WEBVTT 1 00:00:00.390 --> 00:00:02.000 Euclidean Geometry 2 00:00:02.530 --> 00:00:03.510 Logic and Problem-Solving 3 00:00:04.291 --> 00:00:09.651 Logical Thinking in Euclidean Geometry for Digital Content Creators 6 Breaking the Mold of the Right Answer - Problem Solving Techniques 1 4 00:00:30.960 --> 00:00:32.520 Hello, this is Park Jongha 5 00:00:32.520 --> 00:00:37.660 In this lesson, I will introduce problem-solving techniques 6 00:00:37.661 --> 00:00:40.101 such as collecting information and imagining in various ways 7 00:00:40.661 --> 00:00:44.041 In this situation, it is difficult and often useless 8 00:00:44.041 --> 00:00:46.061 to talk about 9 00:00:46.621 --> 00:00:49.900 applying these problem-solving techniques in a short story 10 00:00:49.900 --> 00:00:51.960 The problems we face are 11 00:00:51.960 --> 00:00:56.360 unlike the math problems we are learning now, so they are not immediately apparent 12 00:00:56.820 --> 00:01:01.420 However, I think it is very important to think about the need for these things in everyday life 13 00:01:01.880 --> 00:01:06.660 In that sense, it would be good to remember that collecting sufficient information in this lesson is necessary 14 00:01:06.660 --> 00:01:09.820 You should think about various cases 15 00:01:09.821 --> 00:01:11.481 in various ways 16 00:01:11.840 --> 00:01:14.160 In particular, in the digital content market 17 00:01:14.161 --> 00:01:17.864 where new technologies and services are created daily 18 00:01:17.864 --> 00:01:21.500 you should always be careful when collecting new information 19 00:01:21.790 --> 00:01:26.430 Transforming Given Information into Needed Information 20 00:01:26.990 --> 00:01:30.960 When solving a problem, collecting all the necessary information is important 21 00:01:31.100 --> 00:01:33.401 The more information you have 22 00:01:33.401 --> 00:01:37.001 the easier and faster it is to approach problem-solving 23 00:01:37.141 --> 00:01:40.321 If you lack information, you cannot solve the problem 24 00:01:40.321 --> 00:01:42.701 no matter how many ideas you try 25 00:01:42.881 --> 00:01:47.520 Therefore, it is effective to approach the problem by indicating information 26 00:01:47.520 --> 00:01:50.780 that will likely be used to solve the problem 27 00:01:51.200 --> 00:01:53.700 For example, let's look at this question 28 00:01:53.700 --> 00:01:57.961 In the following figure, what is x's angle 29 00:01:57.961 --> 00:02:00.244 when RT and TU are equal? 30 00:02:01.120 --> 00:02:04.481 Under the condition of the problem that RT is TU 31 00:02:04.481 --> 00:02:08.064 triangle RTU is an isosceles triangle 32 00:02:08.420 --> 00:02:13.120 Since 180 degrees - 106 degrees is 74 degrees 33 00:02:13.121 --> 00:02:18.260 the sizes of the remaining two angles of an isosceles triangle are 37 degrees each 34 00:02:18.680 --> 00:02:21.680 Let's write this down in the problem 35 00:02:21.680 --> 00:02:24.461 If you write down the angles in the problem like this 36 00:02:24.461 --> 00:02:30.240 By the exterior angle theorem 37 00:02:30.240 --> 00:02:33.701 angle x in triangle SVU is 32 degrees plus 37 degrees 38 00:02:33.701 --> 00:02:36.519 equals 69 degrees 39 00:02:37.119 --> 00:02:40.780 Math problems are not solved by eye but by hand 40 00:02:40.780 --> 00:02:43.800 The first thing you should do by moving your hand 41 00:02:43.801 --> 00:02:46.121 is to indicate the necessary information 42 00:02:46.321 --> 00:02:49.500 As you display the information necessary for problem-solving in this way 43 00:02:49.720 --> 00:02:53.040 you will often have the experience of seeing the information you have displayed with your eyes 44 00:02:53.040 --> 00:02:54.681 moving your head 45 00:02:54.681 --> 00:02:58.020 and coming up with ideas 46 00:02:58.021 --> 00:03:00.147 necessary for problem-solving 47 00:03:00.500 --> 00:03:03.140 I recommend that those of you who deal with digital content 48 00:03:03.140 --> 00:03:06.600 also hold a ballpoint pen in your hand on a practice sheet 49 00:03:06.600 --> 00:03:10.400 like an analog and express your thoughts in various ways 50 00:03:10.960 --> 00:03:15.500 Writing something on paper with your hand often gives you ideas 51 00:03:15.500 --> 00:03:17.900 so I recommend trying it at least once 52 00:03:18.440 --> 00:03:20.380 Let's solve one more math problem 53 00:03:21.140 --> 00:03:27.660 Question 1. Find the area of ​​the triangle given below 54 00:03:27.660 --> 00:03:31.960 If we fill in a little more information that we know, it is as follows 55 00:03:31.960 --> 00:03:34.880 First, we can see that the given triangle is isosceles 56 00:03:34.880 --> 00:03:39.800 because the sum of the internal angles of the triangle is 180 degrees 57 00:03:40.040 --> 00:03:43.560 Therefore, the length of two sides of the triangle is 5 58 00:03:44.700 --> 00:03:49.260 Also, if we think of a right triangle with lengths of 3:4:5 59 00:03:49.260 --> 00:03:51.958 if the base is 5 60 00:03:51.958 --> 00:03:55.040 we can see that the triangle's height is 3 61 00:03:55.320 --> 00:04:00.861 Therefore, the area of ​​the triangle is 5 x 3 x 1/2 62 00:04:00.861 --> 00:04:03.005 which is 15/2 63 00:04:03.600 --> 00:04:07.020 There is a saying, "There's more than one way to skin a cat" 64 00:04:07.020 --> 00:04:10.640 You may think that you just have to solve the problem somehow 65 00:04:10.640 --> 00:04:12.860 Rather than solving a complicated and difficult problem 66 00:04:12.860 --> 00:04:16.240 it is better to solve the problem easily and simply 67 00:04:16.240 --> 00:04:20.640 That way, you can reduce calculation errors and solve the problem quickly 68 00:04:21.000 --> 00:04:24.220 To proceed with the problem-solving process easily and simply 69 00:04:24.220 --> 00:04:26.620 you must use the information effectively 70 00:04:26.980 --> 00:04:30.660 To do that, it is important to understand the necessary information 71 00:04:30.920 --> 00:04:34.620 Let's take a look at a problem to figure out the necessary information 72 00:04:35.360 --> 00:04:42.160 Question 2. Calculate the area of ​​the shaded right triangle given the following 73 00:04:42.500 --> 00:04:46.040 Students who learned the Pythagorean theorem in school 74 00:04:46.040 --> 00:04:49.000 will probably start with complicated calculations 75 00:04:49.000 --> 00:04:51.520 such as calculating the length of the hypotenuse of a large right triangle 76 00:04:51.860 --> 00:04:53.820 However, I told you that blindly calculating 77 00:04:53.820 --> 00:04:56.921 does not lead to good results 78 00:04:56.921 --> 00:05:01.240 Before calculating, let's take a look at the length ratio 79 00:05:01.700 --> 00:05:05.620 The length ratio of the simplest right triangle made of natural numbers is 80 00:05:05.621 --> 00:05:07.300 3:4:5 81 00:05:07.300 --> 00:05:10.160 Therefore, all triangles with a 3:4:5 ratio 82 00:05:10.161 --> 00:05:12.341 are right triangles 83 00:05:12.600 --> 00:05:18.000 The given large right triangle also has two sides of 9 and 12 84 00:05:18.000 --> 00:05:21.600 This means that 9:12 is 3:4 85 00:05:21.800 --> 00:05:26.320 Therefore, even without applying the Pythagorean theorem to calculate the hypotenuse 86 00:05:26.320 --> 00:05:29.400 we can see that it is 9:12:15 87 00:05:29.640 --> 00:05:31.180 If we solve the problem 88 00:05:31.180 --> 00:05:34.166 first, we can see that the length of the hypotenuse of the large right triangle is 89 00:05:34.166 --> 00:05:36.760 9:12:15 90 00:05:36.761 --> 00:05:38.601 which is 15 91 00:05:38.601 --> 00:05:40.381 If we write this 92 00:05:40.381 --> 00:05:45.441 we can see that the length of the hypotenuse of the small right triangle is 10 93 00:05:45.820 --> 00:05:50.420 If the lengths of the two sides of the small right triangle are a and b 94 00:05:50.420 --> 00:05:56.300 then the right triangle is also a right triangle with a ratio of 3:4:5 95 00:05:56.300 --> 00:06:00.121 so a:b:10 is 6:8:10 96 00:06:00.121 --> 00:06:04.560 Therefore, the area of ​​the small right triangle can be calculated like this 97 00:06:04.560 --> 00:06:07.320 1/2 x a x b is 98 00:06:07.321 --> 00:06:10.142 1/2 x 6x 8 99 00:06:10.142 --> 00:06:12.051 which is 24 100 00:06:12.051 --> 00:06:18.780 Question 3. Find the size of the given angle x in a regular octagon as follows 101 00:06:19.880 --> 00:06:23.860 Let's approach this problem in two ways 102 00:06:23.860 --> 00:06:29.340 The first way is to figure out the given angles without thinking about the idea 103 00:06:29.740 --> 00:06:34.640 First, let's think about the size of the interior angles of a regular octagon 104 00:06:34.640 --> 00:06:37.080 We can think of six triangles 105 00:06:37.081 --> 00:06:40.381 inside a regular octagon as follows 106 00:06:40.381 --> 00:06:45.121 Therefore, the sum of the interior angles of a regular octagon is 180 degrees x 6 107 00:06:45.121 --> 00:06:46.757 which is 1080 108 00:06:48.020 --> 00:06:53.080 Since the sizes of the eight interior angles of a regular octagon are the same 109 00:06:53.080 --> 00:06:56.400 if we display the size of the angles in the given problem 110 00:06:56.400 --> 00:07:03.040 we can display all the values ​​of the angles 111 00:07:03.060 --> 00:07:06.600 If four angles make a square 112 00:07:06.600 --> 00:07:14.300 45 + 225 + 45 + x is 360 113 00:07:14.300 --> 00:07:17.300 Therefore, x is 45 degrees 114 00:07:17.600 --> 00:07:20.860 Let's approach this problem with an idea 115 00:07:20.860 --> 00:07:24.320 Let's think about the following triangle 116 00:07:24.980 --> 00:07:28.360 If we think about a triangle that contains angle x like this 117 00:07:28.360 --> 00:07:31.761 we can see that the two remaining angles 118 00:07:31.761 --> 00:07:37.221 except x, are angles that divide the interior angles of a regular octagon in half 119 00:07:38.540 --> 00:07:40.521 Because just as we can think of the center of a circle 120 00:07:40.521 --> 00:07:45.360 we can think of the center of a regular octagon as a picture 121 00:07:45.360 --> 00:07:50.060 So in the triangle we drew, the other two angles, excluding x 122 00:07:50.060 --> 00:07:57.140 are angles that divide the 135 degrees angle of a regular octagon in half 123 00:07:57.140 --> 00:08:00.400 Therefore, the sum of the two angles is 135 degrees 124 00:08:00.400 --> 00:08:03.660 and x is 180 degrees minus 135 degrees 125 00:08:03.661 --> 00:08:05.931 which is the answer of 45 degrees 126 00:08:06.300 --> 00:08:11.160 Thinking Beyond the Obvious 127 00:08:11.300 --> 00:08:13.120 One of the charms of Euclidean geometry is that 128 00:08:13.120 --> 00:08:17.220 you can solve problems by imagining things that are not directly revealed in the problem 129 00:08:17.500 --> 00:08:19.580 The problems given to us are 130 00:08:19.581 --> 00:08:22.701 often part of the overall situation 131 00:08:22.701 --> 00:08:26.200 So, just by drawing extension lines or auxiliary lines 132 00:08:26.201 --> 00:08:29.181 and expanding the situation a little 133 00:08:29.181 --> 00:08:32.200 you can often get ideas for solving the problem 134 00:08:32.660 --> 00:08:34.320 Drawing lines that do not exist 135 00:08:34.321 --> 00:08:38.495 sometimes copying the shape of the problem 136 00:08:38.495 --> 00:08:40.055 and moving it to the side 137 00:08:40.275 --> 00:08:44.220 is an active and positive approach that often gives you ideas for solving the problem 138 00:08:44.500 --> 00:08:47.740 Let's look at the related content while solving the problem 139 00:08:48.500 --> 00:08:55.300 Question 4. Find the area of ​​the pentagon ABCDE given below 140 00:08:55.300 --> 00:08:58.160 The sum of the interior angles of a pentagon is 540 degrees 141 00:08:58.161 --> 00:09:00.041 This is 180 degrees x 3 142 00:09:00.400 --> 00:09:03.600 Therefore, if we find the value of angle E 143 00:09:03.601 --> 00:09:10.202 it is 540 - (90 x 3 + 135) 144 00:09:10.203 --> 00:09:12.046 which is 135 degrees 145 00:09:12.460 --> 00:09:16.700 Also, 180-135 equals 45 degrees 146 00:09:16.920 --> 00:09:20.020 If we understand this relationship 147 00:09:20.021 --> 00:09:23.628 we can think of two right isosceles triangles 148 00:09:23.628 --> 00:09:27.220 Since the two sides of an isosceles right triangle are the same length 149 00:09:27.221 --> 00:09:30.223 we can determine the following information 150 00:09:30.500 --> 00:09:35.700 The large triangle FGD is also an isosceles right triangle 151 00:09:35.700 --> 00:09:38.040 Since the side length is 14 152 00:09:38.041 --> 00:09:40.527 the area of ​​the large isosceles right triangle 153 00:09:40.528 --> 00:09:43.315 is 14 x 14x 1/2 154 00:09:43.315 --> 00:09:44.796 which is 98 155 00:09:44.796 --> 00:09:47.860 and the areas of the two smaller isosceles triangles are 156 00:09:47.861 --> 00:09:49.841 4 x 4 x 1/2 157 00:09:49.842 --> 00:09:50.528 which is 8 158 00:09:50.528 --> 00:09:52.608 6 x 6x 1/2 159 00:09:52.608 --> 00:09:53.669 is 18 160 00:09:53.669 --> 00:09:56.532 Therefore, the answer to the area of ​​the pentagon 161 00:09:56.533 --> 00:10:00.313 is 98 - (8 + 18) 162 00:10:00.314 --> 00:10:02.514 and the answer is 72 163 00:10:02.960 --> 00:10:05.020 One thing we need to be careful about is that 164 00:10:05.020 --> 00:10:07.700 we should not use our imagination 165 00:10:07.701 --> 00:10:11.937 and do illogical calculations by eye 166 00:10:12.457 --> 00:10:15.104 For example, some people solve this problem 167 00:10:15.105 --> 00:10:17.393 by drawing a line connecting B and E 168 00:10:17.394 --> 00:10:23.696 and thinking that BCDE is a right triangle 169 00:10:24.036 --> 00:10:27.740 If you set it up like this, it is not logical 170 00:10:27.740 --> 00:10:32.600 This means ABE is an isosceles right triangle 171 00:10:32.600 --> 00:10:35.141 AB = AE = 6 172 00:10:35.141 --> 00:10:38.221 and BE = 10 173 00:10:38.221 --> 00:10:42.441 This is not logical if you think about the ratio of the lengths of the sides 174 00:10:42.441 --> 00:10:44.522 of an isosceles right triangle 175 00:10:44.980 --> 00:10:49.160 Please remember that you cannot solve math problems 176 00:10:49.160 --> 00:10:50.161 with your logic 177 00:10:50.700 --> 00:10:56.500 Question 5. What is the size of each x in the following figure? 178 00:10:57.700 --> 00:11:03.360 So that the rectangle BCDE becomes a square 179 00:11:03.361 --> 00:11:05.694 Let's take point E 180 00:11:05.940 --> 00:11:10.644 Then, since angle ADC is 150 degrees 181 00:11:10.645 --> 00:11:16.211 we can see that angle ADE in triangle ADE is 60 degrees 182 00:11:16.212 --> 00:11:20.932 and that the lengths of sides AD and DE are the same 183 00:11:21.212 --> 00:11:25.100 Therefore, triangle ADE is an equilateral triangle 184 00:11:25.101 --> 00:11:27.556 with all angles at 60 degrees 185 00:11:28.220 --> 00:11:34.320 Since triangle AED is an equilateral triangle, AE is 2B 186 00:11:34.320 --> 00:11:38.970 and triangle AEB is an isosceles triangle 187 00:11:38.970 --> 00:11:44.200 Therefore, angle BAE is angle ABE 188 00:11:44.201 --> 00:11:46.021 and is 15 degrees 189 00:11:46.021 --> 00:11:48.740 We are looking for the size of angle x 190 00:11:48.741 --> 00:11:50.561 to be 60 - 15 191 00:11:50.561 --> 00:11:52.402 which is 45 degrees 192 00:11:52.582 --> 00:11:54.940 If we express it as calculating and imagining 193 00:11:54.941 --> 00:11:58.190 we should be good at calculating and imagining 194 00:11:58.360 --> 00:12:01.360 If we only focus on calculating quickly 195 00:12:01.361 --> 00:12:05.347 and think of the process of imagining leisurely as a waste of time 196 00:12:05.348 --> 00:12:08.727 it will be difficult to learn truly fun math 197 00:12:09.220 --> 00:12:11.640 Our math skills will not improve much either 198 00:12:11.641 --> 00:12:13.520 Our daily lives are the same 199 00:12:13.520 --> 00:12:17.940 It is necessary to do things quickly and efficiently 200 00:12:17.941 --> 00:12:22.590 but it is also necessary to think bigger and imagine new things by looking around 201 00:12:23.010 --> 00:12:28.364 If you are working hard but are not getting much better than average results 202 00:12:28.365 --> 00:12:31.388 you need to take your time and imagine more 203 00:12:31.389 --> 00:12:35.372 rather than blindly making plans to do things faster 204 00:12:35.740 --> 00:12:42.240 Question 6. Find the area of ​​the given rectangle ABCD 205 00:12:42.240 --> 00:12:43.980 Let's solve this problem 206 00:12:43.980 --> 00:12:45.741 How would you solve it? 207 00:12:46.780 --> 00:12:50.220 It seems like we could solve this problem by quickly calculating 208 00:12:50.220 --> 00:12:56.220 However, since we haven't learned much about calculation methods in this course 209 00:12:56.221 --> 00:12:58.405 let's solve this problem by imagining 210 00:12:58.920 --> 00:13:03.440 Let's copy the given shape and paste it 211 00:13:03.960 --> 00:13:07.920 If we paste four of the same shapes like this 212 00:13:07.921 --> 00:13:15.147 we can see that it becomes a square with a side length of 5 213 00:13:15.147 --> 00:13:18.488 Therefore, the square area is 25 214 00:13:18.489 --> 00:13:20.888 and since we pasted four shapes 215 00:13:20.888 --> 00:13:25.703 the area of ​​one shape we are looking for is 4/25 216 00:13:26.076 --> 00:13:29.980 Some people like to paste, and some people like to divide 217 00:13:29.980 --> 00:13:32.260 If you like to divide 218 00:13:32.260 --> 00:13:35.691 draw a line segment BD 219 00:13:35.691 --> 00:13:39.961 Then, cut triangle ABD 220 00:13:39.961 --> 00:13:44.557 and paste it so that AD becomes CD 221 00:13:45.220 --> 00:13:49.880 Since angle A plus angle C is 180 degrees 222 00:13:50.309 --> 00:13:53.729 DBB' becomes a triangle 223 00:13:53.730 --> 00:14:00.870 and B' becomes an isosceles right triangle, which is B'D 224 00:14:01.790 --> 00:14:04.380 All we need now is to find the area 225 00:14:04.381 --> 00:14:09.012 of ​​an isosceles right triangle whose hypotenuse is of length 5 226 00:14:09.220 --> 00:14:11.621 The area of ​​an isosceles right triangle 227 00:14:11.621 --> 00:14:16.194 can be calculated by attaching four of the same ones 228 00:14:16.594 --> 00:14:20.820 If we attach the four isosceles right triangles 229 00:14:20.821 --> 00:14:24.570 we will create a square with a side length of 5 230 00:14:24.570 --> 00:14:27.220 Therefore, the area of ​​the square is 25 231 00:14:27.221 --> 00:14:30.694 and the area of ​​one isosceles right triangle that we are looking for 232 00:14:30.694 --> 00:14:32.459 is 25/4 233 00:14:32.719 --> 00:14:37.300 In other words, the area of ​​the shape we want to calculate is 25/4 234 00:14:38.000 --> 00:14:41.680 We see that difficult problems can be solved 235 00:14:41.681 --> 00:14:44.326 by simply examining them logically one by one while using various imaginations 236 00:14:44.327 --> 00:14:46.727 without doing difficult calculations 237 00:14:46.727 --> 00:14:49.600 This is how we experience the fun of math 238 00:14:50.220 --> 00:14:57.580 Question 7. In the following figure, find the area of ​​the shaded area when AB is 6 239 00:14:58.240 --> 00:15:02.700 Let's approach this problem not by calculating but by imagining 240 00:15:03.980 --> 00:15:09.060 Let's think of the given shapes as tiles on the floor 241 00:15:09.420 --> 00:15:12.280 If we think of them as tiles on the floor 242 00:15:12.280 --> 00:15:16.113 we can attach them, as you can see on the screen now 243 00:15:16.753 --> 00:15:19.680 If we attach four shapes like this 244 00:15:19.681 --> 00:15:24.400 we can see a square with a side length of 6 is created 245 00:15:24.400 --> 00:15:27.940 The area of ​​the given shape is 6 x 6 246 00:15:27.941 --> 00:15:32.773 A quarter of a square with a side length of 6 247 00:15:32.774 --> 00:15:34.712 so the answer is 9 248 00:15:35.800 --> 00:15:38.960 his problem can also be solved using another analytical method 249 00:15:38.960 --> 00:15:42.380 Since solving one problem in various ways is a good way to study 250 00:15:42.381 --> 00:15:44.461 let's solve it using another method 251 00:15:44.680 --> 00:15:47.080 In this problem, the area of ​​the shaded area 252 00:15:47.081 --> 00:15:50.980 can be thought of as a square and a right triangle 253 00:15:51.560 --> 00:15:54.620 It can be interpreted as the area of ​​3/2 + 1 254 00:15:54.621 --> 00:15:57.860 5/2 squares 255 00:15:58.480 --> 00:16:02.980 Therefore, if you find the area of ​​one square with the given condition 256 00:16:02.981 --> 00:16:06.310 you can find the area of ​​the shaded area 257 00:16:06.311 --> 00:16:09.588 by finding 5/2 of it 258 00:16:10.008 --> 00:16:14.560 The area of ​​one square can be found as follows 259 00:16:15.020 --> 00:16:20.060 The area of ​​the shaded triangle is 3/2 260 00:16:20.060 --> 00:16:23.780 If you attach the shape on the left as on the right 261 00:16:23.781 --> 00:16:28.558 you will create a large square with a side length of 6 262 00:16:28.558 --> 00:16:34.440 which has 4 shaded triangles and 4 squares 263 00:16:34.980 --> 00:16:38.020 Therefore, if we call the area of ​​one square A 264 00:16:38.021 --> 00:16:40.161 we can set up the following equation 265 00:16:40.601 --> 00:16:47.840 If we follow the equation you just saw, A will be 18/5 266 00:16:47.840 --> 00:16:49.220 We said that the area of ​​the shaded area 267 00:16:49.221 --> 00:16:53.394 is 5/2 times A 268 00:16:53.394 --> 00:16:58.960 Since the area of ​​a square of the value of A is 18/5 269 00:16:58.961 --> 00:17:04.270 the answer to 5/2a is 5/2 x 18/5 270 00:17:04.271 --> 00:17:05.551 which is 9 271 00:17:06.071 --> 00:17:09.080 I solved the problem by imagining rather than calculating 272 00:17:09.080 --> 00:17:11.420 After learning the Pythagorean theorem 273 00:17:11.421 --> 00:17:14.386 and solving quadratic equations in middle school 274 00:17:14.386 --> 00:17:17.386 many problems are approached by calculating 275 00:17:17.646 --> 00:17:21.480 Calculating this way can solve problems more easily 276 00:17:21.481 --> 00:17:26.421 but sometimes, you will experience such calculations blocking your imagination 277 00:17:26.901 --> 00:17:29.261 Since this course aims to solve all problems 278 00:17:29.261 --> 00:17:32.121 with 4th-grade knowledge 279 00:17:32.301 --> 00:17:36.040 I solved the problem by imagining rather than calculating 280 00:17:36.040 --> 00:17:41.300 I hope this can provide insight to you, the digital content creators 281 00:17:41.300 --> 00:17:44.541 Rather than quickly coming up with an answer 282 00:17:44.541 --> 00:17:47.348 you can think of something else with various imaginations 283 00:17:47.529 --> 00:17:49.089 Converting given information into necessary information The more information you have, the easier and faster you can approach problem-solving 284 00:17:49.089 --> 00:17:50.510 It is an effective way to approach the solution by displaying information that is likely to be used in problem-solving To proceed with the problem-solving process easily and simply, you must use the given information effectively 285 00:17:50.510 --> 00:17:52.131 Imagine a wider part The problems given to us are often part of the overall situation 286 00:17:52.131 --> 00:17:53.352 We can get an idea for solving the problem just by drawing an extension line or auxiliary line and expanding the situation a little However, we should not use our imagination and make illogical calculations by guessing 287 00:17:53.352 --> 00:17:54.573 Even without making difficult calculations, we can solve difficult problems by exercising our imagination and logically considering them