ARITHMETIC 201 Fifteen children together have Rs. 1120 and 50 paisa. How much will each get if they share it equally ? Solution : Given, total amount = Rs. 1120 and 50 paisa No. of friends = 15 Now, Divide : 2874.69 ÷ 63 Solution : Here, 45.63 63 2874.69 –252 354 –315 396 –378 189 –189 × \ 2874.69 ÷ 63 = 45.63 15.4 Division of Rupees and Paisa Read, Think and Learn 74 70 15 Rs. P 1120 50 –105 70 50 –60 10 50 –10+1000 × 1050 –1050 00 74 .70 15 1120.50 –105 70 –60 105 –105 × 0 0 × Divide like as whole numbers. Put the dot (.) in the quotient exactly in the position of the dividend. 10 < 15. So, Rs. 10 converts into paisa. Alternatively
202 The Leading Mathematics - 4 1. Divide : EXERCISE 15.4 Your mastery depends on practice. Practice like you play. 2. Divide : (a) Rs. 47.65 ÷ 5 (b) Rs. 164.50 ÷ 7 (c) Rs. 370.8 ÷ 9 (d) Rs. 242.40 ÷ 8 (e) Rs. 46.13 ÷ 7 (f) Rs. 141.60 ÷ 6 (g) Rs. 2273.20 ÷ 4 (h) Rs. 2574.90 ÷ 3 (i) Rs. 999.60 ÷ 12 (j) Rs. 377.60 ÷ 16 (k) Rs. 1573.60 ÷ 28 (l) Rs. 1924.40 ÷ 34 (m) Rs. 8102.90 ÷ 23 (n) Rs. 2086.40 ÷ 32 (o) Rs. 9668.40 ÷ 42 (p) Rs. 1482.25 ÷ 55 (q) Rs. 6609.20 ÷ 26 (r) Rs. 5033.10 ÷ 38 3. Divide : (a) Rs. 4.5 by 3 (b) Rs. 226 by 5 (c) Rs. 366.10 by 7 (d) Rs. 215.10 by 9 (a) Rs. P (b) (c) 3 24 60 Rs. P 6 38 88 Rs. P 12 50 64 (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) Rs. P 15 80 40 Rs. P 45 654 75 Rs. P 36 839 16 Rs. P 58 2529 38 Rs. P 21 29 82 Rs. P 58 773 72 Rs. P 42 950 46 Rs. P 62 3163 58 Rs. P 32 63 68 Rs. P 33 824 34 Rs. P 55 980 65 Rs. P 45 9654 75
ARITHMETIC 203 Study the bill of some goods. List the total amount and price of each item from the amount of more items. Prepare a report and present it in your classroom. PROJECT WORK (e) Rs. 278.30 by 11 (f) Rs. 782.40 by 12 (g) Rs. 2499.20 by 16 (h) Rs. 9569.30 by 17 (i) Rs. 7814.40 by 24 (j) Rs. 7176.60 by 27 (k) Rs. 9432.50 by 35 (l) Rs. 6102 by 45 15.5 Word Problems on Multiplication and Division of Money Read, Think and Learn The cost of a calculator is Rs. 290.49. Find the cost of 7 such calculators. Solution : Here, Cost of 1 calculator = Rs. 290.45 \ Cost of 7 calculators = Rs. 290.45 × 7 = Rs. 2033.15 Thus, the cost of 7 calculators is Rs.2033.15. Distribute Rs. 525 and 75 paisa among 5 students. Solution : Here, Total amount = Rs. 525 P75 = Rs. 525.75 No. of students = 5 2 6 0 3.3 Rs. 290.45 × 7 Rs. 2033.15
204 The Leading Mathematics - 4 1. If the cost of a copy is Rs. 26.75, what is the cost of 5 copies ? 2. A book costs Rs. 125.75. Find the cost of 1 dozen of books. 3. How much amount is paid for 17 balls at the rate of Rs. 750.50 ? 4. The cost of 3 kg of floor is Rs. 78.75. Find the cost of 1 kg of floor. 5. 7 books cost Rs. 669.34. Find the cost of each book. 6. Distribute Rs.225 and 75 paisa among 15 students. How much does each student get? 7. There are Rs. 2275 and 70 paisa in a bag. If 35 boys equally distribute the money among them, how much does each boy get ? 8. The cost of 5 geometric boxes is Rs. 1228.25. Find the cost of each geometric box. Also, find the cost of 8 such geometric boxes. 9. If Kajal paid Rs. 1795.15 for 7 plates, how much did each plate cost ? Also, how much did she pay for such 18 plates ? Rs. 105.15 5 Rs. 525.75 – 5 025 – 25 07 –5 25 – 25 × EXERCISE 15.5 Your mastery depends on practice. Practice like you play. Now, Hence, each student gets Rs. 125P15.
ARITHMETIC 205 CHAPTER 16 Distance Lesson Topics Pages 16.1 Measurement of Length 206 16.2 Conversion of Length 208 16.3 Addition of Length 210 16.4 Subtraction of Length 213 16.5 Verbal Problems on Length 215 Can you read the scales in the above ruler in mm, cm and inches? What is the length of a pencil in cm and inches ? Which units of distance are used to measure the length of rope? Which units of distance are used to measure the length of highway? What is the length of BP Highway from Dhulikhel to Bardibas ? If you travel this BP Highway in 4 hours by non-stop driving by bike, how much distance do you travel in 1 hour ? What are the conversion of the units of distance between mm, cm, m and km? WARM-UP Measurement 137 The length of the pencil is more than 4 cm and less than 5 cm. To find the more accurate measure; we have to see the smaller units in which the cm is divided. A cm is divided into 10 equal parts. Each part is a millimetre (mm). Using cm and mm we can say approximately that the length of the pencil is 4 cm 7 mm. We have, 10 mm = 1 cm To measure the length of a rope, it will not be wise to use mm or cm. We use a meter scale to measure it. In metre, there are 100 cm. Therefore we have, 100 cm = 1 m To measure the distance from Pokhara to Kathmandu; it would be unwise to use a meter scale. We use a longer scale called the kilometre (km). 1 kelometre has 1000 m. 1000 m = 1 km CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6 5 4 2 3 1 Inches If we see through magnifying glass CHAPTER - 5.3 : DISTANCE 5.3 (A) Measurement of Length Read, Think and Speak 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6 5 4 3 2 1
206 The Leading Mathematics - 4 The length of the pencil is more than 4 cm and less than 5 cm. To find the more accurate measure; we have to see the smaller units in which the cm is divided. A cm is divided into 10 equal parts. Each part is a millimetre (mm). Using cm and mm we can say approximately that the length of the pencil is 4 cm 7 mm. We have, 10 mm = 1 cm To measure the length of a rope, it will not be wise to use mm or cm. We use a meter scale to measure it. In metre, there are 100 cm. Therefore we have, 100 cm = 1 m To measure the distance from Pokhara to Kathmandu; it would be unwise to use a meter scale. We use a longer scale called the kilometre (km). 1 kelometre has 1000 m. 1000 m = 1 km Measurement 137 The length of the pencil is more than 4 cm and less than 5 cm. To find the more accurate measure; we have to see the smaller units in which the cm is divided. A cm is divided into 10 equal parts. Each part is a millimetre (mm). Using cm and mm we can say approximately that the length of the pencil is 4 cm 7 mm. We have, 10 mm = 1 cm To measure the length of a rope, it will not be wise to use mm or cm. We use a meter scale to measure it. In metre, there are 100 cm. Therefore we have, 100 cm = 1 m To measure the distance from Pokhara to Kathmandu; it would be unwise to use a meter scale. We use a longer scale called the kilometre (km). 1 kelometre has 1000 m. 1000 m = 1 km CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6 5 4 2 3 1 Inches If we see through magnifying glass CHAPTER - 5.3 : DISTANCE 5.3 (A) Measurement of Length Read, Think and Speak Measurement 137 The length of the pencil is more than 4 cm and less than 5 cm. To find the more accurate measure; we have to see the smaller units in which the cm is divided. A cm is divided into 10 equal parts. Each part is a millimetre (mm). Using cm and mm we can say approximately that the length of the pencil is 4 cm 7 mm. We have, 10 mm = 1 cm To measure the length of a rope, it will not be wise to use mm or cm. We use a meter scale to measure it. In metre, there are 100 cm. Therefore we have, 100 cm = 1 m To measure the distance from Pokhara to Kathmandu; it would be unwise to use a meter scale. We use a longer scale called the kilometre (km). 1 kelometre has 1000 m. 1000 m = 1 km CM 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6 5 4 2 3 1 Inches If we see through magnifying glass CHAPTER - 5.3 : DISTANCE 5.3 (A) Measurement of Length Read, Think and Speak If we see through magnifying glass 16.1 Measurement of Length Read, Think and Learn 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6 5 4 3 2 1
ARITHMETIC 207 Determine the scale that you will use to measure the following objects and estimate their lengths : EXERCISE 16.1 Your mastery depends on practice. Practice like you play. 138 Allied Mathematics-4 Determine the scale that you will use to measure the following objects and estimate their lengths : (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) EXERCISE 5.3 (A) Your mastery depends on practice. Practice like you play. (a) (d) (g) (j) (b) (e) (h) (k) (c) (f) (i) (l)
208 The Leading Mathematics - 4 We know, 10 mm = 1 cm \ 1 mm = 1 10 cm = 1 ÷ 10 cm = 0.1 cm 100 cm = 1m \ 1 cm = 1 100 m = 1 ÷ 100 m = 0.01 m 1000 m = 1km \ 1 m = 1 1000 km = 1 ÷ 1000 km = 0.001 km Conversion chart of length 16.2 Conversion of Length Read, Think and Learn 1m = 100 cm 1cm = 10 mm × 1000 km m 100÷ × 100 cm 100 ÷ × 10 mm 10 ÷ 1km = 1000 m Example 1 Convert 3.55 m into cm. Solution : Here, 3.55m = 3.55 × 1m = 3.55 × 100 cm = 355.00 cm = 355 cm CLASSWORK EXAMPLES
ARITHMETIC 209 1. Convert the following scales into millimetres. (a) 4 cm (b) 7 cm (c) 5.2 cm (d) 8.9 cm (e) 9 cm 3 mm (f) 12 cm 7 mm (g) 1 cm 9 mm (h) 10 cm 6 mm (i) 3 m (j) 4 m 12 cm (k) 7 m 57 cm (l) 15 m 79 cm (m) 2 m 4 cm 3 mm (n) 5 m 14 cm 5 mm (o) 12 m 7 cm 9 mm 2. Convert into centimetres : (a) 3 m (b) 12 m (c) 1 2 m (d) 15 1 2 m (e) 25.25 m (f) 137.37 m (g) 1 km (h) 5 km 257 m (i) 1400 mm (j) 2040 mm (k) 12460 mm (l) 245375 mm (m) 20 cm 4 mm (n) 45 cm 7 mm (o) 67 cm 9 mm 3. Convert the following length into metres : (a) 2 km (b) 13 km (c) 1 2 km (d) 141 2 km (e) 17.25 km (f) 23.015 km (g) 27 km 7 m (h) 35 km 375 m (i) 1400 cm EXERCISE 16.2 Your mastery depends on practice. Practice like you play. Example 1 Change 52750 m into km and m. Solution : Here, 52750 m = 52750 ÷ 1000 km = 52 km 750 m Thus, changing 52750 m = 52.750 km 52km 1000 52750 –5000 2750 –2000 750m
210 The Leading Mathematics - 4 16.3 Addition of Length Read, Think and Learn (j) 27500 cm (k) 34 m 27 cm (l) 42 m 15 cm (m) 232000 mm (n) 354300 mm (o) 436700 mm 4. Convert the following length into kilometres and metres: (a) 3000 m (b) 23000 m (c) 25400 m (d) 5275 m (e) 17235 m (f) 102501 m (g) 253493 m (h) 324657 m (i) 200000 cm (j) 245000 cm (k) 356008 cm (l) 435678 cm (m) 5647382 cm (n) 6453827 cm (o) 7684658 cm (p) 8765836 cm (q) 3000000 mm (r) 4500000 mm Measure the length and breadth of any ten materials in m, cm and mm that are found in your bedroom. Convert these measures in cm. Prepare a report and present it in your classroom. PROJECT WORK Add 245 m 46 cm and 136 m 87 cm. Solution : Alternatively Here, 245 m 46 m = 245.46 m 136 m 87 cm = 136.87 m Now, m cm 1 0 11 0 245 4 6 +136 8 7 381 133 + 1 –100 382 3 3 11 1 245.46m + 136.87m 382.33m
ARITHMETIC 211 Find the sum of 46 cm 7 mm and 25 cm 6 mm. Solution : Here, Alternatively cm mm 1 1 4 6 7 + 25 6 7 2 3 Find the sum of 4 km 675 m, 3 km 950 m and 2 km 865 m Solution : Here, km m 0 11 0 4 675 3 950 + 2 865 9 2490 + 2 –2000 11 490 Alternatively km m 2 11 0 4 675 3 950 + 2 865 11 490 cm mm 4 6 7 + 25 6 7 1 1 3 + 1 –10 7 2 3 1. Add the following : EXERCISE 16.3 Your mastery depends on practice. Practice like you play. cm mm 4 7 + 3 3 cm mm 1 4 6 + 7 5 cm mm 3 5 8 +17 5 cm mm 9 2 7 +15 4 (a) (b) (c) (d)
212 The Leading Mathematics - 4 m cm 1 4 5 4 + 4 2 1 m cm 2 6 7 8 + 5 2 4 m cm 5 8 9 2 +17 2 8 m cm 6 7 8 4 +39 2 9 km m 5 250 + 4 021 m cm 2 7 275 + 8 826 m cm 5 8 378 +28 825 m cm 425 480 +348 675 34.5cm + 18.8cm 47.7cm + 12.7cm 58.4cm + 27.7cm 94.7cm + 4.6cm 24.36m +17.74m 87.84m +67.24m 92.99m +7.02m 127.98m +87.87m 2. Find the sum of : (a) 37 cm 4 mm and 19 cm 7 cm (b) 27 cm 4 mm, 18 cm 3 mm and 9 cm 9 mm (c) 78 m 4 cm and 69 m 25 cm (d) 144 m 27 cm, 27 m 85 cm and 77 m 27 cm (e) 12 km 270 m and 9 km 875 m (f) 578 km 270 m, 107 km 50 m and 95 km 5m (e) (f) (g) (h) (m) (n) (o) (p) (i) (j) (k) (l) (q) (r) (s) (t)
ARITHMETIC 213 16.4 Subtraction of Length Read, Think and Learn Subtract 25 m 27 cm from 42 m 12 cm. Solution : Here, Find the difference of 40 cm 8 mm and 29 cm 9 mm. Solution : Here, Find the difference of 712 km 2 m and 585 km 875 m. Solution : Here, m cm 3 110 1012 4 2 1 2 –25 2 7 1 6 8 5 Alternatively Here, 25 m 27 cm = 25.27 m 42 m 12 cm = 42.12 m Now, Alternatively Alternatively 311 1012 42.12m – 25.27m 16.85m cm mm 9 310 18 4 0 8 –29 9 1 0 9 40 cm 8 mm = 40.8 cm 29 cm 9 mm = 29.9 cm 712 km 2 m = 712.002 km 585 km 875 m = 585.875 km 319 18 12 40.8 cm 29.9 cm 10.9 cm km m 6 10 110 9 9 12 712 002 –585 875 126 127 6 10 1109 9 1212 712.002 km 585.875 k m 2 m is 002 m in 126.127 k m three digits.
214 The Leading Mathematics - 4 1. Subtract the following : cm mm 4 7 – 3 3 cm mm 1 4 6 – 7 5 cm mm 3 5 8 – 1 7 5 cm mm 9 2 7 – 1 5 4 m cm 1 4 5 4 – 4 2 1 m cm 2 6 7 8 – 5 2 4 m cm 5 8 9 2 – 17 2 8 m cm 6 7 8 4 – 3 9 2 9 km m 5 250 – 4 021 m cm 2 7 275 – 8 826 m cm 5 8 378 – 28 825 m cm 425 480 – 348 675 34.5cm – 18.8cm 47.7cm – 12.7cm 58.4cm – 27.7cm 94.7cm – 4.6cm 24.36m – 17.74m 87.84m – 67.24m 92.99m – 7.02m 127.98m – 87.87m EXERCISE 16.4 Your mastery depends on practice. Practice like you play. (a) (b) (c) (d) (e) (f) (g) (h) (m) (n) (o) (p) (i) (j) (k) (l) (q) (r) (s) (t)
ARITHMETIC 215 If 2 m and 28 cm long rope and 1 m and 95 cm long rope are joined, how long will be the rope in all ? Solution : Here, Line up m and cm and add them. m cm 10 1 2 2 8 + 1 9 5 4 2 3 \ The rope will be 4 m 23cm 16.5 Verbal Problems on Length Read, Think and Learn Measure the length, breadth height of your bedroom in m, cm and mm. Which is longer and which is shorter by how many? Find it. Also, prepare a report by sketching the room and present it in your classroom. PROJECT WORK 2. Find the difference of : (a) 37 cm 4 mm and 19 cm 7cm (b) 27 cm 4 mm, 18 cm 3 mm and 9 cm 9 mm (c) 78 m 4 cm and 69 m 25 cm (d) 144 m 27 cm, 27 m 85 cm and 77 m 27 cm (e) 12 km 270 m and 9 km 875 m (f) 578 km 270 m, 107 km 50 m and 95 km 5 m
216 The Leading Mathematics - 4 1. Study the following questions : (a) What is the total length of pencil and pen ? (b) Find the total length of book and geometric box when they attach together. (c) Find the sum of the length of stick and thread. (d) How long is the thread then the stick ? 146 Allied Mathematics-4 16cm 3mm 62cm 6mm 14cm 2mm 62cm 6mm 22cm 1mm 85cm 1mm 1. Study the following questions : (a) What is the total length of pencil and pen ? (b) Find the total length of book and geometric box when they attach together. (c) Find the sum of the length of stick and thread. (d) How long is the thread then the stick ? (e) How length of much geometric box is less than the book ? (f) Which is longer pencil or pen and by how much? 2. A wall is 25m and 75cm long and another wall is 18m and 87 cm long. What is the total length of the two walls ? 3. The length of a wall is 256 m and 56 cm long. If 95m and 79cm of the wall is cracked by the earthquake of 12 Bhaishakh, 2072, how long of the remaining of the wall is good ? 4. The length of Siddhartha highway from Lumbini to Pokhara is 181 km 20 m and the length of Prithvi Highway from Pokhara to Kathmandu is 203 km 423m. How long road does a man travel to reach Kathmandu from Lumbini ? 5. Karnali Highway is 232 kilometre long to be constructed 156 kilometre and 256 metre is constructed by Nepal Army in 2014. How long of the highway is left to contracted ? 6. The altitudes of Mt. Everest and Mt. Kanchanjunga are 8 km 849 m and 8km 586 m from the sea level respectively. How high is the Mt. Everest than the Mt. Kanchanjunga ? EXERCISE 5.3 (E) Your mastery depends on practice. Practice like you play. EXERCISE 16.5 Your mastery depends on practice. Practice like you play. Out of 15 m and 20 cm long high tower, 9 m and 75 cm is painted. What length of tower is not painted ? Solution : Here, Subtracting m from m and cm from cm after line up m and cm 14 11 10 15m 20cm –9m 75cm 5 m 45cm \ The 5 m 45 cm of the tower is not painted. The length of 3 pieces of ropes are 3 m 25 cm, 5 m 37 cm and 6 m 77 cm respectively. Find the total length of the ropes. Solution : Here, 1 1 3 m 25cm 5 m 37cm + 6m 77cm 15m 39cm 139 cm = 100 cm + 39 cm = 1 m 39 cm
ARITHMETIC 217 (e) How much shorter is the geometric box than the book ? (f) Which is longer pencil or pen and by how much? 2. A wall is 25 m and 75 cm long and another wall is 18 m and 87 cm long. What is the total length of the two walls ? 3. The length of a wall is 256 m and 56 cm long. If 95m and 79cm of the wall is cracked by the earthquake of 12 Bhaishakh, 2072, how long of the remaining of the wall is good ? 4. The length of Siddhartha highway from Lumbini to Pokhara is 181 km 20 m and the length of Prithvi Highway from Pokhara to Kathmandu is 203 km 423m. How long road does a man travel to reach Kathmandu from Lumbini ? 5. A highway is 232 kilometre long to be constructed 156 kilometre and 256 metre is constructed by a construction company. How long of the highway is left to construct? 6. The altitudes of Mt. Everest and Mt. Kanchanjunga are 8 km 849 m and 8 km 586 m from the sea level respectively. How much higher is the Mt. Everest than the Mt. Kanchanjunga ?
218 The Leading Mathematics - 4 Measurement 147 How much water contains in each of the adjoining vessels ? Can you estimate ? We cannot exactly estimate. So we use the measuring vessels or cylinders as below : 25ml 50ml 100ml 200ml 500ml 1000ml Measuring Vessels Now, pour the water containing in the Karuwa in the measuring vessel of 1l. The water rises in the vessel up to 855ml. So, the Karuwa contains 855 ml of water. This is the capacity of Karuwa. You can identify the capacity of other remaining vessels. The amount of liquid containing in an object is called capacity. CHAPTER - 5.4 : CAPACITY 5.4 (A) Introduction to Capacity Read, Think and Speak CHAPTER 17 Capacity Lesson Topics Pages 17.1 Introduction to Capacity 219 17.2 Conversion of Litre and Millilitre 221 17.3 Addition and Subtraction on Capacity 222 17.4 Word Problems on Capacity 224 What is capacity ? Read the measures of the measuring vessels. How many milliliters are in the measuring jar of 1 litre ? How much water can the above jug contain ? How can we find it? What is the capacity of mineral water bottle ? Do you know, how many bottles of water are needed to fill the empty jar ? What is the capacity of water jar ? What is the capacity of rectified spirit drum ? How much water can the tank of the dimension 1 m × 1 m × 1 m contain ? 1 m 1 m 1 m WARM-UP
ARITHMETIC 219 Measurement 147 How much water contains in each of the adjoining vessels ? Can you estimate ? We cannot exactly estimate. So we use the measuring vessels or cylinders as below : 25ml 50ml 100ml 200ml 500ml 1000ml Measuring Vessels Now, pour the water containing in the Karuwa in the measuring vessel of 1l. The water rises in the vessel up to 855ml. So, the Karuwa contains 855 ml of water. This is the capacity of Karuwa. You can identify the capacity of other remaining vessels. The amount of liquid containing in an object is called capacity. CHAPTER - 5.4 : CAPACITY 5.4 (A) Introduction to Capacity Read, Think and Speak 17.1 Introduction to Capacity Read, Think and Learn How much water is contained in each of the given vessels ? Can you estimate ? We cannot exactly estimate. So, we use the measuring vessels or cylinders as below : Now, pour the water containing in the Karuwa in the measuring vessel of 1l. The water rises in the vessel up to 855ml. So, the Karuwa contains 855 ml of water. This is the capacity of Karuwa. You can identify the capacity of other remaining vessels. The amount of liquid containing in an object is called capacity.
220 The Leading Mathematics - 4 Fill the following vessels or pots with water and estimate the water holding in each vessels or pot. Also, measure accurately the capacity of these vessels or pot by using measuring vessels or cylinders and check your estimation : EXERCISE 17.1 Your mastery depends on practice. Practice like you play. 148 Allied Mathematics-4 Fill the following vessels or pots with water and estimate the water holding in each vessels or pot. Also, measure accurately the capacity of these vessels or pot by using measuring vessels or cylinders and check your estimation : (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (m) (n) (o) (p) EXERCISE 5.4 (A) Your mastery depends on practice. Practice like you play. (a) (e) (i) (q) (m) (b) (f) (j) (r) (n) (c) (g) (k) (s) (o) (d) (h) (l) (t) (p)
ARITHMETIC 221 We know that, 1000 ml = 1 l \ 1 ml = 1 1000 l = 1 ÷ 1000 l = 0.001 l. Conversion chart of capacity Convert 3l and 245ml into millilitre. We know that, 1l = 1000ml , 3l 245 ml = 3 × 1000ml + 245ml = 3000ml + 245ml = 3245ml Convert 3750 ml into litre and millilitre. Solution : We know that 1ml = 1 1000 l \ 3750ml = 3750 × 1 1000 l = 3.750 l = 3l 750ml. × 1000 millilitre litre 1000 % 1l = 1000ml 17.2 Conversion of Litre and Millilitre Read, Think and Learn Example 1 Convert 3l and 245ml into millilitre. Solution : We know that, 1 l = 1000 ml Now, 3 l 245 ml = 3 × 1000 ml + 245 ml = 3000 ml + 245 ml = 3245 ml Example 2 Convert 3750 ml into litre and millilitre. Solution : We know that 1 ml = 1 1000 l \ 3750 ml = 3750 × 1 1000 l = 3.750 l = 3l 750 ml. 3 1000 3750 –3000 750 CLASSWORK EXAMPLES EXERCISE 17.2 Your mastery depends on practice. Practice like you play. 1. Convert the following litres and millilitres into millilitre : (a) 3 l (b) 7 l (c) 2 l (d) 17 l (e) 2 l 273 ml (f) 6 l 205 ml
222 The Leading Mathematics - 4 Add 12l 275ml, 9l 486ml and 6l 879 ml. Solution : Here, l ml 1 1 2 2 1 2 279 9 486 + 6 879 28 644 Arrange l and ml invertical. Add ml and l. Subtract 18l 879ml from 22l 150ml. Solution : Here, 1 11 10 14 10 22 l 150ml –18 l 879ml 03 l 271ml 1 11 10 14 10 22 .150 l –18 . 879 l 03 .271 l "Alternatively" \ The sum is 28l 644ml. \ The difference is 3l 271ml. 17.3 Addition and Subtraction on Capacity Read, Think and Learn Measure the actual capacity of any three vessels in l and ml found at your home. Convert these quantities in to ml. Also, prepare a report about it and present it in your classroom. PROJECT WORK (g) 9 l 180 ml (h) 12 l 50 ml (i) 17 litre 20 ml (j) 23 litre 200 ml (k) 25 litre 9 ml (l) 29 litre 5 ml 2. Change the following capacities into litre and millilitre : (a) 4000 ml (b) 6000 ml (c) 8000 ml (d) 15000 ml (e) 2453 ml (f) 8578 ml (g) 12880 ml (h) 17800 ml (i) 19080 ml (j) 20001 ml (k) 22005 ml (l) 20085 ml
ARITHMETIC 223 1. Add the following : l ml 7 255 + 2 123 l ml 1 6 255 + 8 123 l ml 2 5 815 +14 025 l ml 3 6 205 +29 006 9l 550ml +7l 785ml 19l 455ml +4l 238ml 28l 237ml +3l 873ml 78l 478ml +27l 808ml EXERCISE 17.3 Your mastery depends on practice. Practice like you play. (a) (b) (c) (d) (a) (b) (c) (d) (e) (f) (g) (h) (e) (f) (g) (h) (i) (j) (k) (l) 5.875l + 4.378l 9.279l + 6.125l 12.567l + 3.456l 78.905l + 10.095l 2. Subtract the following : l ml 7 255 – 2 123 l ml 1 6 255 – 8 123 l ml 2 5 815 –14 025 l ml 3 6 205 –29 006 9l 550ml – 7l 785ml 19l 455ml – 4l 238ml 28l 237ml – 3l 873ml 78l 478ml – 27l 808ml
224 The Leading Mathematics - 4 5.875l – 4.378l 9.279l – 6.125l 12.567l – 3.456l 78.905l – 10.095l 3. Find the sum of : (a) 5 l 427 ml and 3 l 879 ml (b) 83 l 87 ml and 73 l 783 ml (c) 6 l 250 ml, 5 l 783 ml and 4 l 375 ml (d) 18 l 758 ml, 15 l 849 ml and 8 l 357 ml 4. Find the difference of : (a) 7 l 155 ml and 3 l 256 ml (b) 89 l 58 ml and 88 l 178 ml (c) 56 l 458 ml and 36 l 897 ml (d) 267 l 75 ml and 258 l 147 ml 17.4 Word Problems on Capacity Read, Think and Learn (i) (j) (k) (l) 4 l 650 ml of water is mixed with 10 l 775 ml of pure milk. Find the total quantity of the mixture of milk and water. Solution : Here, 1 1 Quantity of pure milk = 10l 775ml Quantity of water = + 4l 650ml 15l 425ml \ The quantity of mixture is 15l 425ml.
MEASUREMENT 225 A jumbo cold drink bottle has 2 litre 250 ml of cold drink. 5 boys drink 1 litre 175 ml of the drink. Find the remaining quantity of cold drink. Solution : Here, 1 14 10 Total quantity of drink = 2l 250ml Drank quantity of drink = – 1l 175ml 1l 075ml \ Remaining quantity of drink = 1l 075 ml EXERCISE 17.4 Your mastery depends on practice. Practice like you play. 1. In a vessel, there is 5 l 375 ml of soyabean oil. An oil seller mixes 3 l 835 ml of mixed mustard oil in it. Find the total quantity of mixture of oil. 2. A kettle has the capacity of 3 liter 587 ml and another jug has the capacity of 2 liter and 758 ml. Kettle and jug with full of water are poured into a bucket. What is the amount of water in the bucket ? 3. Look at the vessels and their capacities in the figure. Answer the questions. (a) What is the capacity of the tea cup and juice bottle ? (b) How much more capacity does the kettle have than juice glass ? (c) How many glasses of juice could be served from the juice bottle ? (d) If the juice from the bottle is poured into the cup and glass, what amount of juice is left in the bottle ? (e) If the tea from the kettle is poured into the empty cup, glass and bottle, what amount of tea is left in the kettle ? 152 Allied Mathematics-4 1. In a vessel, there is 5l 375ml of soyabean oil. An oil seller mixes 3l 835ml of mixed mustard oil in it. Find the total quantity of mixture of oil. 2. A kettle has the capacity of 3 litre 587ml and another jug has the capacity of 2 litre and 758 ml. Kettle and jug with full of water are poured into a bucket. What is the amount of water in the bucket ? 3. Look at the vessels and their capacities in the figure. Answer the questions ? (a) What is the capacity of the tea cup and juice bottle ? (b) How much more capacity does the kettle have than juice glass ? (c) How many glasses of juice could be served from the juice bottle ? (d) If juice from the bottle is poured into the cup and glass, what amount of juice is left in the bottle ? (e) If tea from the kettle is poured into the empty cup, glass and bottle, what amount of tea is left in the kettle ? 4l 650ml of water is mixed with 10l 775ml of pure milk. Find the total quantity of the mixture of milk and water. Solution : Here, 1 1 Quantity of pure milk = 10l 775ml Quantity of water = + 4l 650ml 15l 425ml A jumbo cold drink bottle has 2 litre 250 ml of cold drink. 5 boys drink 1 litre 175ml of the drink. Find the remaining quantity of cold drink ? Solution : Here, 0 14 10 Total quantity of drink = 2l 250ml Drank quantity of drink = – 1l 650ml 1l 075ml \ The quantity of mixture is 15l 425ml. \ Remaining quantity of drink = 1l 075ml 225ml 379ml 1l 125ml 2l 25ml 5.4 (D) Word Problems on Capacity Read, Think and Learn EXERCISE 5.4 (D) Your mastery depends on practice. Practice like you play.
226 The Leading Mathematics - 4 CHAPTER 18 Weight Lesson Topics Pages 18.1 Introduction to Weight 227 18.2 Relation between Gram and Kilogram 229 18.3 Relation Between Kilogram and Quintal 231 18.4 Addition and Subtraction of Weight 234 18.5 Verbal Problems on Weight 237 Do you see the instruments of the measuring weight? Do you know their name? Can you guess the weight of your bag with books, note copies, pencil and other goods? Which weight is used to find the weight of the bag with goods? Which weight is used to find the weight of a book ? Which weight is used to find the weight of the note copy and pencil ? Can you measure the weight of the above goods ? Which instruments are used to measure the weights of the goods ? What is your weight ? What is your friend’s weight ? Who is heavier ? WARM-UP
MEASUREMENT 227 Measurement 153 What are the heaviness of the adjoining objects ? We can estimate by taking our hand. Bag is more heavier than others and pencil is lighter than others. From this we cannot take exact heaviness. For calculating exact heaviness of the objects, we use the following weighing blocks and bean balance : 10 gm 20 gm 50 gm 100 gm 200 gm 500 gm 1 kg 2 gm 5 gm 10 kg 20 kg The heaviness of the given objects is given below : The heaviness of the object are called weight. We can measure the weight of the goods by using the following devices; 200 gm 500 gm 20gm 50gm 100 gm The quantity of matter contained in an object when rising upward against the downward force is called weight of the object. CHAPTER - 5.5 : WEIGHT 5.5 (A) Introduction to Weight Read, Think and Speak 10 gm Measurement 153 What are the heaviness of the adjoining objects ? We can estimate by taking our hand. Bag is more heavier than others and pencil is lighter than others. From this we cannot take exact heaviness. For calculating exact heaviness of the objects, we use the following weighing blocks and bean balance : 10 gm 20 gm 50 gm 100 gm 200 gm 500 gm 1 kg 2 gm 5 gm 10 kg 20 kg The heaviness of the given objects is given below : The heaviness of the object are called weight. We can measure the weight of the goods by using the following devices; 200 gm 500 gm 20gm 50gm 100 gm The quantity of matter contained in an object when rising upward against the downward force is called weight of the object. CHAPTER - 5.5 : WEIGHT 5.5 (A) Introduction to Weight Read, Think and Speak 10 gm 18.1 Introduction to Weight Read, Think and Learn What are the weights of the adjoining objects ? We can estimate by taking in our hand, but not the sack of white sugar. The sack of white sugar is more heavier than others and pencil is lighter than others. From this we cannot know exact heaviness. For calculating exact heaviness of the objects, we use the following weighing blocks and bean balance : The heaviness of the given objects is given below : The heaviness of the object is called weight. The quantity of matter contained in an object when rising upward against the downward force is called weight of the object. We can measure the weight of the goods by using the following devices
228 The Leading Mathematics - 4 Estimate the weight of the following objects and check their weight by using bean balance and weighting blocks : EXERCISE 18.1 Your mastery depends on practice. Practice like you play. 154 Allied Mathematics-4 Estimate the weight of the following objects and check their weight by using bean balance and weighting blocks : (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) EXERCISE 5.5 (A) Your mastery depends on practice. Practice like you play. (a) (d) (g) (j) (m) (b) (e) (h) (k) (n) (c) (f) (i) (l) (o)
MEASUREMENT 229 Using beam balance, try to discover (a) How many units of 100 gm will balance 1 kg ? (b) How many units of 200 gm will balance 1 kg ? (c) How many units of 500 gm will balance 1 kg ? In each case, how many grams are balancing a kilogram ? 1000 gm = 1 kg \ 1 gm = 1 1000 gm = 1 ÷ 1000 gm = 0.001 kg CLASSWORK EXAMPLES Example 1 Convert 12 kg 729 gm into gram. Solution : Here, 12 kg 729 gm = 12 × 1000 gm + 729 gm = 12000 gm + 729 gm = 12729 gm Example 2 A laptop weighs 2275 gm. How many kilograms are there ? Solution : Here Weight of laptop = 2275 gm = 2275 1000 kg = 2.275 kg \ There is 2.275 kg weight of the laptop. × 1000 kg gm 1000 ÷ 1000gm = 1kg 18.2 Relation between Gram and Kilogram Read, Think and Learn
230 The Leading Mathematics - 4 1. Change the following into grams : (a) 7kg (b) 135kg (c) 3 1 2 kg (d) 18 3 4 kg (e) 5.250 kg (f) 9.729 kg (g) 12.500kg (h) 127.070kg (i) 10.325 kg 2. Convert the following into kilograms : (a) 5000 gm (b) 12000 gm (c) 45780 gm (d) 37858 gm (e) 45354 gm (f) 57624 gm (g) 65487 gm (h) 52417 gm (i) 69858 gm (j) 65989 gm (k) 12455 gm (l) 68543 gm 3. Change the following into kg and gm: (a) 3500 gm (b) 37800 gm (c) 58790 gm (d) 28928 gm (d) 35428 gm (d) 56974 gm (g) 68547 gm (h) 12548 gm (i) 88954 gm (j) 98546 gm (k) 85621 gm (l) 15470 gm EXERCISE 18.2 Your mastery depends on practice. Practice like you play. Collect any five things that are found in your home and measure their weight in kg and gm. Convert them into gm. Also, prepare a report and present it in your classroom. PROJECT WORK
MEASUREMENT 231 18.3 Relation Between Kilogram and Quintal At the end of this topic, the students will be able to: ¾ identify the relation between the kilogram and quintal. Learning Objectives Read, Think and Learn Observe the weights of the following objects in the digital balance. 20 kg 450 kg 160 kg What are the weights of rice, flex machine and motor cycle? If the weight of the objects is more than 100 kg, what is it called ? Do you know? In international systems (SI) units, the mass is measured in kilogram (kg). Kilogram is the base unit of mass. The weight of the object is usually taken to be the force on the object due to gravity. We know that, 1 gm = 1000 mg 1 kg = 1000 gm 100 kg = 1 quintal 1000÷ ×1000 kg ×100 100÷ Quintal 1000÷ ×1000 gm mg
232 The Leading Mathematics - 4 CLASSWORK EXAMPLES Example 1 Convert 3000 kg in quintal. Solution: Here, 3000 kg = 3000 100 = 30 quintals Example 2 Convert 4.5 quintals into kg. Solution: Here, 4.5 quintals = 4.5 × 100 kg = 450 kg Example 3 Observe the weight of the fat man standing on the digital balance. (a) What is the weight of the man shown in the digital balance? (b) Convert his weight into quintal and kg. (c) How many grams is his weight? Solution: (a) From the digital balance, the weight of the man is 118 kg. (b) Now, 118 kg = 100 kg + 18 kg = 1 quintal 18 kg ∴ His weight is 1 quintal 18 kg. (c) 118 kg = 118 × 1000 = 118000 gm. ∴ His weight is 118000 gm. 118 kg
MEASUREMENT 233 PRACTICE 18.3 Your mastery depends on practice. Practice as you play. 1. Convert the following weights in kg: (a) 2 quintals (b) 6 quintals (c) 15 quintals (d) 3.5 quintals (e) 18.2 quintals (f) 21.5 quintals (g) 14.8 quintals (h) 18.9 quintals (i) 42.25 quintals 2. Convert the following weights in quintals: (a) 200 kg (b) 500 kg (c) 81000 kg (d) 92000 kg (e) 72500 kg (f) 3250 kg (g) 41550 kg (h) 78880 kg (i) 12058 kg 3. Convert the following weights in quintal and kg. (a) 238 kg (b) 3725 kg (c) 18570 kg (d) 98420 kg (e) 28256 kg (f) 82530 kg 4. Observe the weight of the car. (a) How many kg are there in 1 quintal? (b) Write the weight of the car on the digital balance. (c) Convert the weight of the car in quintal. (d) Convert the weight of the car in gram only. 625 kg
234 The Leading Mathematics - 4 1. Add the following : k g g m 5 250 + 4 127 k g g m 2 7 275 + 19 825 k g g m 8 5 875 + 27 347 18.4 Addition and Subtraction of Weight EXERCISE 18.4 Your mastery depends on practice. Practice like you play. Read, Think and Learn k g g m 1 1 1 100 3 7 257 + 2 5 827 6 3 084 1 1 000 1 100 37kg 275gm +25kg 827gm 63kg 084gm "Alternatively" Add 37 kg 257 gm and 25 kg 827 gm. Solution : Here, Subtract 125 kg 485 gm from 385 kg 125 gm. Solution : Here, 7 14 10 120000 385.125kg –125.485kg 259.640kg "Alternatively" k g g m 7 1 4 10120 385 125 –125 485 259 640 (a) (b) (c)
MEASUREMENT 235 2. Subtract the following : k g g m 435 205 + 215 008 7kg 745gm +3kg 857gm 12q 82kg + 11 q 35kg 92kg 700gm +53kg 285gm 452kg 175gm +234kg 857gm 9.275kg +2.438kg 35kg 615gm 27kg 212gm +15kg 137gm 68kg 218gm 45kg 485gm +78kg 247gm 78kg 875gm 35kg 789gm +55kg 120gm k g g m 7 340 – 5 127 k g g m 2 9 265 – 19 825 k g g m 9 7 775 – 26 347 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l)
236 The Leading Mathematics - 4 3. Add the following : (a) 15 kg 275 gm, 9 kg 178 gm and 95 kg 25 gm (b) 78 kg 348 gm, 34 kg 84 gm and 35 kg 7 gm (c) 37 kg 387 gm and 19 kg 9 gm (d) (e) (f) (g) (h) (i) 92kg 700gm – 53kg 285gm 452kg 175gm –234kg 857gm 9.27q –2.43q k g g m 435 205 – 215 008 9kg 645gm –5kg 757gm 54kg 125gm –48kg 258gm Measure the weights of your bag and your friend's bag having books and note copies in kg and gm. What is the total weight ? Which bag is heavier or lighter by how much? Prepare a report and present it in the classroom. PROJECT WORK
MEASUREMENT 237 1. Sita bought 5kg 209 gm of potatoes and 2kg 785gm of tomatoes. How much weight of the vegetables did she buy in all ? 2. Samayak sold 135kg 250 gm of rice in the first day and 97kg 785 gm in the second day. How much rice did he sold in these two days ? 3. Mr. Agrawal sold 1q 75 kg of paper in the first day and 2985 kg of paper in the second day. How much paper did he sell in the two days? 4. How much weight should be added in 9kg 75kg to make it 15 kg ? 5. How much weight should be added in 2q 20 kg to make it 3q 15 kg? 6. Jasmin bought 7 kg 500gm of apples and out of them 2kg 758 gm were rotten. What is the weight of good apples ? 7. Raju bought 30kg 750 gm of Sujee from a shop A and 19kg 425 gm from shop B. If he used 35kg 850 gm of Sujee for making ‘Halwa’, how much Sujee is left with him ? 8. Mr. Ananda bought 875 kg of rice. He distributed 237 kg 250 gm of rice in camp A and 375kg 875gm in camp B of earthquake victims. How much rice is left with him ? 18.5 Verbal Problems on Weight EXERCISE 18.5 Your mastery depends on practice. Practice like you play. Read, Think and Learn Ram bought 2 kg 250 gm of apples and 3kg 778 gm of mangoes. How much weight of fruits did he buy altogether ? Solution : Here, Weight of apples = 2 kg 250 gm Weight of mangoes = 3 kg 778 gm Total weight = 6kg 028gm 1 1 \ He bought 6 kg 28 gm of fruits altogether.
238 The Leading Mathematics - 4 CHAPTER 19 Perimeter, Area and Volume Lesson Topics Pages 19.1 Introduction to Perimeter 239 19.2 Perimeter of Rectangle and Square 241 19.3 Introduction to Area 243 19.4 Area of Rectangle and Square and Square 245 19.5 Finding Length or Breadth of Rectangle From Area 248 19.6 Introduction to Volume 250 19.7 Volume of Cuboid and Cube 251 Measure the length, breadth and thickness of your Maths book. What is the height of your bench ? Measure the length and breadth of your bedroom. What are the length and breadth of your school’s ground ? What is the length of your school’s wall ? What are the faces of the book or note copy ? What are the faces of the dice ? WARM-UP
MEASUREMENT 239 The shape of the book is rectangular. It has length and breadth. It has four edges AB, B C, CD and AD on its surface. Pairs of opposite edges are the same in length. Can you measure the length of its edges ? Length of edge AB = 23.8 cm Length of edge BC = 17.6 cm Length of edge CD = 23.8 cm Length of edge AD = 17.6 cm Now, add all the edges, AB + BC + CD + AD = 23.8cm + 17.6cm + 23.8 cm + 17.6cm = 82.8cm This sum is the total boundary of the book. This is called perimeter of the book. The length of around the plane figure is called perimeter of the figure. Find the perimeter of the given shape in the adjoining graph : Solution : Length of PQ = 4 units Length of QR = 5 units Length of RS = 4 units Length of PS = 5 units \ Perimeter of PQRS = Sum of lengths around PQRS = PQ + QR + RS + PS = 4 + 5 + 4 + 5 = 18 units. P Q R S 2 2 23.8cm 17.6cm 23.8cm 17.6cm 82.8cm 19.1 Introduction to Perimeter Measurement 159 The shape of the book is rectangular. It has length and breadth. It has four edges AB, B C, CD and AD on its surface. Pairs of opposite edges are same in length. Can you measure the length of its edges ? Length of edge AB = 23.8 cm Length of edge BC = 17.6 cm Length of edge CD = 23.8 cm Length of edge AD = 17.6 cm Now, add the all edges, AB + BC + CD + AD = 23.8cm + 17.6cm + 23.8 cm + 17.6cm = 82.8cm This sum is the total boundary of the book. This is called perimeter of the book. Find the perimeter of the given shape in the adjoining graph : Solution : Length of PQ = 4 units Length of QR = 5 units Length of RS = 4 units Length of PS = 5 units \ Perimeter of PQRS = Sum of lengths around PQRS = PQ + QR + RS + PS = 4 + 5 + 4 + 5 = 18 units. A B C D 2 2 23.8cm 17.6cm 23.8cm 17.6cm 82.8cm P Q R S The length of around the plane figure is called perimeter of the figure. CHAPTER - 5.6 : PERIMETER 5.6 (A) Introduction to Perimeter Read, Think and Speak Read, Think and Learn
240 The Leading Mathematics - 4 1. Calculate the perimeter of the figures or shapes in the adjoining graph. 2. Find the perimeter of the following shapes : (a) (b) (c) (d) (e) (f) (g) (h) 3. Find the perimeter of the following objects which you have : (a) English Book (b) Nepali copy (c) Instrument box (d) Handkerchief (e) Eraser (f) Desk (g) Room (h) White/Black board A D B C I J K M L N E F Q P Q T SR V U H G 11 cm 11 cm 6 cm 6 cm P Q R S 6 cm 6 cm 5 cm 5 cm A B C D 10 cm 10 cm 7 cm 7 cm K L M N 8 cm Y W X Z 8 cm 8 cm 8 cm A B C 7 cm 8 cm 6 cm P Q R 11 cm 12 cm 7 cm 12 cm 13 cm 8 cm 7 cm 5 cm 5 cm F E C D A B 5 cm 6 cm 6 cm 8 cm 8 cm 10 cm 10 cm 8 cm 8 cm 5 cm 5 cm 5 cm EXERCISE 19.1 Your mastery depends on practice. Practice like you play.
MEASUREMENT 241 ABCD is a rectangle with four sides AB, BC, CD and AD. The, perimeter of rectangle ABCD = AB + BC + CD + AD = 12 + 8 + 12 + 8 = 40 cm In the rectangle ABCD, the length AB and CD are equal, say l and the breadth AD and BC are also equal, say b. Now, Perimeter (P) = AB + BC + CD + AD = l + b + l + b = 2l + 2b = 2(l + b) units. Here, l = 12 cm and b = 8 cm \ Perimeter of ABCD (P) = 2(l + b) = 2(12 + 8) = 2 × 20 = 40 cm In the square, all sides are equal. i.e., l = b. So, its perimeter (P) = 2(l + l) = 4l units. 12 cm 8 cm 8 cm A D C B 12 cm l l b b 1. Find the perimeter of the following rectangles by using formula p = 2(l + b) and check by adding all sides around them : (a) (b) (c) 9 cm 7 cm 11 cm 5 cm 13 cm 6 cm 19.2 Perimeter of Rectangle and Square EXERCISE 19.2 Your mastery depends on practice. Practice like you play. Read, Think and Learn
242 The Leading Mathematics - 4 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 162 Allied Mathematics-4 2. Find the perimeter of the following squares by using formula p = 4l and check by adding their sides : (a) (b) (c) 6cm 8cm 5cm 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l+ b). (a) (b) 73m 171m 75m 110m (c) (d) 140m 44m 32m 180m (e) (f) 44m 15m 19m 4m Project Work-5.13 Measure the length and breadth of your classroom and bedroom. Identify which is square or rectangular and draw their sketch. Find the perimeters of both rooms by adding all sides and using formulae. Also, which room has more perimeter and which room has less perimeter by how much ? Prepare a report and present in the classroom. 2. Find the perimeter of the following squares by using formula p = 4 l and check by adding their sides : (a) (b) (c) 3. Calculate the perimeter of the following rectangular shapes by using formula P = 2(l + b). 6 cm 5 cm 8 cm (a) (b) (c) (d) (e) (f) 140 m 44 m 15 m 19 m 4 m 44 m 32 m 180 m
MEASUREMENT 243 Place the Math Book - 4 on the surface of a white/black board and draw the boundary around it. Remove the book and name the rectangle ABCD. What is ABCD ? It is the occupied region by the Math Book - 4. This region is called area of the Math Book - 4. The occupied region by an object or the plane surface is called its area. Now, measure the length by the rectangle ABCD. Where AB = 14cm and BC = 20cm. Draw the squared grid 14 units on AB and 20 units on BC. How many small squares are there in the rectangle ABCD ? By counting the square grids, it is 280 sq. grids. This is the area of the rectangle ABCD. i.e., Area of the rectangle ABCD = 280 cm2 CLASSWORK EXAMPLES Count the squared grid in the coloured portion and find its area. Assume each small square is 1 cm2 . Solution : In the given squared grid, 19 sq. grids are coloured. Therefore, The area of the given portion = 19 cm2 . 1cm = 1cm2 14cm 20cm D A C B 1cm 19.3 Introduction to Area Read, Think and Learn
244 The Leading Mathematics - 4 Count the squared grid in the coloured portion and find its area. Assume each small square is 1 cm2 . 1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4. (a) (b) (c) (d) EXERCISE 19.3 Your mastery depends on practice. Practice like you play.
MEASUREMENT 245 In the adjoining figure, ABCD is a rectangle with length AB (l) = 5 cm and breadth BC (b) = 3 cm. Divide the length AB in 5 equal sections and the breadth BC in 3 equal sections as shown in the adjoining figure. Then each small square unit means 1 square centimetre (cm2 ). \ Area of rectangle ABCD (A) = Total no. of squares in rectangle ABCD = 15 cm2 = 5 cm × 3 cm = l × b Hence, Area of rectangle (A) = l × b sq. units. In square PQRS, PQ = QR = RS = PS = l. i.e., l = b \ Area of square (A) = l × b = l × l = l2 sq. units. CLASSWORK EXAMPLES Example 1 Find the area of the rectangle with length 18 cm and breadth 14 cm. Solution : Here, in the rectangle, Length (l) = 18 cm Breadth (b) = 14 cm 18 cm 14 cm D 5 cm 3 cm A B C l l S P R Q Square 19.4 Area of Rectangle and Square and Square Read, Think and Learn
246 The Leading Mathematics - 4 Now, we know that, Area (A) = l × b = 18 cm × 14 cm= 252 cm2 \ The area of the rectangle is 252 cm2 . Example 2 Find the area of the square with side 12cm. Solution : Here, in the square, Length (l) = 12 cm Now, we have, Area (A) = l2 = 122 = 144 cm2 \ The area of the square is 144 cm2 . EXERCISE 19.4 Your mastery depends on practice. Practice like you play. 1. Find the area of the following rectangles : (a) (b) (c) (d) (e) (f) (g) (h) 2. Find the area of the rectangles having the following length and breadth: (a) length 35 cm and breadth 27 cm (b) length 124 cm and breadth 74 cm 9 cm 18 cm 11 cm 11 cm 12 cm 18 cm 14 cm 17 cm 11 cm 18 cm 20 cm 12 cm 17 cm 25 cm 22 cm 11 cm 12 cm
MEASUREMENT 247 3. Find the area of the following squares : (a) (b) (c) (d) (e) (f) (g) (h) 4. Find the area of the squares having the following length: (a) Length 43 cm (b) Length 56 cm. Measure the length and breadth of your Maths book and long note copy, and draw their sketch. Find their areas and copy by adding all sides and using formulae. Which thing has more space and which has less space by how much ? Prepare a report and present it in the classroom. PROJECT WORK 8 cm 12 cm 16 cm 10 cm 11 cm 15 cm 12.6 cm 19.5 cm
248 The Leading Mathematics - 4 The area in the given rectangle ABCD having length 8 cm is 32 cm2 . What would be its breadth ? We know that, A = l × b or, 32cm2 = 8cm × b or, 4 32 cm2 8 cm = 8 cm × b 8 cm or, b = 4 cm. Hence, the breadth of the rectangle ABCD is 4 cm. l × b = A \ b = A l and l = A b CLASSWORK EXAMPLES Example 1 The area of a rectangle is 168 cm2 . If its breadth is 12cm, find the length of the rectangle. Solution : Given, Area of the rectangle (A) = 168 cm2 Breadth of the rectangle (b) = 12 cm Length of the rectangle (l) = ? Now, we know that, l = A b = 168 12 = 14 cm \ The length of the rectangle is 14 cm. A = 168 cm2 b = 12 cm l = ? 14 12 168 –12 48 –48 × D l = 8cm b = ? A B C Read, Think and Learn 19.5 Finding Length or Breadth of Rectangle From Area
MEASUREMENT 249 1. Find the length of the missing side of the following rectangles : (a) (b) (c) (d) 2. Find the breadth of the rectangle in the following conditions : (a) Area (A) = 30 cm2 and length (l) = 6 cm (b) Area (A) = 84 cm2 and length (l) = 14 cm (c) Area (A) = 126 cm2 and length (l) = 9 cm (d) Area (A) = 598 cm2 and length (l) = 26 cm 3. Find the length of the rectangle in each condition : (a) Area (A) = 40 cm2 and breadth (b) = 5 cm (b) Area (A) = 78 cm2 and breadth (b) = 6 cm (c) Area (A) = 210 cm2 and breadth (b) = 15 cm (d) Area (A) = 672 cm2 and breadth (b) = 21 cm 4. (a) If the area of a rectangular pond with length 40 m is 1280 m2 , find its breadth. (b) The area of a rectangular book is 408 sq. cm. If its length is 24 cm, find the breadth of the book. (c) The breadth of an A4 paper with area 630 cm2 is 21 cm. What is its length ? (d) What is the length of a garden with area 1960 m2 and breadth 35 m ? A = 84 cm2 ? 7 cm A = 168 cm2 12 cm ? A = 162 cm2 18 cm ? A = 288 cm2 ? 16 cm EXERCISE 19.5 Your mastery depends on practice. Practice like you play.
250 The Leading Mathematics - 4 When a stone is sinked in the glass full of water, what happens ? The amount of water displaced by the stone in the glass is the volume of the stone. The volume is the total capacity of the whole body. The occupied space by an object is called the volume of the object. How much length or dimension of the given solid is there ? It has length (l), breadth (b) and height (h). It is cuboid. How many units are there in the length ? Capacity means the holding amount in an object. But volume means occupied space by the whole object. Liquid is the capacity of cup. Matter which is used to make cup is the volume of the cup. I know, 1 unit length, breadth and height of a block means 1cu. unit volume. 1 1 1 1 1. Count the cubic blocks in each of the following objects and find their volume : (a) (b) (c) (d) (e) (f) (g) (h) 19.6 Introduction to Volume EXERCISE 19.6 Your mastery depends on practice. Practice like you play. Read, Think and Learn Similarly, in the breadth and height ? How many small cubic blocks are there in the cuboid ? There are 24 cubic blocks called the volume of the cuboid.