d) 144 cm3 2. (a) 1440 cm2 (b) 744 cm3 3. (a) 216 cm2 (b) 551.37 cm3 (c) 753.57 cm3 4. (a) 7 cm (b) 13 cm (c) 8.8 cm 5. (a) 11250 l (b) 6801.265l 6. (a) 6 cm (b) 12 cm (c) 10 cm 7. (a) 18 cm (b) 8 cm (c) 300 cm 8. (a) 1350 ft3 (b) 432 cm3 9. (a) 6 cm (b) 4 cm Project Work 9.3 Measure the length, breadth and thickness of cubical tin found you home or neighbour. How much water can they contains? Present this in your classroom.
166 The Leading Mathematics - 7 9.4 Circumference of Circle At the end of this topic, the students will be able to: ¾ establish the relations of the circumference and diameter of a circle. Learning Objectives What are the length of diameter and circumference of the lid of the water bottle ? Length of circumference (c) = 13.19 cm Length of diameter (d) = 4.2 cm Now, what is the ratio of the circumference (c) and diameter (d) of the lid ? Ratio of c and d, c:d = 13.19 cm 4.2 cm = 3.14 Again, we can find the ratio of the circumference (c) and diameter (d) of the circular bangle ? What would be the value of the ratio of c and d ? It is the same as 3.14. Similarly, we can try other circular surfaces. In all cases we find the value of the ratio of the circumference (c) and diameter (d) of a circle is always 3.14, which is constant. This constant number 3.14 is denoted by a mathematical symbol π, read as Pi. The value of π is not rational number which is not terminating and repeating decimal number. So, π is an irrational number, called transcendental number. But the value of π is approximately equal to 22 7 , not exact. i.e. π = 3.141592654 ......... ≈ 22 7
MENSURATION 167 Relation among Circumference (c) and Diameter (d) of a Circle The circle is the locus of a moving point from a fixed point with constant distance in which the fixed point is called centre and the constant distance is called radius of the circle. The chord that passes through the centre is called diameter of the circle. The length of the diameter is twice the length of the radius of the circle. The roundabout of the circle is the circumference of the circle. In another word, the total length roundabout the circle is its perimeter or circumference which is obtained as, c = 2πr or, c = πd. Why ? Where, r = radius d = diameter and, π = 22 7 or, π = 3.14 Hence, the ratio of circumference and diameter of a circle is always constant that is called π (pi). π is irrational number whose value is 3.141592654 ...... and approximately equal to 22 7 , not exact. Then, c = πd and d = c π Also, we know that the length of the diameter is twice of the radius of a circle. So, c = 2πr and r = c 2π CLASSWORK EXAMPLES Example 1 Find the circumference of a circle of radius 5.6 cm . Solution: Given, radius (r) = 5.6 cm ∴ Circumference (c) = 2πr = 2 × 22 7 × 5.6 = 35.2 cm2 c d d d 1 7 of d c ≈ d + d + d + 1 7 × d = 3d + 1 7 × d = 22 7 d. ∴ c d = π ≈ 22 7 (suppose) d c r r
168 The Leading Mathematics - 7 Example 2 A circular plate rolls on the ground for 15 revolutions about the circumference. If the diameter of the plate is 14 cm, find the distance covered by it. Solution: Diameter of the plate (d) = 14 cm ∴ Circumference of the plate (c) = πd = 22 7 × 14 = 44 cm ∴ Distance traveled in 15 revolutions = 15 × 44 cm = 660 cm = 6.6 m Example 3 Find the radius of the circle if its circumference is 44 cm . Solution: Circumference (c) = 44 cm 2 or, 2πr = 44 or, 2 × 22 7 × r = 44 cm or, r = 44 × 7 2 × 22 cm or, r = 7 cm ∴ The radius of the circle is 7 cm. EXERCISE 9.4 Your mastery depends on practice. Practice like you play. 1. Find the circumference of the following circles with the given radius. Take π = 22 7 (a) 7 cm (b) 14 cm (c) 2.1 cm (d) 8.4 cm 2. Find the circumference of the following circles with the given diameter. π = 22 7 (a) 14 cm (b) 2.8 cm (c) 5.6 cm (d) 4.2 cm 3. Find the circumference of the circles having the following information : (π = 3.14) (a) radius = 14 cm (b) diameter = 2.8 cm (c) radius = 2.1 cm (d) diameter = 2.1 cm 4. Find the radius of the circles having the given circumference. π = 22 7 (a) 44 cm (b) 176 cm (c) 35.20 cm (d) 39.60 cm 5. Find the diameter of the circles having the given circumference. π = 22 7 (a) 11 cm (b) 352 cm (c) 17.6 cm (d) 277.20 cm
MENSURATION 169 6. A circular plate has a diameter of 21 cm. Find; (a) Write the relation between the radius and diameter of a circle. (b) Find the radius of the plate (c) Find its circumference (d) Find the distance covered by it in 14 revolution about the circumference. 7. A one rupee coin has the diameter of 3.5 cm. Find; (a) its radius (b) its circumference (c) the distance covered by it is 50 revolution about the circumference. 8. (a) The circumference of the trunk of a tree is 396 cm. Find the diameter of the trunk. (b) The circumference of the trunk of a tree is 704 cm. Find the radius of the trunk. 9. (a) When the wheel of a chariot rotates 25 times on the road and covers the distance of 275 m, what is the length of the radius of the wheel? Find it. (b) Hira runs a circular pond 15 times and covers the distance of 1 km 320 m. What is the diameter of the pond? Find it. 10. (a) How many times does a wheel of the radius 70 cm rotate to cover the distance 440 m? (b) How many times does a wheel of the diameter 8.75 mm rotate to cover the distance 1 m 37 cm 5 mm? ANSWERS 1. (a) 44 cm (b) 88 cm (c) 13.2 cm (d) 52.8 cm 2. (a) 44 cm (b) 8.8 cm (c) 17. 6 cm (d) 13.2 cm 3. (a) 87.92 cm (b) 8.79 cm, (c) 13.19 cm (d) 6.59 cm 4. (a) 7 cm (b) 28 cm (c) 5.60 cm (d) 6.30 cm 5. (a) 3.5 cm (b) 112 cm (c) 5.60 cm (d) 88.20 cm 6. (b) 11.50 cm (c) 66 cm (d) 924 cm or 9.24 m 7. (a) 1.75 cm (b) 11 cm (c) 550 cm or 5.5 m 8. (a) 126 cm (b) 112 cm 9. (a) 1.75 m (b) 28 m 10. (a) 100 times (b) 50 times Project Work 9.4 Find the length of the circumferences of two circular objects found at your home by measuring its diameter. Also, find the ratios of circumference and diameter of both of the objects and establish the relation between the circumference and diameter of a circle. Finally, present the result in your classroom.
170 The Leading Mathematics - 7 Read, Understand, Think and Do 1. There is a triangular pond with edges 13 m, 15 m and 20 m in the middle part of the jungle. A man wants to round on the edges 5 times of the pond by wire to safe from falling babies. (a) Write the formula for finding the perimeter of the triangle. (b) Find the perimeter of the triangular pond. (c) How long wire does he need? (d) Find the cost of wire at Rs. 25/m. 2. Three cubes as shown in the figure below, each of volume 64 cm3 are joined end to end to form a cuboid. (a) Write the formula for finding the total surface area of the cuboid. (b) Find the length of the side of a cube. (c) Prove that the total surface area of cuboid is 224 cm2 . (d) Is the volume of three cubes equal to the volume of cuboid? Justify it. 3. A man is constructing a wall 20 m long, 2 m high and 0.75 m thick by using a brick having 25 cm long, 10 cm wide and 8 cm thick. (a) Write the formula for finding the surface area of a cuboid. (b) How many bricks will be needed to construct the wall? (Assuming no space between two bricks) (c) If bricks are sold at Rs. 2000 per hundreds, how much will it cost to build the wall? (d) Is the total surface area of bricks 1060 cm2 ? Calculate it. a a a MIXED PRACTICE–III
MENSURATION 171 4. Two cubical boxes having lengths 1m and 10 cm are given alongside. (a) Write the formula for finding the total surface area of a cube. (b) How many cubes of 10 cm can be put in a cubical box of 1 m edge? (c) Find the total surface area of the small cubical box. (d) How many times more is the total surface area of large cubical box than the small cubical box? Calculate it. 5. A small cube with side 6 cm can be placed in the cuboid of dimensions 60 cm × 54 cm × 30 cm. (a) Write the formula of the total surface area of cuboid (b) Find the volume of a small cube. (c) How many cubes can be put in a cuboidal object? Find it. (d) Are the total surface area of one cube and its volume equal? 6. A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 18 cm, 14 cm and 9 cm respectively. (a) How much cubic cm of a liquid can be placed in it? Find it. (b) Find the volume of the wood used in it. (c) Find the differences between total surface area of the outer part and inner part of the wood. 7. Hirakaji runs a circular pond 15 times and covers the distance of 1 km 320 m. (a) Write the relationship between the radius and diameter of the circle. (b) How many distance is covered by him, when he runs once at a time. (c) Find the radius of the circular pond. (d) When radius is doubled, what would happen its circumference? 1 10 cm 2 cm 18 cm 9 cm 14 cm
172 The Leading Mathematics - 7 8. In the given circular pond, a tree is standing at the center of the pond. The distance between the tree and the bank of the pond and its outer part are 7 m and 10.5 m respectively. (a) Write the formula for finding the circumference of the circle. (b) Find the circumference of the pond. (c) Find the outer circumference of the pond having radius 10.5 m. (d) The value of π is 3.14. How can you calculate from the circle? Justify it. ANSWERS 1. (b) 48 m (c) 240 m (d) Rs. 6000 2. (a) 4 cm 3. (b) 15000 pcs. (c) Rs. 300000 (d) Yes 4. (b) 1000 pcs. (c) 600 cm2 (d) 100 times 5. (b) 216 cm2 (b) 450 pcs. (c) Yes 6. (a) 700 l (b) 1568 cm3 (c) 560 cm2 7. (b) 88 m (c) 14 m (d) 176 m (double) 8. (b) 44 m (c) 66 m (d) c d
MENSURATION 173 1. Observe the adjoining triangular garden and answer the following questions. (a) Write the formula to find the perimeter of triangle. [1] (b) Find the perimeter of the given triangular garden. [1] (c) How much wood does the house owner need to bound around it if the width and height of the wooden flak are 2 inches and 12 inches respectively. [2] (d) Calculate the cost of the wood at Rs. 4500 per cu. feet. [1] 2. Two cubes overlapped one another to form a cuboid as shown in the adjoining figure having volume 729 cm3 are joined end to end. (a) Write the formula for finding the total surface area of cuboid. [1] (b) Find the length of the side of a cube. [1] (c) Find the total surface area of a cube. [1] (d) Is the total surface area of one cube half of the total surface area of the given cuboid ? Justify it. [2] 3. A worker is constructing a 10 m long, 1.5 m high, and 0.5 m thick wall by using bricks of the size 25 cm long, 10 cm wide, and 7.5 cm thick. (a) What is the formula for finding the volume of a brick? [1] (b) How many steps of bricks from thickness are contained at the height of the wall? [1] (c) How many bricks will be needed to construct the wall? (Assuming no space between two bricks) [2] (d) If the cost of a brick is Rs. 19, what will it cost to build the wall ? Find it. [1] 4. The given picture is the stadium with semi-circles on both breadths of the rectangular football playing ground. The length and breadth of the football playing ground are 100 m and 75 m respectively. (a) Write the formula to find the circumference of a circle.[1] (b) Find the radius of the semi-circles. [1] (c) Find the perimeter of the ground with semi-circles. [2] (d) Calculate the cost of constructing railing at Rs. 555 per meter around it to safety for the audiences [1] a a a a Attempt all the questions. 10 ft. 12 ft. 15 ft. FM : 20 Time : 40 Min. CONFIDENCE LEVEL TEST III MENSURATION
174 The Leading Mathematics - 7 COMPETENCY Development of skill for solving the simple problems related to algebraic expression and equation. CHAPTERS 10. Indices 11. Algebraic Expressions 12. Equation, Inequality and Graph LEARNING OUTCOMES simplify the indices. multiply and divide the algebraic expressions. establish and use the formulae of (a ± b)2 . show the equation on the graph. solve the inequality and show it on the number line. ALGEBRA UNIT IV
Algebra 175 CHAPTER 10 Indices Lesson Topics Pages 10.1 Introduction to Indices 176 What are the squares of 2, 3, 5, ...... ? What is the cube of 2, 3, 5, ...... ? Can you add or subtract the numbers? Discuss with examples. Can you multiply the different numbers? Discuss with examples. Can you multiply the same numbers? Discuss with examples. Can you multiply the same alphabets? Discuss with examples. Can you simplify the numbers on operations? Discuss with examples. Can you find the value of the expression when the values of variables are given? Discuss with examples. WARM-UP
176 The Leading Mathematics - 7 10.1 Introduction to Indices At the end of this topic, the students will be able to: ¾ simplify the expression by using laws of indices. Learning Objectives Indices (Index, in Singular Form) Suppose you need to multiply a number (or variable/ constant) by itself a number of times (say ten times). For example; (a) 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (b) x × x × x × x × x × x × x × x × x × x It takes a long time, as well as it follows much space. So mathematicians invent new notation which involves writing down the number (variable) being multiplicand together with the number of times the number is multiplied by itself. In this new notation the products is written as follows: 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 210 x × x × x × x × x × x × x × x × x × x = x10 In general, we write an to mean "a is multiplied by itself n times", which is called a power of 'a'. The power an is read as a to the power or"a raised to the nth power". The number 'a' which is generally called the factor or base. The number of repeated times 'n' the number multiples is called the index or exponent. The laws of exponents or Rules of combining indices Law I. For any number a and for any two positive integers m and n, am × an = a m + n Powers and roots of algebraic expression may be defined in the same way as the powers and roots of numbers, literal numbers and variables. They can be added or subtracted in the same way as in the case like terms. Their products can be found from the first principle or by direct multiplication. They can be handled nicely with the help of certain rules or laws known as the Laws of Indices or Exponents. Their proofs depend upon
Algebra 177 the associative and distributive laws of multiplication of numbers. Similar laws apply to algebraic expressions of various forms. For instance, 22 × 23 = 4 × 8 = 32 = 25 = 22 + 3 Alternatively, 22 × 23 = 2 × 2 × 2 × 2 × 2 = 25 = 2 2 + 3 Similarly, (– a)3 (– a)4 =(– a)3 + 4 = (– a)7 . Hence, am × an = am + n Law II. For any number a ≠ 0 and for any two positive integers m and n, am ÷ an = am – n if m > n = 1 if m = n = 1 an – m if n > m For instance, 25 ÷ 23 = 25 23 = 32 8 = 4 = 22 = 25 – 3 Alternatively, 25 ÷ 23 = 25 23 = 2 × 2 × 2 × 2 × 2 2 × 2 × 2 = 2 × 2 = 22 = 25 – 3 Similarly, a5 ÷ a2 = a5 a2 = a5 – 2 = a3 a2 ÷ a6 = a2 a6 = 1 a6 – 2 = 1 a4 Hence, am ÷ an = am – n, m > n; am ÷ am = am – m = a0 = 1, m = n and am ÷ an = an – m, m < n Law III. For any numbers a and b and for any two positive integers m and n, (ab)m = ambm; a b m = am bm, b ≠ 0; (am)n = amn For instance, (2 × 3)2 = 62 = 36 = 2 × 2 × 3 × 3 = 22 × 32
178 The Leading Mathematics - 7 Similarly, 4 5 3 = 43 53 (5a2 )3 = 53 (a2 )3 = 125 a2 × 3 = 125 a6 . Hence, (ab)m = ambm; a b m = am bm, b ≠ 0; (am)n = amn Now, we shall extend the above Laws of Indices to those cases where m and n may be negative integers or fractions. (i) We have taken the value of a0 to be equal to 1 or a0 = 1 for any non-zero number a. This is justified because a0 = an – n = an an = 1. (ii) We shall take a–mas the reciprocal of am for any non-zero number a. i.e, a–m = 1 am This is justified because am – m = a0 = 1. or, am.a–m = 1 or, a–m = 1 am as also am = 1 a–m . (iii) The nth root of a, i.e. a n is that quantity which multiplied with itself n times gives a. So, we agree to write a n as a 1 n because a 1 n.a 1 n.a 1 n ... n times = a 1 n n = an n = a1 = a Hence, we can define a quantity like am n by am n = a 1 n m. Furthermore, we have a m n = am 1 n = am n Thus, the four laws of indices stated above are valid when the indices are any rational numbers. i.e., positive or negative, integer or fraction. In addition to the above laws we agree to accept that "If am = an , then m = n, if a ≠ 0."
Algebra 179 CLASSWORK EXAMPLES Example 1 Write the given number in the exponential form: 625 Solution: 625 = 5 × 5 × 5 × 5 = 54 . Example 2 Rewrite the given number without using radical sign ( ): 243 5 Solution: 243 5 = 35 5 = 3. Example 3 Rewrite the following using the radical sign : (a) 2 1 2 (b) 8 – 1 3 (c) a 2 3 Solution: (a) 2 1 2 = 2 (b) 8 – 1 3 = 8-1 = 1 8 = 1 8 (c) a 2 3 = a2 3 Example 4 Simplify : (a) 8– 2 3 (b) 8 27 – 2 3 Solution: (a) 8 – 2 3 = 2 3 × – 2 3 = 2–2 = 1 22 = 1 4 (b) 8 27 – 2 3 = 23 33 – 2 3 = 2 3 3 × (– 2 3) = 2 3 – 2 = 2 – 2 3– 2 = 32 22 = 9 4 Example 5 Simplify each of the following by removing negative index and radical signs. (a) (x–2y3 z0 )–1 (b) x–2y5 × x–3y–2 (c) ( 27 3 ) × ( 81 4 ) (d) x4 Solution: (a) (x–2y3 z0 )–1 = x–2 × (–1) y3 × (–1) z0 × (–1) = x2 y–3z0 = x2 × 1 y3 × 1 = x2 y3 (b) x–2y5 × x–3y–2 = x–2 – 3 y5 – 2 = x–5 y3 = y3 x5 (c) ( 27 3 ) × ( 81 4 ) = 33 3 × 34 4 = 3 × 3 = 9 (d) x4 = (x2 )2 = x2 = x
180 The Leading Mathematics - 7 Example 6 Prove that: (a) x–1y y–1z z–1x = 1 (b) ab2 3 a2 b4 3 a–3b–6 3 = 1 (c) xa xb 2 xb xc 2 xc xa 2 = 1 Solution: (a) x–1y y–1z z–1x = x–1y × y–1z × z–1x = x–1 + 1 y1 – 1 z1 – 1 = x0 y0 z0 = 1 × 1 × 1 = 1 = 1 = RHS Proved. (b) ab2 3 a2 b4 3 a–3b–6 3 = ab2 × a2 b4 × a–3b–6 3 = a1 + 2 – 3 b2 + 4 – 6 3 = a0 b0 3 = 1 × 1 3 = 1 3 = 1 = RHS Proved. (c) xa xb 2 xb xc 2 xc xa 2 = x2a x2b × x2b x2c × x2c x2a = x2a + 2b + 2c x2b + 2c + 2a = x2a + 2b + 2c – 2b – 2c – 2a = x0 = 1 = R.H.S Proved.
Algebra 181 EXERCISE 10.1 Your mastery depends on practice. Practice like you play. 1. Write the following in the exponential form (i.e., by using index notation). (a) 3 × 3 × 3 × 3 (b) 81 (c) a × a × a × a 2. Write the following in the product form. (a) 8a3 (b) 64 (c) x2 y3 3. Rewrite the following without using radical sign. (a) 9 (b) 27 3 (c) 32 5 4. Rewrite the following using the radical sign. (a) 3 – 1 2 (b) a – 1 3 (c) b 2 3 5. Simplify: (a) 8 – 1 2 (b) 27– 1 3 (c) 8 27 – 2 3 6. Write simplified equivalent expressions using positive indices only. (a) x3 x–4 (b) a2 x2 y–2 b–2xy– 4 (c) 2–13–240 7. Simplify: (a) 22 .23 (b) a2 .a3 (c) a2x.a3 (d) (x + y)2 (x + y)3 (e) (2x + 3y)2m (2x + 3y)3n (f) 22 .23 .24 . (g) (3a)2 (h) (xy)3 (i) (ab)2x (j) 3 2 2 (k) a b 2 (l) x + y x – y 3 (m) (52 )3 (n) (a2 )3 (o) [(a + b)2 ]3 . 8. Simplify each of the following by removing negative index and radical signs. (a) (a–2)3 (b) (x–3)–4 (c) (x–2y3 z0 )–1 (d) x–2y5 x–3y–2 (e) 16 –3 2 (f) 2 –3 . 8 –3 (g) x4 y–2 (h) p4 (i) xy2 3 × x2 y 3 (j) x– 2 3 ÷ y– 1 2 6 (k) 4 –5 2 10 27 3 × 81 4
182 The Leading Mathematics - 7 9. Prove that: (a) a2 b–2 3 × b2 c–2 3 × c2 a–2 3 = 1 (b) ab2 × a3 b4 × a–4b–6 = 1 (c) xa + b xb + c 2 xb + c xc + a 2 xc + a xa + b 2 = 1 (d) xa xb a + b xb xc b + c xc xa c + a = 1 (e) (ax ÷ a–y)x – y × (ay ÷ a–z)y – z × (az ÷ a–x)z – x = 1 ANSWERS 1. (a) 34 (b) 34 (c) a4 2. (a) 2 × 2 × 2 × a × a × a (b) 6 × 6 × 6 × 6 (c) x × x × y × y × y 3. (a) 3 (b) 3 (c) 2 4. (a) 1 3 (b) 1 a 3 (c) b3 2 5. (a) 1 2 2 (b) 1 3 (c) 9 4 6. (a) 1 x (b) a2 b2 xy2 (c) 1 18 7. (a) 32 (b) a5 (c) a2x + 3 (d) (x + y)5 (e) (2x + 3y)2m + 3n (f) 512 (g) 9a2 (h) x3 y3 (i) a2xb2x (j) 9 4 (k) a2 b2 (l) (x + y)3 (x – y)3 (m) 15625 (n) a6 (o) (a + b)6 8. (a) 1 a6 (b) 1 x12 (c) x2 y3 (d) xy7 (e) 1 8 (f) 1 64 (g) x2 y (h) p (i) xy (j) y3 x4 (k) 9
Algebra 183 CHAPTER 11 Algebraic Expressions Lesson Topics Pages 11.1 Review on Algebraic Expression 184 11.2 Multiplication of Algebraic Expressions 190 11.3 Division of Algebraic Expressions 199 11.4 Establishment of (a ± b)2 205 How many mountains more than 8000 m are there in the world ? How many highest peaks are in the world ? What are the constant and variable ? What are the sums of a + a + a + a + a and a + b + a + b + a + b? What are the products of a × a × a × a a × a × a and a × b × a × b × a × b? What is the meaning of 4a ? What is the meaning of a4 ? What is the value of 2p if p = 3 ? WARM-UP
184 The Leading Mathematics - 7 11.1 Review on Algebraic Expression At the end of this topic, the students will be able to: ¾ define the algebraic expression and its basic terminology. Learning Objectives Numbers are used to quantify ideal or concrete objects. They are manipulated with the help of fundamental operations called four simple rules – Addition, Subtraction, Multiplication and Division denoted by the signs +, −, × and ÷ respectively. Arithmetic is mostly concerned with numbers and the four simple rules. Raising to a power and extraction of roots of a number also come under arithmetic. In school Arithmetic, we commonly use Hindu-Arabic numerals. We may use letters such as a, b, …, x, y, z also to denote numbers written in the Hindu-Arabic numerals. Such numbers are called literal numbers. Some of these numbers are fixed or known or constant; while others are not fixed or unknown or variable. In algebra, we put our expressions in terms of numbers, symbols and signs. A combination of variables, numbers and operations is known as a mathematical or algebraic expression. Algebra deals with algebraic expressions and relations between such expressions. Notation and Terminology Numbers and literal numbers are the building blocks of algebra. They may be constants or variables. A single number or constant or variable is the simplest of all the algebraic expressions. (a) Term A term is a single number or literal number or the product or quotient of two numbers of any kind. For instance, 3, c, x, 3x, –3ax and a/x are some simplest types of terms. A term is actually a monomial ("monos" means "alone", "nomen" means "name"). A term may begin with a positive or negative sign. If there is no sign, it is taken to be positive.
Algebra 185 (b) Like and unlike terms The terms of an algebraic expression are said to be like if they differ only in their numerical coefficients, otherwise, they are said to be unlike. For example, we have Like Terms (monomials) Unlike terms (monomials) 2a and –7a 2a and 2b 2a2 and –7a2 2a2 and 2a 2ax2 and –7ax2 2xy and 2xy2 (c) Degree When two or more literal numbers are multiplied together, we get their product. In a product of literal numbers, each literal number is called a factor of the product. The degree of a literal factor (non-numerical factor) in a term is the number of times it occurs as factors in the term. Thus, the term abc has three literal factors a, b, c, each of degree one. But, the term as a whole is of degree three. It is the sum total of the degrees of all literal factors. Note that the numerical factor is not considered while considering the degree of a term. For instance, each of the terms 4x3 and 2a2 b has the same degree. It is 3. (d) Algebraic expression Any combination of variables, numbers and operations using signs and symbols is called an algebraic expression. For example: a, 2a, a + b, 3a – 4b, x2 + 2xy + y2 , (x – y)(x + y), (x – y)2 , ax by and ax2 + bx – c a'y2 – b'y + c'. An algebraic expression may have one or more terms combined together or separated by + and/or – sign. An expression having i. two terms is called a binomial (bi = two, nomen = names), e.g., 1 + a, a + b, ax – by, etc. ii. three terms is called a trinomial (tri = three, nomen = names), e.g., 1 + a + b, a + b + c, ax – by + cz, etc. iii. more than one term is called a multinomial (bi = two, nomen = names), e.g., a + b + c + d, a x3 + b x2 + c x1 + d etc. (e) Polynomials A polynomial in a variable x is an expression each of whose term is a constant or a power of the variable or a product of constant and a power of the variable but without any term
186 The Leading Mathematics - 7 containing the variable in the denominator. The degree of the polynomial is the same as the degree of the term containing the highest power of x. A workable definition of a polynomial in a single variable is as follows: An expression of the form anxn + an − 1 x n − 1 + an − 2 x n − 2 + ... + a2 x2 + a1 x1 + a0, (an ≠ 0) where a0, a1, ..., an−2 , an−1, an are constants, x is a variable and n is a non-negative integer, is called a polynomial of degree n in the variable x . We denote such a polynomial by Pn(x). In particular, i. If n = 0, the above polynomial Pn(x) becomes P0(x) = a0. For simplicity, the constant a0 may be denoted by c, so that P0(x) = c. In such a case, the polynomial P0(x) = c is called a constant polynomial or a polynomial of zero degree. ii. If n = 1, the above polynomial Pn(x) becomes P1(x) = a1 x1 + a0. For simplicity, the constant a1 may be denoted by a and the constant a0 by b, so that P1(x) = a x1 + b. In such a case, the polynomial P1(x) = ax1 + b = ax + b (a≠ 0) is called a polynomial of degree 1 in the variable x. Quite often, such a polynomial is known as a linear polynomial in the variable x. Note that x and 3x are also linear in the variable x. iii. If n = 2, the above polynomial Pn(x) becomes P2(x) = a2 x2 + a1 x1 + a0, (a2 ≠ 0). For simplicity, the constant a2 may be denoted by a, the constant a1 by b and the constant a0 by c, so that P2(x) = a x2 + b x1 + c, (a ≠ 0). In such a case, the polynomial P2(x) = a x2 + b x1 + c = a x2 + b x + c (a≠ 0) is called a polynomial of degree 2 in the variable x. Quite often, such a polynomial is known as a quadratic polynomial in the variable x. Note that x2 , 3x2 + 2 and 4x2 + 3x are also quadratic in the variable x. iv. If n = 3, the above polynomial Pn(x) becomes P3(x) = a3 x3 + a2 x2 + a1 x1 + a0, (a3 ≠ 0) For simplicity, the constant a3 may be denoted by a, a2 by b, a1 by c and a0 by d, so that P3(x) = a x3 + b x2 + c x1 + d, (a ≠ 0).
Algebra 187 In such a case, the polynomial P2(x) = a x3 + b x2 + c x1 + d, (a ≠ 0) is called a polynomial of degree 3 in the variable x. Quite often, such a polynomial is known as a cubic polynomial in the variable x. Note that x3 , 3x3 + 2, 4x3 + 3x, 4x3 + 3x2 , 4x3 + 3x2 + x, 4x3 + 3x2 + 2 and 4x3 + 3x + 2 are all cubic in the variable x. Polynomials may be in more than one variable. Below we define some polynomials in two variables x and y. i. If a polynomial is of degree 1 in each of the variables x and y, and there is no term containing the product of the variables, it is called a linear polynomial in the two variables x and y. Some such polynomials are: x + y, ax + y, x + by and ax + by. ii. If a polynomial has no term of degree greater 2 in the variables x and y, it is called a quadratic polynomial in the two variables x and y. Some such polynomials are: x2 + y2 , ax2 – y, x + by2 and ax2 + 2hxy + by2 A polynomial in one or more variables may be zero for certain value or values of the variables. Such a value or values of the variables are called the zeros of the polynomial. For instance, i. The linear polynomial P(x) = x is 0 when x = 0; ii. The linear polynomial P(x) = x – 1 is 0 when x = 1; iii. The quadratic polynomial P(x) = x2 is 0 when x = 0; iv. The quadratic polynomial P(x) = x2 – 1 is 0 when x = 1 and x = –1. v. The quadratic polynomial P(x, y) = x2 – y2 is 0 for all values of x and y such that x = y. CLASSWORK EXAMPLES Example 1 State the degree of each of the polynomials in x and y : (a) x3 + x2 − x + 1 (b) 1 + 2y + 3y4 (c) ax2 + 2hxy + by2 (d) xy3 + x3 y Solution: (a) 3 (b) 4 (c) 2 (d) 4
188 The Leading Mathematics - 7 Example 2 Find the values of the following polynomials for the given values of the variables. State which of the values are the zeros of the given polynomials: (a) x + 3 at x = 1, −3 (b) x2 − 3x + 2 at x = 1, 2, 1 (c) x2 − 2xy + y2 at x = 1, y = 1 and x = 1, y = −1 Solution: (a) At x = 1, x + 3 = 1 + 3 = 4 At x = −3, x + 3 = −3 + 3 = 0. So, x = −3 is a zero of x + 3. (b) At x = 1, x2 − 3x + 2 = 1 − 3 + 2 = 0. So, x = 1 is a zero of x2 − 3x + 2. At x = − 1, x2 − 3x + 2 = 1 + 3 + 2 = 6 So, x = − 1 is not a zero of x2 − 3x + 2. At x = 2, x2 − 3x + 2 = 4 − 6 + 2 = 0. So, x = 2 is a zero of x2 − 3x + 2. (c) At x = 1, y = 1, x2 − 2xy + y2 = 1− 2 + 1 = 0. So, the pairs of values x = 1 and y = 1 form a zero pair of x2 − 3x + 2. At x = 1, y = −1, x2 − 2xy + y2 = 1 + 2 + 1 = 4. So, the pairs of values x = 1 and y = −1 do not form a zero pair of x2 − 3x + 2. EXERCISE 11.1 Your mastery depends on practice. Practice like you play. 1. Tick mark the correct answer: (a) a in a3 i. index ii. exponent iii. power iv. base (b) a2 + 2ab + b2 is a i. monomial ii. binomial iii. Trinomial iv. polynomial
Algebra 189 (c) ax2 + bx − ca'x3 − b'x + c' is a i. rational fraction ii. linear polynomial in x iii. polynomial in x iv. quadratic polynomial in x 2. State the degree of each of the polynomials in the variables x and y: (a) x4 − 4x3 + 6x2 − 4x +1 (b) 1 + 2y + 3y4 (c) ax2 + 2hxy + by2 + 2gx + 2fy + c (d) x2 y2 + x2 y (e) x2 yx2 + 3x2 y2 + 2xyz (f) x2 y + 2x2 yz3 + 3x2 z 3. Find the values of the following polynomials for the given values of the variables. State which of the values are the zeros of the given polynomials: (a) x − 3 at x = 1, 3 (b) x2 + 3x + 2 at x = 1, 2, −1 (c) x2 − 2xy + y2 at x = 2, y = 2 and x = 2, y = −2. (d) x3 – 2x2 + y at x = 1, y = 1 and x = – 1, y = – 2 (e) xy2 – 2xy + y2 at x = 2, y = 1 and x = 0 and y = 0 ANSWERS 2. (a) 4 (b) 4 (c) 2 (d) 4 (e) 5 (f) 6 3. (a) –2, 0 (b) 6, 12, 0 (c) 0, 16 (d) 0, – 5 (e) 0, – 1
190 The Leading Mathematics - 7 11.2 Multiplication of Algebraic Expressions At the end of this topic, the students will be able to: ¾ multiply the algebraic expressions. Learning Objectives Review on Addition and Subtraction of Algebraic expressions Now, we shall see how the four simple rules together with the laws of indices can be applied in simplifying various complicated expressions and solve given problems. We denote the variables by the letters x, y and z, unless otherwise stated. Working Rules for Addition and Subtraction I. Addition of expressions containing terms having the same or different literal factors, i.e., like and unlike factors Rule: Add the corresponding numerical factors including signs, i.e. find the algebraic sum of the corresponding coefficients. II. Subtraction of expressions containing terms having the same or different literal factors, i.e., like and unlike factors Rule: Change the sign of every term of the subtrahend (first expression) and then find the algebraic sum of the corresponding coefficients. Multiplication of Algebraic expressions I. Multiplication by a monomial Rule: Consider one term of the multiplicand at a time. Multiply numerical coefficients and literal coefficients separately and then find their product. To multiply the powers of the same base, keep the base as it is and add the exponents. Take a = a1 , b = b1 and so on. Basic idea behind the multiplication by a monomial depends upon the distributive properties of numbers, viz. a (b + c) = ab + ac. a a
Algebra 191 II. Multiplication by a binomial Rule: Multiply the multiplicand by each term of the multiplier as in Rule I above; and then find the sum. In particular, if A is an expression to be multiplied by the binomial a + b, we apply A (a + b ) = A a + A b and (a + b ) A = a A + b A. III. Multiplication of binomial expressions Category A: Multiplication of two binomial expressions 1. (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2 . [ ab = ba, why?] = a2 + 2ab + b2 . So, (a + b)2 = a2 + 2ab + b2 . 2. (a − b)(a − b) = a(a − b) − b(a − b) = a2 − ab − ba + b2 , a > b = a2 − 2ab + b2 . So, (a − b)2 = a2 − 2ab + b2 . 3. (a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba −b2 = a2 – b2 So, (a + b)(a − b) = a2 − b2 . Category B: Multiplication of three binomial expressions 1. (a + b)(a + b)(a + b) = (a + b)(a + b)2 = (a + b)(a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 . That is, (a + b)3 = a3 + 3a2 b + 3ab2 + b3 . b c a b + c I J A a2 b2 ab ab a + b a + b a a a a b b b b E F D B G C I J A b2b(a – b) a – b a – b a – b (a – b)2 a – b (a – b)b b b b b F E D B G C a a
192 The Leading Mathematics - 7 2. (a − b)(a − b) (a − b) = (a − b)(a − b)2 = (a − b)(a2 − 2ab + b2 ) = a3 − 2a2 b + ab2 − ba2 + 2ab2 − b3 = a3 − 3a2 b + 3ab2 − b3 That is, (a − b)3 = a3 − 3a2 b + 3ab2 − b3 IV. Multiplication of a Trinomial by binomial Category A: Multiplication of trinomial by binomial 1. (a + b)(a + b + c) = a(a + b + c) + b(a + b + c) = a2 + ab + ac + ba + b2 + bc = a2 + 2ab + b2 + ca + bc So, (a + b)(a + b + c) = a2 + 2ab + b2 + ca + bc. 2. (a − b)(a − b − c) = a(a − b − c) − b(a − b − c), a > b > c = a2 − ab − ac − ba + b2 + bc = a2 − 2ab + b2 − ca + bc So, (a − b)(a − b − c) = a2 − 2ab + b2 − ca + bc. Category B: Multiplication of two trinomials 1. (a + b + c)(a + b + c) = a(a + b + c) + b(a + b + c) + c(a + b + c) = a2 + ab + ca + ab + b2 + bc + ca + bc + c2 = a2 + b2 + c2 + 2ab + 2bc + 2ca So, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca. 2. (a − b − c)(a − b − c) = a(a − b − c) − b(a − b − c) − c(a − b − c) = a2 − ab − ca − ab + b2 + bc − ca + bc + c2 = a2 + b2 + c2 − 2ab + 2bc − 2ca So, (a − b − c)2 = a2 + b2 + c2 − 2ab + 2bc − 2ca. b a a b c a + b a + b + c b c b a a – b a – b a – b – c a b c a a b c a + b + c a + b + c b c b c a a – b – c a – b – c a
Algebra 193 CLASSWORK EXAMPLES Example 1 Multiply : (a) 3x3 − 4y2 and 2x2 y (b) (− x3 )(x4 − 2x2 + 3x − 1) Solution: (a) Horizontal or row form: Multiply each term of 3x3 – 4y2 by 2x2 y and find the algebraic sum, 2x2 y × (3x3 − 4y2 ) = 2x2 y × 3x3 + 2x2 y × (−4y2 ) = 6x5 y + (– 8)x2 y3 [ 2x2 y × 3x3 = 6x5 y] = 6x5 y − 8x2 y3 . [ 2x2 y (−4y2 ) = (– 8)x2 y3 ] Vertical or column form: 3x3 – 4y2 × 2x2 y 6x5 y − 8x2 y3 (b) Horizontal or row form: Multiply each term of (x3 – 2x2 + 3x − 1) by − x2 and find the algebraic sum, (−x2 )(x3 − 2x2 + 3x −1) = (−x2 )(x3 ) + (−x2 )(−2x2 ) + (−x2 )(3x) + (– x2 )(−1) = −x5 + 2x4 − 3x3 + x2 Vertical or column form: x3 – 2x2 + 3x – 1 × – x2 –x5 + 2x4 – 3x3 + x2 Example 2 Multiply : (a) x + 2 and x + 3 (b) x + 2 and x + 2 (c) x + a and x + b (d) 3x + y and x + 3y Solution: (a) (x + 2)(x + 3) = x(x + 3) + 2(x + 3) = x.x + 3x + 2x + 2 × 3 = x2 + (2 + 3)x + 6 = x2 + 5x + 6
194 The Leading Mathematics - 7 (b) (x + 2)(x + 2) = x(x + 2) + 2(x + 2) = x.x + 2x + 2x + 2 × 2 = x2 + (2 + 2)x + 2 × 2 = x2 + 2.2x + 22 = x2 + 4x + 4. (c) (x + a)(x + b) = x(x + b) + a(x + b) = x.x + bx + ax + a×b = x2 + ax + bx + ab = x2 + (a + b)x + ab (d) (3x + y)(x + 3y) = 3x(x + 3y) + y(x + 3y) = 3x.x + 3x × 3y + yx + y × 3y = 3x2 + 9xy + xy + 3y2 = 3x2 + 10xy + 3y2 Example 3 Find the product of : (a) x − 2 and x − 3 (b) x − 2 and x − 2 (c) x − a and x − b (d) 3x − y and x − 3y Solution: (a) (x − 2)(x − 3) = x(x − 3) − 2(x − 3) = x.x − 2x − 3x + (−2) × (−3) = x2 – (2 + 3)x + 2×3 = x2 − 5x + 6 (b) (x − 2)(x − 2) = x(x −2) − 2(x −2) = x.x − 2x − 2x + (−2) × (−2) = x2 – (2 + 2)x + 2 × 2 = x2 − 2.2x + 22 = x2 – 4x + 4. (c) (x − a)(x − b) = x(x − b) − a(x − b) = x.x − bx − ax + (−a) × (− b) = x2 – (a + b)x + ab
Algebra 195 (d) (3x − y)(x − 3y) = 3x(x − 3y) − y(x − 3y) = 3x.x – 3x × 3y − yx + (y) × 3y = 3x2 – 9xy – xy + 3y2 = 3x2 – 10xy + 3y2 Example 4 Multiply : (a) x + 2 and x − 3 (b) x + 2 and x − 2 (c) x + a and x − b (d) 3x + y and x – 3y Solution: (a) (x + 2)(x − 3) = x(x − 3) + 2(x − 3) = x.x −3x + 2x + 2×(−3) = x2 + (–2 + 2)x + (−6) = x2 − x − 6 (b) (x + 2)(x − 2) = x(x − 2) + 2(x − 2) = x.x − 2x + 2x + 2 × (−2) = x2 – 2 × 2 = x2 – 4. (c) (x + a)(x − b) = x(x − b) + a(x − b) = x.x − bx + ax + a ×(− b) = x2 + ax – bx – ab = x2 + (a − b)x − ab (d) (3x + y)(x − 3y) = 3x(x − 3y) + y(x − 3y) = 3x.x + 3x × (−3y) + yx + y × (−3y) = 3x2 – 9xy + xy – 3y2 = 3x2 + (– 9 + 1)xy − 3y2 = 3x2 – 8xy − 3y2 Example 5 Find the product of : (a) x − 1 and x2 + x + 1 (b) x + y and x2 − xy + y2 (c) 3x + 2y and x − 3y + 1 (d) ax + by + c and a'x + b'y + c' Solution: (a) (x − 1)(x2 + x + 1) = x(x2 + x + 1) + (− 1)(x2 + x + 1) = x3 + x2 + x − x2 − x −1 = x3 – 1 (b) (x + y)(x2 – xy + y2 ) = x (x2 − xy + y2 ) + y(x2 – xy + y2 ) = x3 – x2 y + xy2 + x2 y − xy2 + y3 = x3 + y3
196 The Leading Mathematics - 7 (c) (3x + 2y)(x − 3y + 1) = 3x (x − 3y + 1) + 2y(x − 3y + 1) = 3x2 − 9xy + 3x + 2yx − 6y2 + 2y = 3x2 − 7xy + 3x + 2y − 6y2 (d) (ax + by + c)(a'x + b'y + c') = ax (a'x + b'y + c') + by(a'x + b'y + c') + c(a'x + b'y + c') = aa'x2 + ab'xy + ac'x + a'bxy + bb'y2 + bc'y + ca'x + cb'y + cc' = aa'x2 + (ab'+ a'b)xy + bb'y2 + (ac' + ca') x + (bc' + cb')y + cc' Example 6 Show that : (a) (x + 1)(x + 1)(x + 1) = x3 + 3x2 + 3x + 1 (b) (x − y)(x − y)(x − y) = x3 – 3x2 y + 3xy2 − y3 Solution: (a) (x + 1)(x + 1)(x + 1) = (x + 1)(x2 + 2x + 1) = x (x2 + 2x + 1) + 1(x2 + 2x + 1) = x.x2 + 2x.x + x + x2 + 2x + 1 = x3 +2x2 + x + x2 + 2x + 1 = x3 + 3x2 + 3x + 1 (b) (x − y)(x − y)(x − y) = (x − y)(x2 − 2xy + y2 ) = x (x2 − 2xy + y2 ) − y (x2 − 2xy + y2 ) = x.x2 – 2x2 y + xy2 − yx2 + 2xy2 – y3 = x3 − 3x2 y + 3xy2 − y3 Example 7 If p – 1 p = 3, find the values of p2 + 1 p2 and p3 – 1 p3. Solution: Given, p – 1 p = 3 Now, p2 + 1 p2 = p – 1 p 2 + 2.p.1 p = 32 + 2 = 9 + 2 = 11
Algebra 197 Again, p3 + 1 p3 = p – 1 p p2 + p.1 p + 1 p2 = 3 p2 + 1 p2 + 1 = 3(11 + 1) = 3 × 12 = 36 EXERCISE 11.2 Your mastery depends on practice. Practice like you play. 1. Multiply : (a) −3x2 y3 by −2x3 y2 (b) (− abx2 y3 )(cdx3 z2 ) (c) (a − b + c) by abc (d) (x5 + 3x3 − 2x2 + x − 1) by (−x2 ) 2. Find the product of : (a) (− abx2 y3 )(cdx3 z2 )(− 6abx2 y3 ) (b) (− abx2 y3 )(cdx3 z2 )(− 6abx2 y3 ) 1 3 b2 x3 y2 . 3. Multiply : (a) x + 3 and x + 4 (b) x + 3 and x + 3 (c) −(x + a) and (x + b) (d) 3x + 2y and 2x + 3y 4. Find the product of : (a) x − 3 and x − 4 (b) x − 3 and x − 3 (c) a − x and b − x (d) 3x − 2y and 2x − 3y 5. Multiply : (a) x + 3 and x − 4 (b) x + 3 and x − 3 (c) x + b and x − a (d) 3x + 2y and 2x − 3y 6. Find the product of (without using the formulae) : (a) x − 2 and x2 + 2x + 4 (b) x + 2 and x2 − 2x + 4 (c) 2x + 3y and 3x − y + 1 (d) ax + by − c and a'x + b'y − c' 7. Show that : (a) (x + 2)(x + 2)(x + 2) = x3 + 6x2 + 12x + 8 (b) (x − 1)(x − 1)(x − 1) = x3 − 3x2 + 3x − 1
198 The Leading Mathematics - 7 8. Use the square and difference of square formulae to find the product of : (a) −(a + b) and (a + b) (b) (a − b) and (a − b) (c) −(a + b) and −(a + b) (d) (a − b) and (b − a) (e) (2x + 3y) and (2x + 3y) (f) (2x − 3y) and (2x − 3y) (g) (2x + 3y) and (2x − 3y) (h) (a − b) (a + b)(a2 + b2 )(a4 + b4 ) 9. Use cube formulae to compute the product: (a) (x + 2y)(x + 2y)(x + 2y) (b) (x − 2y)(x − 2y)(x − 2y). 10. (a) If a + b = 3 and ab = 2, find the values of a2 + b2 and a3 + b3 . (b) If a – b = 5 and ab = 3, find the values of a2 + b2 and a3 – b3 . (c) If p + 1 p = 3, find the values of p2 + 1 p2 and p3 + 1 p3 . (d) If x – 1 x = 5, find the values of x2 + 1 x2 and x3 – 1 x3. ANSWERS 1. (a) 6x5 y5 (b) – abcdx5 y3 z2 (c) a2 bc – ab2 c + abc2 (d) – x7 – 3x5 + 2x4 – x3 + x2 2. (a) 6a2 b2 cdx7 y6 z2 (b) 2a2 b4 cdx10y8 z2 3. (a) x2 + 7x + 12 (b) x2 + 6x + 9 (c) – x2 – (a + b) x – ab (d) 6x2 + 13xy + 6y2 4. (a) x2 – 7x + 12 (b) x2 – 6x + 9 (c) ab – (a + b)x + x2 (d) 6x2 – 13xy + 6y2 5. (a) x2 – x – 12 (b) x2 – 9 (c) x2 – (a – b)x – ab (d) 6x2 – 5xy – 6y2 6. (a) x3 – 8 (b) x3 + 8 (c) 6x2 + 7xy + 2x + 3y – 3y2 (d) aa'x2 + (ab' + a'b)xy – (ca' + c'a)x – (bc' + b'c)y + bb'y2 + cc' 8. (a) – a2 – 2ab – b2 (b) a2 – 2ab + b2 (c) a2 + 2ab + b2 (d) – a2 + 2ab – b2 (e) 4x2 + 12xy + 9y2 (f) 4x2 – 12xy + 9y2 (g) 4x2 – 9y2 (h) a8 – b8 9. (a) x3 + 6x2 y + 12xy2 + 8y3 (b) x3 – 6x2 y + 12xy2 – 8y3 10. (a) 5, 9 (b) 31, 170 (b) 7, 18 (d) 27, 140
Algebra 199 11.3 Division of Algebraic Expressions At the end of this topic, the students will be able to: ¾ divide the algebraic expressions. Learning Objectives Division of a polynomial by another is not that simple as the previous three operations. We shall consider simple cases only. Rule I: Rule for dividing powers Arrange powers having the same base as fraction; Replace the power with smaller exponent by 1; Keep the base with larger exponent and subtract the smaller exponent from the larger; In case the exponents are equal, replace the fraction by 1. For instance, x5 x3 = x5 – 3 = x2 ; x3 x5 = 1 x5–3 = 1 x2; x5 x5 = 1 (a + b)5 (a + b)3 = (a + b)5 – 3 = (a + b)2 ; (a + b)3 (a + b)5 = 1 (a + b)5 – 3 = 1 (a + b)2; (a + b)5 (a + b)5 = 1 The above process reduces the fraction into the lowest term. Rule II: Rule for dividing a monomial by a monomial Collect powers having the same base and arrange them as separate fractions; Divide numerical coefficients; Divide literal coefficients using Rule I; Multiply the results. For instance, x2 y xy = x2 x y y = x.1 = x. A = x5 l = x3 b = ? A = x2 y l = ? b = xy
200 The Leading Mathematics - 7 15x5 y2 3x3 y7 = 15 3 x5 x3 y2 y7 = (5)(x5 − 3) 1 y7 – 2 = 5.x2 . 1 y5 = 5x2 y5 15 (a + b)5 (a – b)2 3(a + b)3 (a – b)7 = 5 3 (a + b)5 (a + b)3 (a – b)2 (a – b)7 = 5(a + b)2 1 (a – b)7 – 2 = 5(a + b)2 (a – b)5 Rule III: Rule for dividing a polynomial by a monomial Divide each term of the polynomial by the monomial applying Rule II and add the result. For instance, 1. 15x2 y2 + 10x3 y4 + 5x3 y2 5x4 y3 = 15x5 y2 5x4 y3 + 10x3 y4 5x4 y3 + 5x3 y2 5x4 y3 = 3.x y + 2.y x + 1 xy 2. (8a3 b − 8a2 b2 + 2ab3 ) ÷ 2ab = 8a3 b − 8a2 b2 + 2ab3 2ab = 8a3 b 2ab + –8a2 b2 2ab + 2ab3 2ab = 4a2 – 4ab + b2 The above division process may be written in the following form (Long division form): 4a2 − 4ab + b2 2ab 8a3 b − 8a2 b2 + 2ab3 (–) ±8a3 b − 8a2 b2 + 2ab3 (–) 8a2 b2 + 2ab3 (–) ± 2ab3 × Divisor d = Quotient Q Dividend D Remainder R Or, Dividend = Quotient × Divisor + Remainder (D = Q × d + R)