WEBVTT

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Euclidean Geometry

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Logic and Problem-Solving

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Euclidean Logical Thinking for Digital Content Creators
3. Finding Ideas with Euclidean Thinking Part 1 : Thinking in Ratios

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Hello, this is Jonghwa Park

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In this session, we’ll explore a method to generate ideas

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by setting a standard, comparing it with other elements

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and considering appropriate ratios

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One effective way to understand the world rationally and logically

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is to think by comparing elements against a standard

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For example, why do we use numbers?

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The reason we use numbers is to compare

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A person who is 180 cm tall is about 10 cm taller than someone who is 170 cm tall

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We use numbers

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to compare, understand, and grasp such differences

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The ancient Greeks seemed to think the same way

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Instead of focusing on absolute values, they often relied on relative ratios

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to make wise judgments

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Today, let’s explore their thinking techniques together

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Applying Ratios and Comparisons to Problem Solving

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In English, there is something called a rational number

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The term “rational”  from rational number originates from

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the Latin word “ratio,” which means proportion or ratio

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Given two integers a and b, a rational number is written as b over a

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A number expressed as a ratio of integers is a rational number

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Around 500 BCE, Pythagoras believed

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that the world was composed of numbers

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However, the only numbers he recognized were rational numbers

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He thought the most logical and rational way to understand the world

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was by perceiving it through numbers

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This belief likely contributed to the dual meaning of “rational,”

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as both a mathematical term and a descriptor of logical reasoning

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Thinking in terms of ratios and proportions is actually a highly effective technique for problem-solving

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Let’s delve deeper by examining some problems

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Problem 1: In the diagram below, what is the length of x

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This problem involves two right triangles

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By analyzing the diagram

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you can observe that triangle ABC is similar to triangle ADE

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If we flip and align triangle ADE, we can confirm their similarity

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Since the two triangles are similar

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Length of  AC : Length of  AB = Length of AE: Length of AD

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Using this ratio, we can calculate the following

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We can conclude that x = 5

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Problem 2: In the diagram below, line segments AF, BE, and CD are parallel

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If AF = 5, AB = 6, BC = 2, and CD = 9

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What is the length of BE?

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To solve this, draw a line segment FG

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 parallel to AC

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From drawing this parallel line FG

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we can determine that AF = BH = CG = 5

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Triangles FHE and FGD are similar

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So, FH:HE=FG:GD

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6:HE=8:4

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So, HE=3

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Finally, BE=BH+HE

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which is 5+3

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The answer is 8

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This problem demonstrates how creating similar triangles

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helps in applying ratios for problem-solving

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Using similarity and applying ratios through comparison

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is a representative technique for generating ideas in problem-solving

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In music, there is the concept of perfect pitch

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It’s the ability to identify a note, such as Sol or Fa

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accurately just by hearing it

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However, not many people possess perfect pitch

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Most people identify notes relatively, using a reference note

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because it is easier and more effective

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Similarly, problem-solving often involves setting a standard, making comparisons

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and applying ratios

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You can experience this process by solving problems

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Problem 3: In the following scenario, a right triangle ABCD is folded at point B

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so that AB aligns perfectly with BC

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What is the area of the shaded region?

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The problem can be approached as follows

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By folding triangle ABE onto triangle BDE

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the two triangles become congruent, meaning they have the same area

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What we are trying to find is the area of triangle CDE

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Since triangles BDE and CDE share the same height

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their areas have a ratio of 3:1

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Thus, the areas of triangles ADE, BDE, and CDE

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are in the ratio 3:3:1

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It's 3/7, 3/7, and 1/7 of the total area

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For the large triangle ABCD, its total area is calculated as 3×4×1/2

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which is 6

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Therefore, the areas of triangles ABE, BDE, and CDE

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are each 18/7, 18/7, and 6/7

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The area of the shaded region is 6/7

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Problem 4: Inside a regular octagon with an area of 40, point P is given

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For triangles ABP and EFP

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The problem asks for the combined area of the two areas

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To determine the area of triangles ABP and EFP

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consider the auxiliary lines shown on the screen

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The combined area of triangles ABP and EFP is half of the area of rectangle ABEF

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1/2 of the area

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The only information provided is the total area of the octagon

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If we can determine what percentage

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of the total octagon’s area the rectangle ABEF occupies

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then half of that area will give us the desired answer

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Now it's time for us to examine the octagon

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The basic information we have of an octagon is

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as can be seen on the screen, we can divide it into

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8 equal triangular sections from the center

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If we draw the rectangle ABEF onto this diagram, this is what we can see

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two of the triangles that divided the rectangle into 8 sections

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also is divided rectangle ABEF into two sections

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Rectangle ABEF spans two of these triangular sections

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which means it occupies a quarter of the total octagon’s area

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Since the area of the two triangles

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is 1/2 of rectangle ABEF

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this is 1/4 of the octagon's total area

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A ratio we can find in problem is like this

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For example, in the given triangle

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let's think about the two triangles' area ratio

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The ratio of the areas of trangle OAC and OBC

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As the two triangles have the same height

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The lines AC and CB ratio would be the area ratio

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Let us look at another one

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When the following triangle's area is 10

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What is the area of b?

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The area of a triangle is the length x width x 1/2

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So, when the length is the same, the height becomes the ratio of the area

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Let us find the ratio of the area

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If we say that the area of the given triangle is S

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The area of A is 3/7 x S

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The area of B + C is 4/7 x S

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And as B and C have the same length and width, the area is the same

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So, the area of B = C = 4/7 x 1/2

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S x 2/7

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As the area of the entire triangle is 10

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The area of B is 20/7

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Problem 5: In the following figure, AD:DB = 1:2

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and AE:EC = 1:1

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The three sections divided from triangle ABC

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What is the ratio of the areas of triangles ADE, BDE, and BCE?

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To determine the ratio of the areas of these three triangles

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we need to establish a base and height as a reference

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Using AC as the base, triangle ABC

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can be divided into triangles ABE and BCE

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The areas of triangles ABE and BCE are equal

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because they share the same base and height

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In the second figure, the ratio of the areas of triangles ADE and BDE is 1:2

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In the second figure, the ratio of the areas of triangles ADE and BDE is 1:2

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To sum up, if we say that the area of triangle ADE is S

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The area of triangle BDE is 2S

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And the area of triangle BCE equals the sum of the areas of triangles ADE and BDE, which is 3S

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Thus, the ratio of the areas of triangles ADE, BDE

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and BCE is 1:2:3

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Let us continue on with the next question

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Problem 6: In the following diagram, AD:DB = 1:2

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and BE:EC = 3:5

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If the area of quadrilateral ADEC is 39

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what is the total area of triangle ABC?

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To solve, divide triangle ABC into segments based on the given ratios

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BE:EC = 3:5, so is we call the area of triangle ABE, S

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The area of triangle AEC is Sx5/8

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the area of triangle ABE is  Sx3/8

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AD:DB = 1:2, so triangles ADE and BDE

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divide the area of eighths of S by a ratio of 1 to 2

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So, the area of triangle ADE is 3S/8 x 1/3

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1S/8

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The area of triangle BDE is 3S/b x 2/3

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1S/4

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In the problem, the area of quadrilateral ADEC is 39

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So we can find S through the following equation

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From this, S=52

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Identifying Information You Want to Compare

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Let me ask you a question

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When given rectangles ABCD and ACEF as follows

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What do you think the area of rectangle ACEF would be?

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Comparing the two rectangles, let’s focus on triangle ACD

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First, the area of rectangle ABCD is twice the area of triangle ACD

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Similarly, the area of rectangle ACEF

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if we divide it into two parts as shown below

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is also twice the area of triangle ACD

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Therefore, the area of rectangle ACEF is equal to the area of rectangle ABCD

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and the area is 4 multiplied by 8

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which equals 32

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This way, comparing based on a specific reference and identifying ratios

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is an idea frequently used in problem-solving

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What we need to find is setting a reference point for comparison

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Sometimes finding a reference not explicitly stated in the problem

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can also be the key point

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Problem 7: In the following figure, the area of rectangle ABCD is 36

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the area of triangle ABP is 6

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and the area of triangle ADQ is 9. Find the area of triangle APQ

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To solve this problem, some people use variables

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and quickly establish relationships

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However, if you blindly create equations and perform complex calculations

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you end up doing difficult computations

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Such difficult calculations often fail to work out

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and frequent mistakes happen during complex calculations

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We need to identify more information

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Additionally, sometimes it’s necessary to find the key points not explicitly shown in the problem

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First, let’s draw a diagonal as follows

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Since the area of rectangle ABCD is 36

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the area of triangle ABC, which is half of the rectangle, is 18

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Since the area of triangle ABP is 6, the area of triangle APC is 12

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and the ratio of BP to PC is 6 to 12

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which simplifies to 1:2

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Let’s express this as a:2a

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Since the area of triangle ADQ is 9

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the area of triangle ACQ is also 9

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and the ratio of CQ to DQ is 1:1

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Let’s write this as b:b

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Looking at this setup, since the area of rectangle ABCD is 36

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3a×2b=36

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and a×b=6

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The area of triangle CPQ is 2a x b x 1/2

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which is a x b

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and a x b = 6

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The area of triangle APQ, which we need to find

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is the total area

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minus the areas of ABP, ADQ, and CPQ

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Therefore, from 36, subtracting 6, 9, and 6

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the result is 15, which is the answer

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Given rectangles ABCE and CEFG as follows

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when the shaded area is 11, find the area of rectangle ABCD

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In the given figure, if we draw a line as shown below

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we see that the areas of triangles ACH and DEH are equal

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Our strategy is to prove that the areas of triangles DGP and EFP are equal

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If we can show this, the shaded area will equal the area of triangle ACD

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and if this area is 11, the area of rectangle ABCD is confirmed as 22

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We can deduce that triangles DGP and EFP are congruent

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because EF and DG are both 2 and angles DPG and EPF are vertical angles and thus equal

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Therefore, the area of rectangle EFGH equals the area of triangle ACH

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and in the end, the shaded area equals the area of triangle ACD

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As observed earlier, the area of rectangle ABCD is 22

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A psychologist once said this:

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If Koreans stopped comparing and stopped worrying about money, they’d be happy

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Those who talk about happiness often point out comparison

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as the root of unhappiness and the starting point of misery

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But remember, comparison is also the root of problem-solving and the starting point for answers

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the root of problem-solving and the starting point for answers

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The key to problem-solving is finding what to compare and thinking in terms of ratios

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For those of you creating digital content

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use comparison not as an emotional concept but as a thinking skill

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and apply it effectively to solve problems

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Applying Comparison and Ratios in Problem-Solving
Rational number: A number that can be expressed as a ratio of two integers, representing proportion and ratio
Similar shapes are often created to compare and use ratios

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When solving problems, it's important to set a reference and use comparison and ratios
When comparing two shapes with the same height, the ratio of their bases becomes the ratio of their areas

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Identifying Information You Want to Compare
Comparison is the root of problem-solving and the starting point for answers

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Focusing on specific aspects to compare and identify ratios
What we need to find is setting a reference point for comparison

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Sometimes finding a reference not explicitly stated in the problem can also be the key point