ARITHMETIC 95 Solution: (a) Here, P = Rs. 500, T = 3 years, R = 4 %, I = ? Using, I = P × T × R 100 We get, I = 500 × 3 × 4 100 = Rs 60. (b) Here, I = Rs 50, T = 5 years, R = 2 %, P = ? Using, P = 100 × I T × R We get, P = 100 × 50 5 × 2 = Rs 500. (c) Here, P = Rs 200, I = Rs 30, R = 3 %, T = ? Using, T = 100 × I P × R We get, T = 100 × 30 200 × 3 = 5 yrs. (d) Here, I = Rs 75, P = Rs 500 , T = 5 years, R = ? Using, R = 100 × I P × T We get, R = 100 × 75 500 × 5 = 3 %. Example : 2 A certain amount of money doubles itself in 5 years. Find the rate percent. Solution: Suppose the principal = Rs x, It doubles itself in 5 years. So, T = 5 years, Amount, A = 2x. We know, I = A – P = 2x – x = x. Using, R = 100 × I P × T We get, R = 100 × x x × 5 = 20 %.
96 The Leading Maths - 8 Example : 3 In what time will the interest on Rs. 1225 be Rs. 183.75 at 31 3% p.a.? Solution: Here, P = Rs. 1225, I = Rs. 183.75 and Rs. = 31 3 %. Using the formula, T = 100 × I P × R , we get or, T = 100 × 183.75 1225 × 3 1 3 or, T = 18375 × 3 1225 × 10 or, T = 9 2 or, T = 4 1 2 years. Example : 4 A certain sum of money lent out at simple interest of 5% amounts to Rs. 1380. Find the principal. Solution: Here, A = Rs. 1380, T = 3 years and R = 5%. By using the formula, P = 100 × A 100 + T × R, we get P = Rs. 100 × 1380 100 + 3 × 5 = Rs. 1200. EXERCISE 6.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is interest ? Define it. (b) How much will interest take on Rs. 12000 if 12% per year? (c) How much will the amount take back on Rs. 50000 at 15% per year when it is deposited in a bank? (d) What is the principal amount? (e) If a girl deposited Rs. a in a bank at b% simple interest for c years, how much interest does she get?
ARITHMETIC 97 2. Circle ( ) the correct answer. (a) Which formula is used to find the simple interest on the principal 'P' at R% per year for T years? i. SI = PTR ii. SI = ATR 100 iii. SI = PTR 100 iv. SI = TR% (b) What is the relation between the simple interest SI, principal (P) and amount (A). i. SI = A + P ii. SI = P – A iii. A = P – SI iv. P = A – SI (c) Which relation is true for simple interest? i. SI = A – P ii. P = 100 SI TR iii. T = 100 SI PR iv. All of above (d) Which is the correct simple interest on Rs. 200 at 10% per year for 2 years? i. Rs. 20 ii. Rs. 40 iii. Rs. 80 iv. Rs. 100 (e) What principal deposited at 15% for 3 years will earn a simple interest of Rs. 36 ? i. Rs. 360 ii. Rs. 108 iii. Rs. 3600 iv. Rs. 80 (f) As certain sum of money doubles itself in a year. What rate is imposed on it? i. 20% ii. 150% iii. 75% iv. 100% (h) In what time will the interest on Rs. 25000 be Rs. 11250 at 15% p.a.? i. 2 years ii. 3 years iii. 4 years iv. 5 years 3. Find the interest, if, (a) P = Rs. 500, T = 3 years, R = 4 %. (b) P = Rs. 30000, T = 5 years, R = 4 %. (c) P = Rs 400000, T = 3 years, R = 5 %. (d) P = Rs 250750, T = 4 years, R = 3 1 2%. 4. Find the principal, if, (a) T = 2 years, R = 5 %, I = Rs.50. (b) T = 10 years, R = 2 %, I = Rs 2500. (c) T = 5 years, R = 10 %, I = Rs. 30500. (d) T = 3 years, R = 3 %, I = Rs 25500.
98 The Leading Maths - 8 5. Find the time, if, (a) R = 10 %, I = Rs 30 , P = Rs 100. (b) R = 5 %, I = Rs. 5000, P = Rs. 20000. (c) R = 5 %, I = Rs 5000, P = Rs 303000. (d) R = 4 %, I = Rs. 2000 , P = Rs. 200000. 6. Find the rate percent per annum, if, (a) I = Rs 50, P = Rs 750 , T = 5 years (b) I = Rs 40, P = Rs. 800 , T = 8 years (c) I = Rs 40000, P = Rs. 100000 , T = 5 years (d) I = Rs. 20.50, P = Rs. 100, T = 2 years 7. (a) Find the simple interest on Rs. 700 in 3 years at 4% per annum. (b) Find the simple interest on Rs. 375500 in 5 years at 3% per annum. (c) Find the interest on Rs. 600650 for 5 years at 6% per annum. 8. (a) Find the principal that will yield an interest of Rs. 750 at 3% per annum in 21 2 years. (b) What principal invested at 5% p.a. for 3 months will yield a SI of Rs. 25100? (c) If the simple interest on a sum of money invested at 21 2% per annum for 3 years is Rs. 60.50, find the sum. 9. (a) A certain amount of money trebles itself in 9 years. Find the rate percent. (b) A certain amount of money is invested at 20% interest rate. Find in how many years it will double itself. (c) A certain amount of money is invested at 25% interest rate. What will be the amount after 8 years? 10. (a) At what rate percent per annum, the interest on Rs. 6250 will be Rs. 750 in 4 years? (b) At what rate percent per annum will Rs. 640 amount to Rs. 764.80 in 3 years? (c) At what rate percent per annum will Rs. 1500 amount to Rs. 1800 in 4 years?
ARITHMETIC 99 11. (a) If the simple interest on Rs. 800 at 7 1 2 % per annum is Rs. 150, find the time. (b) In what time will Rs. 675 amount to Rs. 747 at 3 1 3% ? (c) In what time will Rs. 1000 amount to Rs. 1440 at 10% per annum? 12. (a) Find the amount on Rs. 840 for 5 years at 41 2 % per year. (b) Find the principal that amounts to Rs. 530 in 2 years at 3% per annum. (c) Find the principal that amounts to Rs. 420 at simple interest in 4 years at 3% per annum. ANSWERS 3. (a) Rs. 60 (b) Rs. 6000 (c) Rs. 60000 (d) Rs. 35105 4. (a) Rs. 500 (b) Rs. 12500 (c) Rs. 61000 (d) Rs. 283333 1 3 5. (a) 3 yrs (b) 5 yrs (c) 3 1 3 yrs (d) 2.5 yrs 6. (a) 1 1 3 % (b) 5 8 % (c) 8% (d) 10.25% 7. (a) Rs. 84 (b) Rs. 56325 (c) Rs. 180195 8. (a) Rs. 806 2 3 (b) Rs. 10000 (c) Rs. 167333 1 3 9. (a) 22 2 9% (b) 5 yrs (c) Amount will be thrice. 10. (a) 3% (b) 6.5 % (c) 5% 11. (a) 2.5 yrs (b) 3.2 yrs (c) 4.4 yrs 12. (a) Rs. 1029 (b) Rs. 500 (c) Rs. 375 Project Work 6.1 Study the fixed deposit certificate of five persons in your home or neighbour or financial company and list out the deposited amount, simple interest rate and duration of time in tabulation form. Calculate the simple interest and getting amount after the given time in the same table. Prepare a report and present in your classroom.
100 The Leading Maths - 8 Read, Understand, Think and Do 1. In a year there are 12 months and 365 days. (a) Convert 12 into binary number system. (b) Convert 365 into quinary number system. (c) 12 is a rational number. Justify it. (d) Write 365 into scientific notation. 2. The average distance from the Earth to the Sun is 1.46 × 108 kilometers. The average distance from the Earth to the Moon is 3.84 × 105 kilometers. (a) Find the average distance from the Earth to the Sun in standard form. (b) Find the average distance from the Sun to moon on a full-moon day in scientific notation. 3. Factories A and B produced potato chips. Last year, each factory reported the amount of each ingredient. Factory A reported their amounts in scientific notation and Factory B reported their amounts in standard form. Ingredient Factory A (amount in kg) Factory B (amount in kg) Potato 4.87 × 106 3,309,000 Oil 5.61 × 105 356,000 Salt 2.87 × 105 193,500 (a) Which factory used more oil last year? By how much? (b) What is the total amount of salt used by both factories? (c) What is the total weight of all of the ingredients in Factory A? 4. Out of two water tanks, the capacity of the first tank is 3.4 × 102 liter and the second tank is 4.57 × 103 liter. (a) Find the total capacity of both tanks. Write it in scientific notations. (b) How much water should be added in the first tank to equal with the second tank? Calculate it. (c) If the cost of one-liter water is 10 paisa, how much will be the total cost of the water in the second tank? Find it. MIXED PRACTICE–II
ARITHMETIC 101 5. In the mixture of 35 liters of milk, the ratio of pure milk to water is 5:2. The cost of one-liter milk is Rs. 90. (a) Find the quantity of pure milk and water in the given mixture. (b) How much pure milk can be bought in Rs. 1350? Find it. (c) If the shopkeeper sold 35 liters milk in Rs. 3500, how much profit did he get in one liter? Find it. (d) If the shopkeeper wants to make a profit Rs. 700, what should be the selling price of milk in one liter? Calculate it. 6. A fruit seller bought 500 oranges for Rs. 4000. 50 of them were rotten. He sold the remaining oranges for Rs.10 each. (a) Find the cost of 15 oranges. (b) How many oranges are good for sell? Calculate it. (c) What percentage of oranges were rotten? Find it. (d) Find his percentage of profit or loss. 7. Two friends Ramesh and Mahesh invested Rs. 50,00,000 in a factory in the ratio of 3:2. (a) How much amount did Ramesh invest in the factory? Find it. (b) If they deposited that amount in a bank at the rate of 10% p.a. in 2 years, how much interest they would get after 2 years? (c) Find the difference in amounts that they invested in the factory. 8. A dealer buys a machine for Rs 2500. He marks it at Rs 3300 and then gives a discount of 10% on it. (a) Find the discount price of the machine. (b) Find the selling price of the machine. (c) Find the profit or loss percent made by the dealer. (d) If the dealer wants to sell a machine in Rs. 2500, what should be the rate of discount allowed in the mark price of that machine? 9. A retailer buys an article for Rs 300. He increases its price by 20% and then gives 10% discount on the new price. (a) Find the new price (marked price) of the article. (b) Find the cost of such 20 articles. (c) Find the selling price of the articles. (d) Find the profit percent made by the retailer on selling 20 articles. 10. Ramesh deposited Rs. 1,00,000 in a bank at the rate of 10% per annum for 3 years. (a) Write the formula for finding simple interest. (b) How much interest does Ramesh get after 3 years?
102 The Leading Maths - 8 (c) If Ramesh withdraws Rs. 50,000 at the end of 2 years, how much amount does Ramesh get at last? (d) Ramesh expense the withdrews amount in education and health of his child in the ratio of 3:2. How much amount did he expense in education? Find it. 11. The cost of a table and a chair are in the ratio of 5:2. The cost of a chair is Rs.700. (a) Find the cost of such types of 4 chairs. (b) What should be the cost of 3 tables. Calculate it. (c) If a shopkeeper allowed 10% discount while purchasing a table and a chair, how much amount will require for buying it? Find it. 12. Three friends Ashok, Nabaraj and Shusil have invested Rs. 84,00,000 in the ratio of 3:4:5 to start a company. After 1 year, the company got a loss and sold for Rs 6,00,000 less and distributed. (a) How much amount did Shusil invest in a company? Calculate it. (b) How much amount did Ashok get after 1 year? (c) If Ashok deposited the invested amount in a bank at the rate of 10% p.a , what will be the amount Ashok will get after 1 year? ANSWERS 1. (a) 11002 (b) 24305 (d) 3.65 × 102 2. (a) 146000000 km (b) 1.46384 × 108 km 3. (a) A, 205000 kg (b) 480500 kg (c) 5.718 × 106 km 4. (a) 4910 l, 4.91 × 103 (b) 4230 l (c) Rs. 457 5. (a) 25 l, 10 l (b) 54 l (c) 10% (d) Rs. 110 6. (a) Rs. 1200 (b) 450 pcs. (c) 10% (d) 12.5% 7. (a) Rs. 3000000 (b) Rs. 1000000 (c) Rs. 1000000 8. (a) Rs. 330 (b) Rs. 2970 (c) 18.8% (d) 24 8 33 % 9. (a) Rs. 360 (b) Rs. 6000 (c) Rs. 324 (d) 14% 10. (a) SI = PTR 100 (b) Rs. 30000 (c) Rs. 77000 (d) Rs. 30000 11. (a) Rs. 2800 (b) Rs. 5250 (c) Rs. 1575 12. (a) Rs. 3500000 (b) Rs. 1950000 (c) Rs. 2310000
ARITHMETIC 103 1. The present age of a girl is 24 years. (a) Write any one difference between binary and quinary number systems. [1] (b) Convert her age into the binary number system. [1] (c) How many months are there her age? Express it into the quinary system. [1] (d) 24 is a rational number. Justify it. [1] 2. (a) The distance between the Sun and the Earth is 147.1 million kilometers. Express this distance into full digits and words in national system. Also, convert it into scientific notation. [2] (b) Najir and Sara ate half and one-third of pizza respectively. Express the eaten part of the pizza into decimal forms. Also, find two rational numbers between half and one third. [2] 3. In the mixture of 20 liters of milk, the ratio of pure milk to water is 4:1. The cost of oneliter milk is Rs. 110. (a) Find the quantity of pure milk and water in the mixture of milk. [1] (b) How much milk can be bought for Rs. 1430? Find it. [1] (c) If the shopkeeper sold 25 liters of milk for Rs. 3375, how much profit did he get in one liter? Find it. [1] (d) If the shopkeeper wants to make a profit Rs. 875 by selling 25 liters of milk, what should be the selling price of the milk per liter? Calculate it. [1] 4. A fruit seller bought 100 kg of apples for Rs. 14500. 5 kg of them were rotten. He sold the remaining apples for Rs. 160 per kg. (a) Find the cost of 30 kg of apples. [1] (b) What percentage of apples were rotten? Find it. [1] (c) Find his percentage of profit or loss. [2] 5. A shopkeeper buys an article for Rs. 5400. She increases its price by 20% and then gives 10% discount on the new price. (a) Write the definition of discount. [1] (b) Find the new price (marked price) of the article. [1] (c) Find the selling price. [1] 6. A certain amount of money is invested at a 25% interest rate. (a) How much interest did the depositor get? Find it. [1] (b) Find in how many years it will be four times itself. [1] FM : 21 Time : 40 Min. CONFIDENCE LEVEL TEST II ARITHMETIC Attempt all the questions.
104 The Leading Maths - 8 COMPETENCY solution of the behaviour problems related to perimeter, area and volume CHAPTERS 7. Area of plane shapes LEARNING OUTCOMES find the area of triangle and quadrilateral, establish the formula of finding area of a circle and use it. MENSURATION UNIT III Estimation Teaching Hours - 8 (Th. + Pr.)
MENSURATION 105 CHAPTER 7 AREA OF PLANE SHAPES Lesson Topics Pages 7.1 Area of Triangle and Quadrilateral 106 7.2 Area of Circle 126 What is plane shape? Tell some examples. What is rectangle? What is square? What are the differences between rectangle and square? What is area? What is the shape of the area of your maths book? What are the shapes of the area of chessboard and coin? Measure the length and breadth of your maths book. What are its length and breadth? Can you find its area? WARM-UP
106 The Leading Maths - 8 7.1 Area of Triangle and Quadrilateral At the end of this topic, the student will be able to: ¾ find the area of triangle and quadrilateral. Learning Objectives 1. Area and Perimeter of Plane Geometric Shapes Activity-1 Area and Perimeter of Rectangle D C A B The occupied space by a closed figure on the plane surface is called its area (A). In the adjoining geoboard, a rectangle has occupied the space ABCD whose length, AB = 4 units and its breadth, BC = 3 units. Draw 4 columns and three rows with the same wide. How many square units are there in it? Therefore, the area of the rectangle with length 4 units and breadth 3 units is 12 sq. units. It means, Area of Rectangle (A) = 12 sq. units = 4 units × 3 units = Length × Breadth. Hence, the area of a rectangle with length l units and breadth b units is the product of l and b. i.e., A = l units × b units = l b sq. units. Again, the length around the plane surface is called its perimeter (P). Now, the perimeter of the rectangle ABCD (P) = 4 units + 3 nits + 4 units + 3 units = 14 units. Hence, the perimeter of a rectangle with length l units and breadth b units is l + b + l + b = 2l + 2b = 2(l + b) units. i.e. P = 2(l + b) units. 2. Area and Perimeter of Square We know that square is the rectangle with equal adjacent sides, this means length and breadth are equal in the square. Suppose, the length and breadth of the square is l units. Therefore, the area of the square (A) = l units × l units = l2 sq. units, and the perimeter of the square (P) = 2(l + l) units = 4l units. l l l b
MENSURATION 107 Activity-2 Area and Perimeter of Right-angled Triangle Draw a rectangle on a paper and take out it by cutting. Also, cut the rectangle diagonally as shown in the adjoining figure. What do you get? It forms two right-angled triangles. What may be the area of each right-angled triangle? Since the area of the rectangle is l × b sq. units, so the area of the right-angled triangle is half the area of the rectangle. i.e. Area of the right triangle (A) = 1 2 × Area of rectangle = 1 2 × l × b sq. units In the right-angled triangle, length (l) and the breadth (b) of the rectangle are represented by the base (b) and perpendicular (p) respectively. Hence, Area of the right-angled triangle (A) = 1 2 pb sq. units The perimeter of the right-angled triangle is the sum of perpendicular (p), base (b) and hypotenuse (h). i.e. Perimeter of the right-angled triangle (P) = p + b + h Since, by Pythagoras' theorem, h2 = p2 + b2 so, Perimeter of the right-angled triangle (P) = (p + b + p2 + b2 ) units. Activity-3 Area and Perimeter of Triangle with Base and Height Draw a triangle with base b units and height h units on a paper and take out it by cutting. Also, cut it vertically along its height as shown in the adjoining figure. What do you get? We get two right-angled triangles. Are they equal? For the first right-angled triangle 'I', The perpendicular (p) = h and base (b) = x (suppose). So, the area of first right-angled triangle I (A1) = 1 2 pb = 1 2 hx sq. units. b x h I II b – x
108 The Leading Maths - 8 For the second right-angled triangle II, The perpendicular (p) = h and base (b) = b – x. So, the area of first right-angled triangle II (A2) = 1 2 pb = 1 2 h (b – x) sq. units. Then, the area of the given triangle (A) = Area of the first right-angled triangle I (A1) + Area of the first right-angled triangle II (A2) = 1 2 hx + 1 2 h (b – x) = 1 2 (x + b – x)h = 1 2 bh sq. units. Hence, the area of the given triangle (A) = 1 2 bh sq. units and The perimeter of the scalene triangle (P) = (b + h2 + x2 + h2 + (b – x)2 ) units. Activity-4 Area and Perimeter of Isosceles Triangle Draw an isosceles triangle with equal sides a units and base b units on a paper and take out it by cutting. Also, cut or fold it vertically through its vertex as shown in the adjoining figure. What do you get? We get two right-angled triangles. Are they equal? In each right-angled triangle, since the perpendicular drawn from the vertex to its base bisect the base. So, the length of base b 2 and the hypotenuse is a, then by using Pythagoras' theorem, Perpendicular of right triangle (p) = h2 – b2 = a2 – b 4 2 = a2 – b2 4 = 4a2 – b2 4 = 1 2 4a2 – b2 . a a b a b 2
MENSURATION 109 Therefore, the area of each right-angled triangle (A1) = 1 2 pb = 1 2 × 1 2 4a2 – b2 × b 2 = b 8 4a2 – b2 Hence, the area of the isosceles triangle (A) = 2 × A1 = 2 × b 8 4a2 – b2 = b 4 4a2 – b2 sq. units The perimeter of the isosceles triangle (P) = a + a + b = 2a + b units. Activity-5 Area and Perimeter of Equilateral Triangle Draw an equilateral triangle with equal sides a units on a paper and take out it by cutting. Also, cut or fold it vertically through its vertex as shown in the adjoining figure. What do you get? We get two right-angled triangles. Are they equal? In each right-angled triangle, since the perpendicular drawn from the vertex to its opposite side bisects that side. So, the length of that a 2 and the hypotenuse is a. Then, by Pythagoras' theorem, Perpendicular of the right-angled triangle (p) = h2 – b2 = a2 – a 2 2 = a2 – a2 4 = 4a2 – a2 4 = 3a2 4 = 3a 2 a a 2 a a a
110 The Leading Maths - 8 Therefore, the area of each right-angled triangle (A1) = 1 2 pb = 1 2 × 3a 2 × a 2 = 3a2 8 Hence, the area of the equilateral triangle (A) = 2 × A1 = 2 × 3a2 8 = 3a2 4 Hence, the area of the equilateral triangle (A) = 3a2 4 sq. units. The perimeter of equilateral triangle (P) = a + a + a = 3a units. Activity-6 Area and Perimeter of Parallelogram Draw a parallelogram on a paper and take out it by cutting. Also, cut the rectangle diagonally as shown in the adjoining figure. What do you get? It forms two scalene triangles with same heights. Both triangles are the same shape and size. What is the area of each triangle? Therefore, the area of each triangle (A1) = 1 2 bh Hence, the area of parallelogram (A) = 2 × 1 2 bh = bh sq. units The perimeter of the parallelogram (P) = a + b + a + b = 2a + 2b = 2(a + b) units. Activity-7 Area and Perimeter of Rhombus Draw a rhombus with diagonals d1 and d2 on a paper as shown in the adjoining figure. How many triangles do you get? These four triangles have the same area, each have height d1 2 and base d2 2 . b h b b h h a a d1 l d2 l d1 2 d1 2 d2 2 d2 2
MENSURATION 111 Therefore, the area of the rhombus is A = 4 × Area of one triangle = 4 × 1 2 bh = 4 × 1 2 × d1 2 × d2 2 = 1 2 d1d2. Hence, the area of rhombus (A) = 1 2d1d2 sq. units = Half of the product of its diagonals. The perimeter of the parallelogram (P) = l + l + l + l = 4l units. Activity-8 Area and Perimeter of Trapezium Draw a trapezium with parallel bases b1 and b2 on a paper as shown in the adjoining figure. Draw a diagonal in it and form two triangles. Also, draw the heights for both triangles. Then, the area of the trapezium (A) = Area of triangle with base b1 + Area of triangle with base b2 = 1 2 b1h + 1 2 b2h = 1 2 h(b1 + b2). "Next Method:" b1 b1 h h b2 b2 Hence, the area of trapezium (A) = 1 2 h(b1 + b2) sq. units = Half of the product of height and sum of bases and the perimeter of the parallelogram (P) = (b1 + b2 + l1 + l2) units. Activity-9 Area and Perimeter of Quadrilateral a c d b d h1 h2 Draw a quadrilateral on a paper. Draw a diagonal d in it and form two triangles. Also, draw the heights for both triangles as shown in the adjoining figure. , The area of the quadrilateral (A)= Area of triangle with height h1 + Area of triangle with height h2 = 1 2 dh1 + 1 2 dh2 = 1 2 d(h1 + h2). b1 h h b2 l1 l2
112 The Leading Maths - 8 Hence, area of quadrilateral (A) = 1 2 d(h1 + h2) sq. units = Half of the product of diagonal and the sum of heights The perimeter of the parallelogram (P) = (a + b + c + d) units. Activity-10 Area and Perimeter of Kite x d1 – x a b d2 Draw a kite with vertical diagonal d1 and horizontal diagonal d2 on a paper as shown in the adjoining figure. The horizontal diagonal divides the kite into two triangles. The upper triangle has base d2 and height x (suppose) and the lower triangle has base d2 and height (d1 – x). Then the area of the kite (A) = Area of the upper triangle + Area of the lower triangle = 1 2 × d2 × x + 1 2 × d2 × (d1 – x) = 1 2 (x + d1 – x) d2 = 1 2d1d2 Hence, the area of kite (A) = 1 2d1d2 sq. units = Half of the product of its diagonals and the perimeter of the parallelogram (P) = a + b + a + b = 2a + 2b = 2(a + b) units. Activity-11 Area and Perimeter of Arrow-Head (Tip) Draw the arrow-head with vertical diagonal d1 and horizontal diagonal d2 on a paper as shown in the adjoining figure. The vertical diagonal d1 divides the arrow-head into two congruent triangles, left and right sides. Produce the vertical diagonal d1 up to the horizontal diagonal d2 that are perpendicular to each other. Then, The area of the arrow-head is A = 2 × Area of triangle with height d1 and base d2 2 = 2 1 2 × d2 2 × d1 = 1 2 d1 d2 Hence, the area of arrow-head (A) = 1 2d1d2 sq. units = Half of the product of its diagonals. The perimeter of the parallelogram (P) = a + b + a + b = 2a + 2b = 2(a + b) units. d2 2 d2 2 b a d1
MENSURATION 113 Note down the area and perimeter of the following plain geometric shapes which you have come across in earlier grades. Name Figure Area and perimeter Rectangle l b A = length × breath or A = l × b Perimeter (P) = 2(l + b) Square a d A = (side)2 A = a2 A = 1 2 d2 , d = diagonal Perimeter (P) = 4a Right Triangle b p h Area = 1 2 perpendicular(p)×base(b Perimeter (P) = p + b + h Hypotenuse (h) = p2 + b2 Triangle b h a c Area = 1 2 base × height or, A = 1 2 b × h Perimeter (P) = a + b + c Isosceles Triangle a a b Area = b 4 4a2 – b2 Perimeter (P) = 2a + b Equilateral Triangle a a a Area = 3a2 4 Perimeter (P) = 3a
114 The Leading Maths - 8 Name Figure Area and perimeter Parallelogram b a h Area = base × height or, A = b × h Perimeter P = 2(a + b) Rhombus d1 a d2 Area = 1 2 (product of diagonals) or, A = 1 2 (d1 × d2) Perimeter = 4a, a = length of sides Trapezium b2 a h b1 c Area = 1 2 (sum of parallel sides) × height or, A = 1 2 (b1 + b2) × h Perimeter = a + b1 + b2 + c Quadrilateral h2 h1 d Area = 1 2 diagonals × (sum of heights) or, A = 1 2 d × (h1 + h2) Perimeter (P) = sum of sides Kite d2 a d1 b Area = 1 2 (product of diagonals) or, A = 1 2 (d1 × d2) Perimeter (P) = 2(a + b) Arrow-head d2 a d1 b Area = 1 2 (product of diagonals) or, A = 1 2 (d1 × d2) Perimeter (P) = 2(a + b)
MENSURATION 115 CLASSWORK EXAMPLES Example : 1 Observe the following plane shapes and find their areas.. (a) (b) (c) (d) Solution: (a) The shape is a triangle with base (b) = 8 cm height (h) = 5 cm ∴ Area (A) = 1 2 b × h = 1 2 × 8 × 5 = 20 cm2 (b) The shape is a rectangle with length (l) = 8 cm breadth (b) = 4 cm ∴ Area (A) = l × b = 8 × 4 = 32 cm2 (c) The shape is the combination of triangle and rectangle of (a) and (b) . Area = Area of triangle + Area of rectangle = 20 + 32 = 52 cm2 (d) The shape is a combination of two trapeziums ABEF + EBCD. Now, Area of trapezium ABEF = 1 2 (AB + EF) × AF = 1 2 (7 + 3) × 3 = 1 2 × 10 × 3= 15 cm2 5 cm 8 cm 8 cm 4 cm 5 cm 4 cm 8 cm 8 cm 4 cm 3 cm 5 cm A 3 cm B C D F 7 cm E 5 cm 8 cm 8 cm 4 cm 5 cm 4 cm 8 cm 8 cm 4 cm 3 cm 5 cm A 3 cm B C D F 7 cm E
116 The Leading Maths - 8 Area of trapezium EBCD = 1 2(ED + BC) × CD = 1 2 (5 + 8) × 3 = 1 2 × 13 × 4 = 26 cm2 Area of the shape = Area of (ABEF + EBCD) = (15 + 26) = 41 cm2 Example : 2 Observe the given plane shape and answer the following questions. (a) Write the formula to find the area of isosceles triangle. (b) Find the area of isosceles triangle ABC. (c) Find the area of parallelogram PQRS. (d) Find the area of the shaded portion. (e) If the shaded portion is painted in another colour, calculate the cost of painting charge at Rs. 3.25 per sq. cm. Solution: (a) Area of isosceles triangle (A) = b 4 4a2 – b2 . (b) In isosceles triangle ABC, equal sides AB = BC = a = 10 cm, base side = BC = b = 12 cm. Now, we know that, Area of isosceles triangle ABC (A1) = b 4 4a2 – b2 = 12 4 4 × 102 – 122 = 3 400 – 144 = 3 256 = 3 × 18.87 = 56.6 cm2 .
MENSURATION 117 (c) In parallelogram PQRS, base side QR (b) = 7 cm, height (h) 4 = cm Now, we have Area of parallelogram PQRS (A2) = b × h = 7 × 4 = 28 cm2 (d) Area of the shaded portion (A) = A1 – A2 = 56.6 – 28 = 28.6 cm2 . (e) Rate of painting charge (R) = Rs. 3.25 per sq. cm. ∴ Cost of painting charge (C) = R × A = Rs. 3.25 × 28.6 = Rs. 92.95 Example : 3 ABCD is a rectangle. E and F are mid points of AB and DC, where BC = 8 cm and AB = 6 cm,. (a) Write the formula to calculate the area of rectangle and quadrilateral.. (b) Find the area of the rectangle ABCD. (c) Find the area of the quadrilateral GEHF. (d) Find the area of the shaded part. Solution: (a) Area of rectangle (A) = l × b and, Area of quadrilateral (A) = 1 2 d × (h1 + h2) (a) In the give rectagle ABCD, join EF. Draw GJ ⊥ EF and HI ⊥ EF So, AE = BE = GJ = EI = 6 2 = 3 cm. Now, area of ∆GEF = 1 2 × EF × GJ = 1 2 × 8 × 3 = 12 cm2 A B C D E F G H 8 cm 6 cm I J
118 The Leading Maths - 8 And area of ∆EHF = 1 2 × EF × HI = 1 2 × 8 × 3 = 12 cm2 Hence, area of the shape GEHF = (12 + 12) = 24 cm2 ∴ Area of the rectangle ABCD = BC × AB = 8 × 6 = 48 cm2 ∴ Area of the shaded part = (48 – 24) cm2 = 24 cm2 EXERCISE 7.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is area? (b) Write down the formula to find the area of triangle having base 'b' and height 'h'. (c) What is the relation between the area of the parallelogram and triangle? (d) Write the formula to find the area of quadrilateral and trapezium. (e) What is the area of the rectangle having length 7 cm and breadth 9 cm? (f) Write the area of the triangle having base side 20 cm and height 25 cm. 2. Circle ( ) the correct answers. (a) Which formula is used to calculate the are of parallelogram having base 'b' and height 'h'. i. 1 2 b × h ii. ph iii. b × h iv. 1 2 ab (b) Which formula is used to find the area of rhombus with the length of diagonal d1 and d2 ? i. d1 × d2 ii. d1 + d2 iii. d1 + d2 2 iv. 1 2 d1 × d2
MENSURATION 119 (c) Which formula is used to calculate the area of trapezium with height 'h', parallel bases b1 and b2? i. 1 2 h(p1 + p2) ii. 1 2 h (b1 + b2) iii. 1 2 b1 × b2 × h iv. 1 2 d(h1 + h2) (d) What is the area of a right triangle having the length of shorter sides 10 cm and 12 cm? i. 120 cm2 ii. 50 cm2 iii. 60 cm2 iv. 240 cm2 (e) What is the area of the shaded portion in the given figure? i. 100 cm2 ii. 25 cm2 iii. 125 cm2 iv. 75 cm2 3. Observe teh given shapes. Answer the following questions. (i) A B C D 4 cm (ii) A D B C 9.5 cm (a) Which formula use to find the area of a square? Write it. (b) What is the area of the given square? Find it. (c) What is the perimeter of the given square ? Find it. (d) Compare their area and perimeter. 4. Observe the given shapes. Answer the following questions for each. (i) A D B C 2.5 cm 6 cm (ii) 13 cm 12 cm A B C D (a) What formula is used to find the area of a parallelogram ? Write it. (b) What is the area of the given square ? Find it. (c) What is the perimeter of the given square ? Find it. (d) Compare their area and perimeter. 10 cm 10 cm 5 cm 5 cm
120 The Leading Maths - 8 5. Observe the following triangular shapes and answer the following questions for each object. (i) (ii) (iii) (iv) (v) (vi) (v) (vi) (a) Write the formula use to find the area of a triangle having base 'b' and height 'h'. (b) What is the height of the given shapes? Find it if necessary. (c) What is the area of the given shapes? Find it. (d) Find their perimeters. A B CD h = 8 m 11 m 10 m 4 m 4 m 10 m B D C A B 5 m 3 m D C A 12 ft P Q R Q 12 cm R P 16 cm T 15 cm QP S 6 cm 3 cm Q R P 8 cm N O M 12 3 cm
MENSURATION 121 6. Find the area of the following parallelograms. (a) (b) 7. Observe the following triangular shapes and answer the following questions for each object. (i) (ii) AC = 6 cm and BD = 8 cm AC = 9.5 cm and BD = 5.8 cm (iii) (iv) AO = 5 cm and BO = 4 cm AO = 6.5 cm and BO = 4.5 cm (a) Write the formula to find the area of a rhombus. (b) If the length of the legs of a right triangle are 'a' units and 'b' units, what is the area of the right triangle ? (c) What is the area of the given rhombus ? Find it. 8. Find the area of the following trapeziums. (a) (b) A 4 cm B C D E 3.5 cm A D B F C 3.5 cm 4.8 cm O A D B C O A D B C O A D B C O D A C B 6 cm 6 cm 8 cm 11 cm 4 cm 8 cm
122 The Leading Maths - 8 (c) (d) 9. Find the area of the following quadrilaterals. (a) (b) P1 = 3 cm, P2 = 5 cm and d = 8 cm P1 = 2.5 cm, P2 = 6.3 cm and d = 8.5 cm (c) (d) AD = 5 cm AF = 3 cm AD = 13 cm DF = 12 cm BF = 4 cm EC = 6 cm CE = 6 cm BD = 20 cm 10. Find the area of the following kites. (a) (b) d1 = 10 cm and d2 = 6 cm d1= 10.5 cm and d2 = 6.8 cm 6.8 cm 10.5 cm 7.2 cm 9.5 cm 6 cm 7 cm A B D C p1 p2 E d F E A B D C p2 p1 F d A D B C p1 p2 E F d E A B D p2 p1 F C d A C B D d2 d1 A C B D d2 d1
MENSURATION 123 (c) (d) AC = 14.2 cm and BO = 5.1 cm BD = 13 cm and AO = 5 cm 11. Find the area of the following arrow-heads. (a) (b) d1 = 3 cm and d2 = 6 cm d1 = 5.2 cm and d2 = 8.7 cm (c) (d) d1 = 4 cm and CE = 5 cm DE = 3 cm and DC = 5 cm 12. ABCD is a rectangular field of length 120 m and breadth 80 m. (a) Find the area of the field. (b) Find the perimeter of the field . (c) Find the length of fencing material that goes 4 round about it. (d) Find the cost of fencing at the rate of Rs.9.79 per meter. A B D C O A B D C O A B D d2 d1 C A B D C A B D d1 E C A B D d2 d1 C E A B D C
124 The Leading Maths - 8 13. ABCD is a rectangle of length BC = 40 cm and diagonal BD = 50 cm. (a) Find the area of the rectangle. (b) Find the perimeter of the rectangle. (c) Find the length of thread at 5 round around it. (d) Find the cost of thread at Rs. 0.28 per cm. 14. Find the area of the shaded parts in each of the following figures. Also, calculate the cost of tiling on the shaded area at Rs. 325 per m2. (a) 18 m 12 m 16 m 10 m (b) 20 m 2 m 16 m 2 m (c) 12 m 8 m (d) 12 m 20 m (e) 25 m 2 m 2 m 2 m 18 m 16 m ↓ ↓ (f) 2 m 2 m 20 m 12 m (g) d2 d1 A B C D 12.5 m 18 m (h) h a b 12.5 m 18.9 m d1 = 3 m, d2 = 5 m a = 3 m, b = 5 m, h = 2.5 m A 40 cm 50 cm D B C
MENSURATION 125 15. Find the area of the following shapes. (a) (b) (c) (d) ANSWERS 3. (i) (b) 16 cm2 (c) 16 cm (ii) (b) 45.125 cm2 (c) 26.87 cm 4. (i) (b) 15 cm2 (c) 17 cm (ii) (b) 48 cm2 (c) 28 cm 5. (i) (b) 8 cm (c) 44 m2 (d) 30.43 m (ii) (b) 6 cm (c) 12 cm2 (d) 21.21 cm (iii) (b) 4 cm (c) 10 cm2 (d) 18.94 cm (iv) (b) 10.39 ft. (c) 3 ft2 (d) 6 ft (v) (b) 8.94 cm (c) 71.55 cm2 (d) 40 cm (vi) (b) 12 cm (c) 36 cm2 (d) 33.37 cm (vii) (b) 6.93 cm (c) 27.71 cm2 (d) 24 cm (viii) (b) 12 cm (c) 72 cm2 (d) 40.97 cm 6. (a) 36 cm2 (b) 20 cm2 (c) 14 cm2 (d) 16.8 cm2 7. (a) 24 cm2 (b) 27.55 cm2 (c) 40 cm2 (d) 58.5 cm2 8. (a) 42 cm2 (b) 38 cm2 (c) 62.2 cm2 (d) 46.98 cm2 9. (a) 32 cm2 (b) 37.4 cm2 (c) 36 cm2 (d) 110 cm2 10. (a) 38 cm2 (b) 35.7 cm2 (c) 289.68 cm2 (d) 65 cm2 11. (a) 9 cm2 (b) 22 cm2 (c) 20 cm2 (d) 12 cm2 12. (a) 9600 m2 (b) 400 m (c) 1600 m (d) Rs 15,600 13. (a) 1200 cm2 (b) 140 cm (c) 700 cm (d) Rs 196 14. (a) 56 cm2 , Rs. 18200 (b) 68 cm2 , Rs. 22100 (c) 48 cm2 , Rs. 15600 (d) 120 cm2 , Rs. 39000 (e) 82 cm2 , Rs. 26650 (f) 60 cm2 , Rs. 19500 (g) 217.5 cm2 , Rs. 70687.50 (h) 226.25 cm2 , Rs. 73531.25 15. (a) 39 cm2 (b) 52 cm2 (c) 52 cm2 (d) 226.80 cm2 3 cm 3 cm 4 cm 5 cm 8 cm 8 cm 2cm 2cm 6 cm 8 cm 10 cm 6 cm 2cm 2 cm 2 cm 10 cm 6 cm 4 cm 4 cm 2 cm 2 cm 6 cm 7 cm 5.2 cm 18 cm
126 The Leading Maths - 8 7.2 Area of Circle At the end of this topic, the student will be able to: ¾ derive the formula to calculate the area of circle. ¾ find the area of circle using formula. Learning Objectives What is the shape of bangle, ring, base of marker and water bottle? What is the roundness i.e. perimeter of the circle? Do you remember, where are the centre, radius and diameter of the circle? What are arc and semi-circle? What are segment and sector in the circle? Which chord of a circle is called diameter? A B C D O WARM-UP I. Review on circle ACTIVITY - 1 Take two lemons or oranges and cut down them vertically and horizontally. What do you see? Discuss the different parts of circular face in horizontally cutting part. 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 round about of the circle is the circumference of the circle. In another word, the total length round-about the circle is its perimeter or circumference which is obtained as,
MENSURATION 127 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 always constant that is called π (pi). π is irrational number whose values is 3.141592654 ...... and approximately equal to 22 7 , not exact. II. Area of a Circle Activity: 1 Here is a way to find the formula for the area of a circle: Cut a circle into equal sectors (12 in this example). Divide just one of the sectors into two equal parts. Now, we have thirteen sectors as numbered them 1 to 13: 1 2 3 4 5 7 6 8 9 10 11 12 13 Cut one sector in half (Why ?) c c = 2πr c r Rearrange the 1 and 13 sectors like this: r c/2 1 2 3 4 5 6 7 8 9 10 11 12 13 c/2 r Which resembles a rectangle: c d r 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) If the circle is cut down more than more pieces, it will be approximately as a rectangle when these pieces arrange as in the alongside figure.
128 The Leading Maths - 8 r 1 2 3 4 5 6 7 8 9 10 11 12 13 πr πr r What are the (approximate) height and width of the rectangle? The height is the circle’s radius: just look at sectors 1 and 13 above. When they were in the circle, they were “radius” high. The width (actually one “bumpy” edge) is half of the curved parts around the circle. In other words, it is about half the circumference of the circle. We know that: Circumference = 2 × π × r And so the width is about, Half the circumference = 1 2 c = 1 2 × 2πr = πr And so we have (approximately): 1 2 3 4 5 6 7 8 9 10 11 12 13 Radius π × Radius Now, we just multiply the width by the height to find the area of the rectangle: Area of circle (A) Area of rectangle = w × h = π × r = πr2 sq. units We know that r = d 2 = c 2π ∴ A = πr2 Then, A = π 4 d2 sq. units
MENSURATION 129 And, A = πr2 = c 2π 2 = π c2 4π2 = 1 4π c2 sq. units. CLASSWORK EXAMPLES Example : 1 Find the area of the circle if its radius is 15 cm . Solution: Here, in the given circle, radius (r) = 15 cm Now, we have, Area of the circle (A) = πr2 = 3.14 × 15 × 15 = 706.50 cm 2 Example : 2 A circular plate of the diameter 14 cm rolls on the ground for 15 revolutions about the circumference. (a) Find the circumference of the plate. (b) Find the distance covered by it. Solution: Here, diameter of the plate (d) = 14 cm (a) We have, Circumference of the circular plate (c) = πd = 22 7 × 14 = 154 cm (b) Distance traveled in 15 revolutions = 15 × 154 cm = 2310 cm = 23.10 m Example : 3 In eh given circle, the circumference is 44 cm . (a) Find the radius of the circle. (b) Find the diameter of the circle. (c) Find the area of the circle. r = 15 cm 44 cm
130 The Leading Maths - 8 Solution: Here,, in the given circle. (a) Circumference (c) = 44 cm 2 or, 2πr = 44 or, 2 × 22 7 × r = 44 or, r = 44 × 7 2 × 22 ∴ r = 7 cm (b) Diameter of the circle (d) = 2r = 2 × 7 = 14 cm. (c) Area of the circle (A) = πr2 = 22 7 × 7 × 7 = 154 cm 2 Example : 4 Find the radius and the circumference of the circle if it has an area of 114.74 cm2 . (a) What is the radius of the circle ? Find it. (b) Find the length of the diameter of the circle. (c) Find the length of the circumference of the circle. (d) Establish the relation between the circumference and diameter of the circle. Solution: (a) Here, area of the circle (A) = 114.74 cm 2 or πr2 = 114.74 or 22 7 × r2 = 114.74 or r 2 = 114.74 × 7 22 or, r2 = 824.18 22 ∴ r = 824.18 22 = 6.12 cm 44 cm
MENSURATION 131 (b) Diameter of the circle (d) = 2r = 2 × 6.12 = 12.24 cm. (c) Circumference of the circle (c) = 2πr = 2 × 22 7 × 824.18 22 = 38.47 cm. (d) Relation between circumference and diameter of the circle; c d = 38.47 12.24 = 3.1429 = π. Hence, c = πd. EXERCISE 7.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What type of plane shape is called circle? (b) Define circle. Write the name of any five objects in circular shapes. (c) Draw a circle and label the parts center, radius, diameter and circumference. (d) Write the difference between arc and chord of a circle. (e) What is the difference between sector and segment of a circle? (f) What is the area of the circle of radius 7 cm? (g) What is the circumference of the circle with diameter 7 cm? (h) What is the area of the circle having diameter 21 cm? 2. Circle ( ) the correct answers. (a) What is the relation between radius and circumference of a circle? i. d = 2r ii. c = πr iii. 2πr = c iv. c = πr 2 (b) What is the formula to calculate the area of a circle having radius 'r'? i. A = 2πr2 ii. A = πr2 iii. A = πd2 2 iv. 3πr 2 (c) Which formula is need to calculate the area of a circle in term of its diameter 'd'? i. A = 1 4 πd2 ii. A = πd2 iii. A = 4π d2 iv. A = 4πd2 (d) What is the area of the circle with radius 14 cm2 ? i. 288 cm2 ii. 616 cm iii. 44 cm2 iv. 616 cm2 (e) What is the area of the circle with diameter 21 cm? i. 345.6 cm2 ii. 436.5 cm2 iii. 346.5 cm2 iv. 1386 cm2
132 The Leading Maths - 8 3. Observe the radius of the circle given below and answer the following questions. Take π = 22 7 (i) 7 cm (ii) 14 cm (iii) 2.1 cm (iv) 8.4 cm (a) Write the formula to find the area of the circle. (b) Find the length of the circumference of the circle. (c) Calculate the area of the circle. 4. Find the area and circumference of the following circles of given diameter. Take π = 22 7 (i) 14 cm (ii) 2.8 cm (iii) 5.6 cm (iii) 4.2 cm (a) Write the formula to find the area of the circle. (b) Find the length of the area of the circle. (c) Calculate the circumference of the circle. 5. Find the area and circumference of the following circles: (Take π = 3.14) (a) radius = 14 cm (b) diameter = 2.8 cm (c) radius = 2.1 cm (d) diameter = 2.1 cm 6. Find the area of the circle having the given circumference. Take π = 22 7 (a) 44 cm (b) 88 cm (c) 35.20 cm (d) 39.60 cm (e) 30.08 cm (f) 396 cm 7. Find the radius and circumference of a circle if its area is given to be. (Take π = 3.14) (a) 50.24 cm 2 (b) 28.26 cm 2 (c) 2122.64 cm 2 (d) 6.16 cm 2 (e) 1386 cm 2 (f) 13.86 cm 2 8. A circular plate has a diameter of 21 cm. (a) Find the area of the plate. (b) Find its circumference. (c) Find the distance covered by it in 14 revolution about the circumference. 9. A circular one-rupee coin has the diameter of 2.1 cm. (a) Find the area of the coin. (b) Find its circumference. (c) Find the distance covered by it in 50 revolution about the circumference
MENSURATION 133 10. (a) The circumference of the trunk of a tree is 396 cm. Find the area of the circular cross-section of the trunk. (b) The circumference of the trunk of a tree is 704 cm. Find the radius and area of the circular cross-section of the trunk. 11. (a) The circumference of a circular pond is 88 m. What is the area of the pond? (b) The diameter of a circular bangle is 3.5 cm. Find its area. (c) The circumference of the upper and lower roundness of the Nepali stool are 16 inches and 9 inches respectively. Find the difference between both roundness of the stool. ANSWERS 3. (i) (b) 44 cm (c) 154 cm2 (ii) (b) 88 cm (c) 616 cm2 (iii) (b) 13.2 cm (c) 13.86 cm2 (iv) (b) 52.8 cm 221.76 cm 4. (i) (b) 154 cm2 (c) 44 cm (ii) (b) 6.16 cm2 (c) 8.8 cm (iii) (b) 24.64 cm2 (c) 17. 6 cm (iv) (b) 13.86 cm2 (c) 13.2 cm 5. (a) 615.44 cm2 , 87.92 cm (b) 6.15 cm2 , 8.79 cm (c) 13.85 cm2 , 13.19 cm (d) 3.46 cm2 , 6.59 cm 6. (a) 154 cm2 (b) 616 cm2 (c) 98.56 cm2 (d) 124. 74 cm2 (e) 71.97 cm2 (f) 12474 cm2 7. (a) 4 cm, 25.12 cm (b) 3 cm, 18.84 cm (c) 26 cm, 163.28 cm (d) 1.4 cm, 8.8 cm (e) 21 cm, 131.94 cm (f) 2.1 cm, 13.19 cm 8. (a) 1386 cm2 (b) 132 cm (c) 1848 cm or 18.48 m 9. (a) 9.625 cm2 (b) 11 cm (c) 550 cm or 5.5 m 10. (a) 126 cm (b) 112 cm 11. (a) 616 m2 (b) 9.625 m2 (c) 136.92 sq. inches Project Work 7.2 1. Cut down a circle with the radius more than 5 cm from the chart paper using a compass and record its radius. Cut down the circle into 16 parts and also cut one of them half. Now, arrange these all pieces into rectangular form. Derive the formula of the circle using the formula of the area of the rectangle. Justify the area of the circle using recorded radius of the compass is the same area of the rectangle. Prepare a report and present it in your classroom. 2. Take a spherical fruit and cut down it vertically or horizontally in half. Measure the circumference of the cross-section of the half fruit using thread and find its radius. Calculate the area of the cross-section. Prepare a report and present it in your classroom.
134 The Leading Maths - 8 1. In the given figure of circle, there are two triangles, in which ABC is an equilateral triangle and PQR is an isosceles triangle with measurement. (a) Write the formula for finding the area of isosceles triangle. (b) Find the area of ΔABC. (c) How much area of ΔPQR is greater or less than the area of ΔABC? Calculate it. (d) All diameters are chord, but all chords are not diameter. Justify it. 2. In the figure of parallelogram, trapezium is inside in it. (a) Write the formula for finding the area of trapezium. (b) Write any two properties of parallelogram. (c) Find the area of the shaded part. (d) How much area of the parallelogram is greater than the area of trapezium? 3. Here are given two figures; rhombus and parallelogram. The lengths of diagonals in a rhombus are 6 cm and 8 cm respectively Base and height of parallelogram are 6 cm and 4 cm respectively. (a) Write the formula for finding the area of parallelogram. (b) Find the area of parallelogram. (c) Are the area of parallelograms and rhombus equal? Justify it. (d) Why rhombus is not a regular polygon? A B 6 cm C Q 6 cm 7 cm 7 cm P R12.5 cm 18.9 cm 6 cm 8 cm 5 cm A B Rhombus C D Parallelogram A C B D MIXED PRACTICE–III
MENSURATION 135 4. In the given trapezium, ABDC is a parallelogram, where AB = 6 cm, CF = 9 cm, BF = 4 cm. (a) Write the formula for finding the area of trapezium. (b) Find the area of Δ AEC. (c) Find the area of quadrilateral ABDE. (d) How many times the area of trapezium is larger than the area of Δ BDF? 5. In the given figure, O is the center of the circle, ADMB is a trapezium and ABCD is a parallelogram. D E M C A O B Where, OD = 7 cm, DE = 4 cm and DM = 5 cm. (a) Write the formula for finding the area of circle. (b) Find the area of parallelogram ABCD. (c) How much area of trapezium is greater or less than the area of parallelogram? Compare it. (d) ΔODM is an isosceles triangle. Justify it. C E D F 6 cm B 9 cm 4 cm A ANSWERS 1. (b) 15.59 cm2 (c) greater by 3.38 cm2 2. (b) 201.25 cm2 3. (b) 24 cm2 4. (b) 6 cm2 (c) 18 cm2 (d) 5 times 5. (b) 56 cm2 (c) 18 cm2
136 The Leading Maths - 8 FM : 20 Time : 40 Min. CONFIDENCE LEVEL TEST III MENSURATION 1. Observe the adjoining plane field. (a) Which formula do you use to find the area of a square. [1] (b) Find the area of the shape B. [1] (c) How much area does a farmer plough the field ? Find it. [2] (d) Find the ratio of the areas of A and B. [1] 2. The shape of the door of a hall is the shape of a rectangle with a semicircle over it as shown in the figure. (a) Which formula do you use to find the area of a circle? [1] (b) Find the perimeter of the door. [1] (c) How much area does a painter paint the door? Find it. [2] (d) Find the ratio of the area of the rectangular part [1] 3. Observe the adjoining figure. (a) Write the formula to find the area of an isosceles triangle. [1] (b) What is the length of the circumference of the circular shape ? Find it. [1] (c) Write the relation between radius of the circle and side of the equilateral triangle PQR. [1] (d) If the area of the cricket ground is 7.546 × 105 sq. ft, what will be its circumference? Find it. [2] 4. Observe the adjoining figure in which ABCD is a parallelogram and PQRS is a square. (a) Write the formula to find the area of a rhombus. [1] (b) Find the perimeter of the parallelogram ABCD. [1] (c) What is the area of the shaded portion in the figure? Find it. [2] (d) Compare the perimeters of the parallelogram ABCD and the square PQRS.[1] 20 m 14 m A B 20 m 4 ft 6 ft 8 in P Q R O 8 3 in 30 m PS R A B C D Q 6 m 6 m 20 m
ALGEBRA 137 COMPETENCY Solution of the behaviour problems related to algebraic expression CHAPTERS 8. Indices 9 Algebraic Expression 10. Equation, Inequality and Graph LEARNING OUTCOMES simplify by using laws of indices, factorise the algebraic expression, find the Highest Common Factor (HCF) and Lowest Common Multiples (LCM) of the given algebraic expressions, introduce the algebraic fraction, simplify the algebraic expression, solve the simultaneous equation of two variables, solve the quadratic equation. ALGEBRA UNIT IV Estimation Teaching Hours - 40 (Th. + Pr.)
138 The Leading Maths - 8 CHAPTER 8 INDICES Lesson Topics Pages 8.1 Simplification of Indices 139 What is the product of 4 and 8? Can you write it in powering form? What is the value of a × a × a? What is index? What are the values of x3 × x2 and x4 ÷ x2 ? Is a x × ay = ax + y correct? Why ? Give reason with suitable example. WARM-UP Dengue Virus
ALGEBRA 139 8.1 Simplification of Indices At the end of this topic, the student will be able to: ¾ simplify the algebraic expression by using the laws of indices. Learning Objectives I. Introduction 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 invented new notation which involves writing down the number (variable) being multiplicated together with the number of times the number is multiplied by itself. It 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 (or exponent or index) of 'a'. The power an is read as a to the power or"a raised to the nth power". The number 'a' which is called the factor or base. The number of repeated times 'n' the number multiples is called the index (plu. indices) or exponent. The laws of exponents or Rules of combining indices: Law : I Multiplying Indices of the Same or Equal Base am an = am + n, m + n ∈Z Proof: Multiply powers of equal base am .an (m and n are natural numbers) am.. an = m + n times (a.a...a) m times (a.a...a) n times . = am + n When 2 is added at 10 times, what do you find? Discuss about 2 × 10 and 210 n ← Exponent a ← Base
140 The Leading Maths - 8 e.g., (43 ) (42 ) = (3 + 2) times 5 times (4. 4. 4) 3 times (4. 4) 2 times ∴ (43 ) (42 ) = 43+2 = 45 Law : II Dividing Indices of the Same or Equal Base 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 Proof: Since am = a × a × a × .... × a (m factors) and an = a × a × a × .... × a, (n factors) am an = a.a.a...m factors a.a.a...n factors = a.a.a...(m – n) factors = am – n. The proof of the second part is left as an exercise. Again, for n > m am ÷ an = am an = a.a.a...m factors a.a.a...n factors = 1 a.a.a...(n – m) factors = 1 an – m eg, 45 ÷ 43 = 45 – 3 = 42 , 45 ÷ 45 = 45 – 5 = 40 = 1, 43 ÷ 45 = 1 ÷ 45 – 3 = 1 42
ALGEBRA 141 Law : III Distributing power on Different Bases 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. Proof: Clearly, (ab)m = (ab). (ab)...(ab) (m factors) = (a × a × a × ... m factors) (b × b × b × m factors) = ambm. The proofs of the rest are left as exercises. e.g., (– 3a)2 = (– 3)2 (a)2 = 9a2 , 4 5 3 = 43 53, (5a2 )3 = 53 (a2 )3 = 125 a2 × 3 = 125 a6 . We shall now 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–m as 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.a1 n.a1 n... n times = (a 1 n)n = an n = a1 = a.
142 The Leading Maths - 8 Hence, we can define a quantity like am n by a m 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.” CLASSWORK EXAMPLES Example : 1 (a) Find the value of 25 × 23 . (b) Simplify a3 × a4 . Solution: (a) 25 × 23 = 25 + 3 = 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 Example : 2 (a) Simplify: 27 ÷ 23 (b) Divide : a5 ÷ a2 Solution: (a) 27 ÷ 23 = 27 – 3 = 24 = 2 × 2 × 2 × 2 = 16 (b) a5 ÷ a2 = a5 – 2 = a3 Example : 3 (a) Evaluate: (32 )3 (b) Simplify: (a4 )5 Solution: (a) (32 )3 = 32 × 3 = 36 = 3 × 3 × 3 × 3 × 3 × 3 = 729 (b) (a4 )5 = a4 × 5 = a20 (b) a3 × a4 = a3 + 4 = a7
ALGEBRA 143 Example : 4 (a) Evaluate : 2 3 × 6 9 3 (b) Simplify : x y . z x 4 Solution: (a) 2 3 × 6 9 3 = 2 3 3 × 6 9 3 = 2 3 × 2 3 × 2 3 × 6 9 × 6 9 × 6 9 = 64 729 (b) x y . z x 4 = x y 4 z x 4 = x.x.x.x y.y.y.y × z.z.z.z x.x.x.x = z4 y4 = z y 4 Example : 5 (a) Find the quotient of 34 ÷ 34 (b) Evaluate: am ÷ am Solution: (a) 34 ÷ 34 = 34 – 4 = 30 = 1 (b) am ÷ am = am–m = a0 = 1 Example : 6 Divide and express the answer in index form. (a) 37 ÷ 39 (b) m5 ÷ m9 Solution: (a) 37 ÷ 39 = 37 – 9 = 3–2 = 1 32 (b) m5 ÷ m9 = m5 – 9 = m– 4 = 1 m4 Example : 7 (a) Evaluate 27 3 . (b) Find the value of 82 3. Solution: (a) 27 3 = (27) 1 3 = (33 ) 1 3 = 33 × 1 3 = 31 = 3 (b) 8 2 3 = 2 3 × 2 = 22 = 4
144 The Leading Maths - 8 Example : 8 (a) Which is greater 230 or 320 ? (b) Find the value of 23 2 – (23 )2 . Solution: (a) 230 = (23 )10 = 810 320 = (32 )10 = 910 Since, 9 > 8, 910 > 810 ∴ 320 > 230 (b) 23 2 – (23 )2 = 29 – 26 = 512 – 64 = 448 Example : 9 Simplify: xa xb c . xb xc a . xc xa b Solution: Here, xa xb c × xb xc a × xc xa b = xca xbc × xab xca × xbc xab = xca – bc × xab – ca × xbc – ab = xca – bc + ab – ca + bc – ab = x0 = 1 EXERCISE 8.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Write the following in the exponential form by using index notation. i. 3 × 3 × 3 × 3 ii. 81 (b) Write the following in the product form. i. 8a3 ii. 64 (c) What is index? Define.